NCERT Solutions for Class 12 Maths Chapter 5 - Continuity and Differentiability

Continuity and Differentiability are important foundational steps for advanced calculus. Continuity of a function means that the function's graph can be drawn without any break or the graph can be drawn without lifting the pen. Differentiability means the function derivative exists at every point of the given interval, or we can also define differentiability if there is only a tangent to the given point in an interval. In this article, we will see many continuous and differentiable functions, and we will learn many counter-examples.

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1. NCERT Class 12 Maths Chapter 5- PDF Free Download
2. Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae
3. NCERT Continuity and Differentiability Class 12 Questions And Answers (Exercise)
4. NCERT solutions for class 12 maths - Chapter-wise
5. Importance of solving NCERT Class 12 Maths Chapter 5

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These NCERT solutions are created by a team of experts at Careers360. These experts also prepared the Class 12 Maths Chapter 5 continuity and differentiability Notes to clarify the concepts. These solutions for class 12 are designed according to the latest curriculum designed by the CBSE, ensuring that students grasp the concepts efficiently and easily. For a better understanding, you can practice NCERT Exemplar Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability to enhance your knowledge.

NCERT Class 12 Maths Chapter 5- PDF Free Download

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Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae

Continuity: A function $f(x)$ is continuous at a point $x = a$ if:

• $f(a)$ exists (finite, definite, and real).
• $\lim\limits_{x\rightarrow a} f(x)$ exists.
• $\lim\limits_ {x\rightarrow a} f(x) = f(a)$.
• $\lim\limits_{x\rightarrow 0} f(x) =f(0)$
• $\lim\limits _{x \rightarrow a} f(x)$ exists

Discontinuity: $f(x)$ is discontinuous in an interval if it is discontinuous at any point in that interval.

Algebra of Continuous Functions:

The sum, difference, product, and quotient of continuous functions are continuous.

Differentiation:

The derivative of $f(x)$ at $x = a$, denoted as $f’(a)$, represents the slope of the tangent line to the graph.

Chain Rule:

If $f = v o u$, where $t = u(x)$, and if both $\frac{dt}{dx}$ and $\frac{dv}{dx}$ exist, then: $\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$.

Derivatives of Some Standard Functions:

• $\frac{d}{dt}(x^n) = nx^{n-1}$
• $\frac{d}{dt}(\sin x) = \cos x$
• $\frac{d}{dt}(\cos x) = -\sin x$
• $\frac{d}{dt}(\tan x) = \sec^2 x$
• $\frac{d}{dt}(\cot x) = -\csc^2 x$
• $\frac{d}{dt}(\sec x) = \sec x \cdot \tan x$
• $\frac{d}{dt}(\csc x) = -\csc x \cdot \cot x$
• $\frac{d}{dt}(a^x) = a^x \cdot ln(a)$
• $\frac{d}{dt}(e^x) = e^x$
• $\frac{d}{dt}(ln x) = \frac{1}{x}$

Mean Value Theorem:

The mean Value Theorem states that if $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists some $c$ in $(a, b)$ such that: $f’(c) = \frac{(f(b) - f(a))}{(b - a)}$.

Rolle's Theorem:

Rolle's Theorem states that if $f(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists some $c$ in $(a, b)$ such that $f’(c) = 0$.

Lagrange's Mean Value Theorem:

Lagrange's Mean Value Theorem states that if $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists some c in $(a, b)$ such that:

$f’(c) = \frac{(f(b) - f(a))}{(b - a)}$.

NCERT Continuity and Differentiability Class 12 Questions And Answers (Exercise)

Question 1. Prove that the function $f ( x) = 5 x -3$ is continuous at $x = 0, at\: \: x = - 3$ and at $x = 5$

Answer:

Given function is
$f ( x) = 5 x -3$
$f(0) = 5(0)-3 = -3$
$\lim\limits_{x\rightarrow 0} f(x) = 5(0)-3 = -3$
$\lim\limits_{x\rightarrow 0} f(x) =f(0)$
Hence, the function is continuous at x = 0
$f(-3)= 5(-3)-3=-15-3=-18 \Rightarrow \lim\limits_{x \rightarrow -3} f(x) = 5(-3)-3 = -15-3=-18$

$ \Rightarrow \lim\limits_{x \rightarrow -3} f(x) = f(-3)$
Hence, the function is continuous at $x = -3$
$f(5)= 5(5)-3=25-3=22 \Rightarrow \lim\limits_{x \rightarrow 5} f(x) = 5(5)-3 = 25-3=-22$

$\Rightarrow \lim\limits_{x \rightarrow 5} f(x) = f(5)$
Hence, the function is continuous at $x = 5$

Question 2. Examine the continuity of the function $f (x) = 2x ^2 - 1 at x = 3.$

Answer:

Given function is
$f(x) = 2x^2-1$
at $x = 3$
$f(3) = 2(3)^2-1 = 2\times 9 - 1=18-1=17\\$

$ \lim\limits_{x\rightarrow 3}f(x) = 2(3)^2-1=2\times 9-1=18-1=17$
$\lim\limits_{x\rightarrow 3}f(x) = f(3)$
Hence, the function is continuous at $x = 3$

Question 3Examine the following functions for continuity.
$(a) f (x) = x - 5$

Answer:

Given function is
$f(x) = x-5$
Our function is defined for every real number, say k
and value at $x = k$ , $f(k) = k-5$
and also,
$\lim\limits_{x\rightarrow k} f(x) = k -5\\$

$ \lim\limits_{x\rightarrow k} f(x) = f(k)$
Hence, the function $f(x) = x-5$ is continuous at every real number

Question 3 b) Examine the following functions for continuity.

$f (x) = \frac{1}{x-5} , x \neq 5$

Answer:

Given function is
$f(x ) = \frac{1}{x-5}$
For every real number k, $k \neq 5$
We get,
$f(k) = \frac{1}{k-5}\\$

$ \lim\limits_{x\rightarrow k}f(x ) = \frac{1}{k-5}\\$

$ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{1}{x-5}$ continuous for every real value of $x$, $x \neq 5$

Question 3 c) Examine the following functions for continuity.

$f (x) = \frac{x ^2-25}{x+5}, x \neq -5$

Answer:

Given function is
$f(x ) = \frac{x^2-25}{x+5}$
For every real number k, $k \neq -5$
We get,
$f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\$

$ \lim\limits_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}$

$= \frac{(k +5)(k-5)}{k+5} = k-5\\$

$ \lim\limits_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{x^2-25}{x+5}$ continuous for every real value of x , $x \neq -5$

Question 3 d) Examine the following functions for continuity. $f (x) = | x - 5|$

Answer:

Given function is
$f (x) = | x - 5|$
for $x > 5 , f(x) = x – 5$
for $x < 5 , f(x) = 5 – x$
So, there are different cases.
case(i) $x > 5$
for every real number $k > 5$ , $f(x) = x – 5$ is defined
$f(k) = k - 5\\$

$ \lim\limits_{x\rightarrow k }f(x) = k -5\\$

$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, function $f(x) = x – 5$ is continuous for $x > 5$
case (ii) $x < 5$
for every real number $k < 5$ , $f(x) = 5 – x$ is defined
$f(k) = 5-k\\ \lim\limits_{x\rightarrow k }f(x) = 5 -k\\ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, function $f(x) = 5 – x$ is continuous for $x < 5$
case(iii) $x = 5$
for $x = 5$ , $f(x) = x – 5$ is defined
$f(5) = 5 - 5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = 5 -5=0\\ \lim\limits_{x\rightarrow 5 }f(x) = f(5)$
Hence, function $f(x) = x – 5$ is continous for $x = 5$

Hence, the function $f (x) = | x - 5|$ is continuous for every real number.

Question 4. Prove that the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer

Answer:

Given function is
$f (x) = x^n$
The function $f (x) = x^n$ is defined for all positive integer, n
$f(n) = n^n\\$

$\lim\limits_{x\rightarrow n}f(x) = n^n\\$

$ \lim\limits_{x\rightarrow n}f(x) = f(n)$
Hence, the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer

Question 5.Is the function f defined by
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
continuous at x = 0? At x = 1? At x = 2?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x , & if x \leq 1 \\ 5 & if x \geq 1 \end{matrix}\right.$
The function is defined at $x = 0$ and its value is $0$
$f(0) = 0\\$

$ \lim\limits_{x\rightarrow 0}f(x) = f(x) = 0\\$

$ \lim\limits_{x\rightarrow 0}f(x) = f(0)$
Hence, the given function is continuous at $x = 0$
Given function is defined for $x = 1$
Now, for $x = 1$ Right-hand limit and left-hand limit are not equal.
$f(1) = 1\\ \lim\limits_{x\rightarrow 1^-}f(x) = f(x) = 1\\$

$ \lim\limits_{x\rightarrow 1^+}f(x) =f(5) = 5$
R.H.L $\neq$ L.H.L.
Therefore, the given function is not continuously $x =1$
Given function is defined for $x = 2$ and its value at $x = 2$ is $5$
$f(2) = 2\\ \lim\limits_{x\rightarrow 2}f(x) = f(5) = 5\\\lim\limits_{x\rightarrow 2}f(x) = f(2)$
Hence, the given function is continuous at $x = 2$

Question:6.Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} 2x+3 & if x \leq 2 \\ 2x-3 & if x \geq 2 \end{matrix}\right.$
Given function is defined for every real number k
There are different cases for the given function.
case(i) $k > 2$
$f(k) = 2k-3\\ \lim\limits_{x\rightarrow k}f(x) = 2k-3\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 2$

case(ii) $k < 2$
$f(k) = 2k +3\\ \lim\limits_{x\rightarrow k}f(x) = 2k+3\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 2$

case(iii) $x = 2$
$\lim\limits_{x\rightarrow 2^-}f(x) = 2x+3 = 2\times 2 + 3 = 4 + 3 = 7\\ $

$\lim\limits_{x\rightarrow 2^+}f(x) = 2x-3 = 2\times 2-3 = 4-3 = 1$
Right hand limit at x= 2 $\neq$ Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity

Question:7.Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3 <x< 3 & \\ 6x +2 & if \: \: x \geq 3 & \end{matrix}\right.$
Given function is defined for every real number k
There are different cases.
case (i) $k < -3$
$f(k) = -k + 3\\ \lim\limits_{x\rightarrow k}f(x) = -k + 3\\ $

$\lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k < -3

case(ii) $k = -3$
$f(-3) = -(-3) + 3 = 6\\ \lim\limits_{x\rightarrow -3^-}f(x) = -k + 3=-(-3)+3 = 6\\$

$ \lim\limits_{x\rightarrow -3^+}f(x) = -2x = -2(-3) = 6\\$

$ R.H.L. = L.H.L. = f(-3)$
Hence, the given function is continuous for x = -3

case(iii) $-3 < k < 3$
$f(k) = -2k \\ \lim\limits_{x\rightarrow k}f(x) = -2k\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in -3 < k < 3 given function is continuous.

case(iv) $k = 3$
$f(3) = 6x+2 = 6\times3+2 =18+2=20\\$

$ \lim\limits_{x\rightarrow 3^-}f(x) = -2x = -2(3) = -6\\$

$ \lim\limits_{x\rightarrow 3^+}f(x) = 6x+2 = 6\times3+2 = 20\\$

$ R.H.L. = f(3) \neq L.H.L.$
Hence, $x = 3$ is the point of discontinuity

case(v) $k > 3$
$f(k) = 6k+2 \\$

$ \lim\limits_{x\rightarrow k}f(x) = 6k+2 \\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k > 3

Question:8.Find all points of discontinuity of f, where f is defined by

$f (x )= \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$

Answer:

Given function is
$f (x ) \left\{\begin{matrix} \frac{|x|}{x} & if \: \: x \neq 0 \\ 0 & if \: \: x = 0 \end{matrix}\right.$
if x > 0 , $f(x)=\frac{x}{x} = 1$
if x < 0 , $f(x)=\frac{-(x)}{x} = -1$
Given function is defined for every real number k
Now,
case(i) k < 0
$f(k) = -1\\ \lim\limits_{x\rightarrow k }f(x) = -1\\$

$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, the given function is continuous for every value of k < 0
case(ii) k > 0
$f(k) = 1\\ \lim\limits_{x\rightarrow k }f(x) = 1\\$

$ \lim\limits_{x\rightarrow k }f(x) = f(k)$
Hence, the given function is continuous for every value of k > 0
case(iii) x = 0
$f(0) = 0\\ \lim\limits_{x\rightarrow 0^- }f(x) = -1\\$

$ \lim\limits_{x\rightarrow 0^+}f(x) = 1\\ f(0) \neq R.H.L. \neq L.H.L.$
Hence, 0 is the only point of discontinuity

Question:9.Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} \frac{x }{|x|} & if \: \: x < 0\\ -1 & if x \geq 0 \end{matrix}\right.$
if x < 0 , $f (x) =\frac{x }{|x|} = \frac{x}{-(x)} = -1$
Now, for any value of x, the value of our function is -1
Therefore, the given function is continuous for every value of x
Hence, no point in discontinuity

Question:10.Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right. $

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x+1 & if \: \: x \geq 1 \\ x^2 +1 & if x \: \: <1 \end{matrix}\right.$
Given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k+1\\ \lim\limits_{x\rightarrow k}f(x) = k+1\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 1$

case(ii) $k < 1$
$f(k) = k^2 +1\\ \lim\limits_{x\rightarrow k}f(x) = k^2+1\\ $

$\lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 1$

case(iii) $x = 1$

$\lim\limits_{x\rightarrow 1^-}f(x) = x^2+1 = 1^2 + 1 = 1 + 1 = 2\\$

$ \lim\limits_{x\rightarrow 1^+}f(x) = x+1 = 1+1 = 2\\ f(1) = 1^2+1 = 2 \\ $

$R.H.L. = L.H.L. = f(1)$

Hence, at x = 2 given function is continuous.
Therefore, no point of discontinuity

Question 11.Find all points of discontinuity of f, where f is defined by

$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$

Answer:

Given function is
$f ( x) = \left\{\begin{matrix} x^3 -3 & if \: \: x \leq 2\\ x ^2 +1 & if \: \: x > 2 \end{matrix}\right.$
Given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 2$
$f(k) = k^2+1\\ \lim\limits_{x\rightarrow k}f(x) = k^2+1\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 2$

case(ii) $k < 2$
$f(k) = k^3 -3\\ \lim\limits_{x\rightarrow k}f(x) = k^3-3\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 2$

case(iii) $x = 2$
$\lim\limits_{x\rightarrow 2^-}f(x) = x^3-3 = 2^3- 3 = 8- 3 = 5\\$

$ \lim\limits_{x\rightarrow 2^+}f(x) = x^2+1= 2^2+1 = 4+1 = 5\\$

$ f(2) = 2^3-3 = 8 - 3 = 5\\ f(2)=R.H.L.=L.H.L.$
Hence, the given function is continuous at $x = 2$
There is no point of discontinuity

Question 12: Find all points of discontinuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x ^{10} -1 & if x \leq 1 \\ x ^2 & x > 1 \end{matrix}\right.$
Given function is defined for every real number k
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k^2\\ \lim\limits_{x\rightarrow k}f(x) = k^2\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of k > 1

case(ii) $k < 1$
$f(k) = k^{10} -1\\$

$ \lim\limits_{x\rightarrow k}f(x) = k^{10}-1\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of k < 1

case(iii) x = 1
$\lim\limits_{x\rightarrow 1^-}f(x) = x^{10}-1 = 1^{10} - 1 = 1 - 1 = 0\\$

$ \lim\limits_{x\rightarrow 1^+}f(x) = x^2 = 1^2 = 1\\$

$ f(1) = x^{10}-1 = 0\ f(1) = L.H.L. \neq R.H.L.$
Hence, x = 1 is the point of discontinuity.

Question 13. Is the function defined by

$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$

A continuous function?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x+5 & if x \leq 1\\ x-5 & if x > 1 \end{matrix}\right.$
Given function is defined for every real number $k$
There are different cases for the given function.
case(i) $k > 1$
$f(k) = k-5\\ \lim\limits_{x\rightarrow k}f(x) = k-5\\$

$ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k > 1$

case(ii) $k < 1$
$f(k) = k+5\\ \lim\limits_{x\rightarrow k}f(x) = k+5\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for each value of $k < 1$

case(iii) $x = 1$
$\lim\limits_{x\rightarrow 1^-}f(x) = x+5 = 1 + 5 = 1 + 5 = 6\\ \lim\limits_{x\rightarrow 1^+}f(x) = x-5 = 1-5 = -4\\ f(1) = x+5 =1+5= 6 \\ L.H.L. = f(1) \neq R.H.S.$

Hence, $x = 1$ is the point of discontinuity.

Question:14.Discuss the continuity of the function f, where f is defined by

$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$

Answer:

Given function is
$f (x)\left\{\begin{matrix} 3 & if 0 \leq x \leq 1 \\ 4& if 1 < x < 3 \\ 5& if 3 \leq x \leq 10 \end{matrix}\right.$
Given function is defined for every real number k
Different cases are their
case (i) $k < 1$
$f(k) = 3\\ \lim\limits_{x\rightarrow k}f(x) = 3\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k < 1$

case(ii) $k = 1$
$f(1) = 3 \\ \lim\limits_{x\rightarrow 1^-}f(x) = 3\\ \lim\limits_{x\rightarrow 1^+}f(x) = 4\\ R.H.L. \neq L.H.L. = f(1)$
Hence, the given function is discontinuous at $x = 1$
Therefore, $x = 1$ is the point of discontinuity.

case(iii) $1 < k < 3$
$f(k) = 4 \\ \lim\limits_{x\rightarrow k}f(x) = 4\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $1 < k < 3$ given function is continuous.

case(iv) $k = 3$
$f(3) =5\\ \lim\limits_{x\rightarrow 3^-}f(x) = 4\\ \lim\limits_{x\rightarrow 3^+}f(x) =5\\ R.H.L. = f(3) \neq L.H.L.$
Hence, $x = 3$ is the point of discontinuity

case(v) $k > 3$
$f(k) = 5 \\ \lim\limits_{x\rightarrow k}f(x) = 5 \\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k > 3$
case(vi) when $k < 3$

$f(k) = 4 \\ \lim\limits_{x\rightarrow k}f(x) = 4\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $k < 3$ given function is continuous

Question:15 Discuss the continuity of the function f, where f is defined by $f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

Answer:
Given function is satisfied for all real values of $x$
case (i) $k < 0$
Hence, the function is continuous for all values of $x < 0$

case (ii) $x = 0$
L.H.L at $x= 0$

R.H.L. at $x = 0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous at $x = 0$

case (iii) $k > 0$
L.H.L. = R.H.L. = $f(0)$
Hence, the function is continuous for all values of x > 0

case (iv) k < 1

Hence, the function is continuous for all values of x < 1

case (v) k > 1

Hence, the function is continuous for all values of x > 1

case (vi) x = 1
Hence, the function is not continuous at x = 1

Question:16.Discuss the continuity of the function f, where f is defined by

$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$

Answer:

Given function is
$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$
Given function is defined for every real number $k$
Different cases are their
case (i) $k < -1$
$f(k) = -2\\ \lim\limits_{x\rightarrow k}f(x) = -2\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of $k < -1$
case(ii) k = -1
$f(-1) = -2 \\ \lim\limits_{x\rightarrow -1^-}f(x) = -2\\ \lim\limits_{x\rightarrow -1^+}f(x) = 2x = 2(-1) = -2\\ R.H.L. =L.H.L. = f(-1)$
Hence, the given function is continuous at $x = -1$
case(iii) $k > -1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for all values of $x > -1$

case(vi) $-1 < k < 1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of $k$ in $-1 < k < 1$ given function is continuous.

case(v) $k = 1$
$f(1) =2x = 2(1)=2\\ \lim\limits_{x\rightarrow 1^-}f(x) = 2x=2(1)=2\\ \lim\limits_{x\rightarrow 1^+}f(x) =2\\ R.H.L. = f(1) = L.H.L.$
Hence, at x =1 function is continuous

case(vi) $k > 1$
$f(k) = 2 \\ \lim\limits_{x\rightarrow k}f(x) = 2 \\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, the given function is continuous for every value of k > 1
case(vii) when $k < 1$
$f(k) = 2k \\ \lim\limits_{x\rightarrow k}f(x) = 2k\\ \lim\limits_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in $k < 1$ given function is continuous.

Therefore, continuous at all points

Question:17.Find the relationship between a and b so that the function f is defined by
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
is continuous at x = 3.

Answer:

Given function is
$f (x) = \left\{\begin{matrix} ax +1 , &if x < 3 \\ bx +3 & if x > 3 \end{matrix}\right.$
For the function to be continuous at $x = 3$, R.H.L. must be equal to L.H.L.
$\lim\limits_{x\rightarrow 3^-}f(x)= ax + 1 = 3a+1\\ \lim\limits_{x\rightarrow 3^+}f(x) = bx+3=3b+3$
For the function to be continuous
$\lim\limits_{x\rightarrow 3^-}f(x) = \lim\limits_{x\rightarrow 3^+}f(x) \\ 3a + 1= 3b+3\\ 3(a-b)=2\\ a-b = \frac{2}{3}\\ a = b+\frac{2}{3}$

Question:18.For what value of l is the function defined by
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
continuous at x = 0? What about continuity at x = 1?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} \lambda (x^2 -2x) & if x \leq 0 \\ 4x+1 & if x > 0 \end{matrix}\right.$
For the function to be continuous at $x = 0$, R.H.L. must be equal to L.H.L.
$\lim\limits_{x\rightarrow 0^-}f(x) = \lambda (x^2-2x) = 0\\ \lim\limits_{x\rightarrow 0^+}f(x) = 4x+1=1$
For the function to be continuous
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x) \\ 0\neq 1$
Hence, if no value of the function is continuous at $x = 0$

For $x = 1$
$f(1)=4x+1=4(1)+1=5\\ \lim\limits_{x\rightarrow 1}f(x) =4+1=5 \\\ \lim\limits_{x\rightarrow 1}f(x) = f(x)$
Hence, the given function is continuous at $x =1$

Question:19.Show that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points. Here, [x] denotes the greatest integer less than or equal to x.

Answer:

Given function is
$g (x) = x- [x]$
Given is defined for all real numbers k
$\lim\limits_{x\rightarrow k^-}f(x) = k - (k-1) = k-k+1 =1\\ \lim\limits_{x\rightarrow k^+}f(x) = k - k = 0\\ \lim\limits_{x\rightarrow k^-}f(x) \neq \lim\limits_{x\rightarrow k^+}f(x)$
Hence, by this, we can say that the function defined by $g (x) = x- [x]$ is discontinuous at all integral points

Question 20.Is the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$ ?

Answer:

Given function is
$f (x) = x^2 - sin x + 5$
Clearly, the Given function is defined at x = $\pi$
$f(\pi) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = \pi^2-\sin \pi+5 =\pi^2-0+5 = \pi^2+5\\ \lim\limits_{x\rightarrow \pi}f(x) = f(\pi)$
Hence, the function defined by $f (x) = x^2 - sin x + 5$ continuous at x = $\pi$

Question 21.Discuss the continuity of the following functions:
a) $f (x) = \sin x + \cos x$

Answer:

Given function is
$f (x) = \sin x + \cos x$
The given function is defined for all real numbers.
We, know that if two function $g(x)$ and $h(x)$ are continuous then $g(x)+h(x)$ , $g(x)-h(x)$ , $g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x)$

$ = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$ \text{We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$

$ \lim\limits_{h\rightarrow 0}\sin (c+h) =$ $\lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{ We know that }\cos(a+b) = \cos a \cos b + \sin a\sin b$

$\lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that $\sin x$ and $\cos x$ are continuous functions.
So, we can say that
$f(x) = g(x) + h(x) = \sin x + \cos x$ is also a continuous function

Question 21. b)Discuss the continuity of the following functions:
$f (x) = \sin x - \cos x$

Answer:

Given function is
$f (x) = \sin x - \cos x$
The given function is defined for all real numbers.
We know that if two functions g(x) and h(x) are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$

$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$

$ \text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$

$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous discussion of the continuity of the following functions:
$f (x)$ is continuous, we can say that
$f(x) = g(x) - h(x) = \sin x - \cos x$ is also a continuous function.

Question 21. c)Discuss the continuity of the following functions:
$f (x) = \sin x \cdot \cos x$

Answer:

Given function is
$f (x) = \sin x \cdot \cos x$
The given function is defined for all real numbers.
We know that if two functions $g(x)$ and $h(x)$ are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$

$ \lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c, \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x = \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$

$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x are continuous functions.
So, we can say that
$f(x) = g(x).h(x) = \sin x .\cos x$ is also a continuous function

Question:22.Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Answer:

We know that if two functions g(x) and h(x) are continuous, then.
$\frac{g(x)}{h(x)} , h(x) \neq0\text{ is continuous}$

$\frac{1}{h(x)} , h(x) \neq 0\ \text{is continuous} \frac{1}{g(x)} , g(x) \neq0\text{ is continuous}$
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text {We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$

$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x)$

$= \lim\limits_{x\rightarrow c}\cos x $

$= \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that }\cos(a+b) = \cos a \cos b + \sin a\sin b$

$\lim\limits_{h\rightarrow 0}\cos (c+h)$

$=\lim\limits_{h \rightarrow 0} (\cos c \cos h+ \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, the function $h(x) = \cos x$ is a continuous function
We proved independently that $\sin x$ and $\cos x$ a continuous functions.
So, we can say that
cosec x = $\frac{1}{\sin x} = \frac{1}{g(x)}$ is also continuous except at $x=n\pi$
sec x = $\frac{1}{\\cos x} = \frac{1}{h(x)}$ is also continuous except at $x=\frac{(2n+1) \pi}{2}$
cot x = $\frac{\cos x}{\sin x} = \frac{h(x)}{g(x)}$ is also continuous except at $x=n\pi$

Question 23.Find all points of discontinuity of f, where

$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$

Answer:

Given function is
$f (x ) = \left\{\begin{matrix} \frac{\sin x }{x} & if x < 0 \\ x+1 & if x > 0 \end{matrix}\right.$
$\lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1\\ \lim\limits_{x\rightarrow 0^+}f(x) = x + 1 = 1\\ \lim\limits_{x\rightarrow 0^-}f(x) = \lim\limits_{x\rightarrow 0^+}f(x)$
Hence, the function is continuous.
Therefore, no point of discontinuity

Question 24.Determine if f is defined by
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
Is it a continuous function?

Answer:

Given function is
$f (x) = \left\{\begin{matrix} x^2 \sin 1/x & if x \neq 0 \\ 0 & if x = 0 \end{matrix}\right.$
Given function is defined for all real numbers k
when x = 0
$f(0) = 0\\ \lim\limits_{x\rightarrow 0}f(x)=\lim\limits_{x\rightarrow 0}\left ( x^2\sin\frac{1}{x} \right )=\lim\limits_{x\rightarrow 0}\left ( \frac{x.\sin\frac{1}{x}}{\frac{1}{x}} \right ) = 0(1)=0 \ \ \ \ \ \ (\because\lim\limits_{x\rightarrow 0}\frac{\sin x}{x} = 1)$
$\lim\limits_{x\rightarrow 0}f(x) = f(0)$
Hence, the function is continuous at x = 0
when $x \neq 0$
$f(k) = k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k}f(x)=\lim\limits_{x\rightarrow k}\left ( x^2\sin\frac{1}{x} \right )=k^2\sin \frac{1}{k}\\ \lim\limits_{x\rightarrow k} = f(k)$
Hence, the given function is continuous for all points

Question:25. Examine the continuity of f, where f is defined by

$f (x) = \left\{\begin{matrix} \sin x - \cos x & if x \neq 0 \\ -1 & if x = 0 \end{matrix}\right.$

Answer:

Given function is
$f (x) = \sin x - \cos x$
The given function is defined for all real numbers.
We know that if two functions $g(x)$ and $h(x)$ are continuous then $g(x)+h(x), g(x)-h(x), g(x).h(x)$ all are continuous
Lets take $g(x) = \sin x$ and $h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim\limits_{x\rightarrow c}g(x) = \lim\limits_{x\rightarrow c}\sin x = \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{We know that} \sin(a+b) = \sin a \cos b + \cos a\sin b$

$\lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\cos x$

$= \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that} \cos(a+b) = \cos a \cos b + \sin a\sin b$

$\lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that $ \sin x$ and $\cos x$ are continuous functions.
So, we can say that
$f(x) = g(x) - h(x) =$ $\sin x$ - $\cos x$ is also a continuous function

When $x = 0$
$f (0) = (-1) \lim\limits_{x\rightarrow 0^-}f(x)$

$= \sin 0 - \cos 0 = -1$

$ \lim\limits_{x\rightarrow 0^+}f(x) = \sin 0 - \cos 0 = -1 $

$ R.H.L. = L.H.L. = f(0)$
Hence, the function is also continuous at $x = 0$

Question:26.Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} \frac{k \cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right. \: \: \: at \: \: x = \pi /2$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} \frac{k\cos x }{\pi - 2x } & if x \neq \pi/2 \\ 3 & if x = \pi/2 \end{matrix}\right.$
When $x = \frac{\pi}{2}$
$f(\frac{\pi}{2}) = 3\\let\ x=\pi +h\\ \lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= \lim\limits_{h\rightarrow 0}\frac{k\cos\left ( \frac{\pi}{2}+h \right )}{\pi-2\left ( \frac{\pi}{2}+h \right )} = k. \lim\limits_{h\rightarrow 0}\frac{-\sin h}{-2h} = \frac{k}{2}\\$
For the function to be continuous
$\lim\limits_{x\rightarrow \frac{\pi}{2}}f(x)= f(\frac{\pi}{2})\\ \frac{k}{2} = 3\\ k = 6$
Therefore, the value of k so that the function f is continuous at 6

Question:27. Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right. \: \: at \: \: x = 2$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} kx^2 &if x \leq 2 \\ 3 & if x > 2 \end{matrix}\right.$
When $x = 2$
For the function to be continuous
$f(2) = R.H.L. = LH.L.$
$f(2) = 4k\\ \lim\limits_{x\rightarrow 2^-}f(x)= 4k\\ \lim\limits_{x\rightarrow 2^+}f(x) = 3\\ f(2) = \lim\limits_{x\rightarrow 2^-}f(x) = \lim\limits_{x\rightarrow 2^+}f(x)\\ 4k = 3\\ k = \frac{3}{4}$
Hence, the values of k so that the function f is continuous at x= 2 is $\frac{3}{4}$

Question:28. Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right. \: \: at \: \: x = \pi$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} kx + 1 & if x \leq \pi \\ \cos x & if x > \pi \end{matrix}\right.$
When x = $\pi$
For the function to be continuous
, f( $\pi$ ) = R.H.L. = LH.L.
$f(\pi) = k\pi+1\\ \lim\limits_{x\rightarrow \pi^-}f(x)= k\pi+1\\ \lim\limits_{x\rightarrow \pi^+}f(x) = \cos \pi = -1\\ f(\pi) = \lim\limits_{x\rightarrow \pi^-}f(x) = \lim\limits_{x\rightarrow \pi^+}f(x)\\ k\pi+1 = -1\\ k = \frac{-2}{\pi}$
Hence, the values of k so that the function f is continuous at x= $\pi$ is $\frac{-2}{\pi}$

Question:29Find the values of k so that the function f is continuous at the indicated point in Exercises

$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right. \: \: at x = 5$

Answer:

Given function is
$f (x) = \left\{\begin{matrix} kx +1 & if x \leq 5 \\ 3x-5 & if x > 5 \end{matrix}\right.$
When $x = 5$
For the function to be continuous
$f(5) = R.H.L. = LH.L.$
$f(5) = 5k+1\\ \lim\limits_{x\rightarrow 5^-}f(x)= 5k+1\\ \lim\limits_{x\rightarrow 5^+}f(x) = 3(5)-5 = 15-5=10\\ f(5) = \lim\limits_{x\rightarrow 5^-}f(x) = \lim\limits_{x\rightarrow 5^+}f(x)\\ 5k+1 = 10\\ k = \frac{9}{5}$
Hence, the values of k so that the function f is continuous at $x= 5$ is $\frac{9}{5}$

Question:30Find the values of a and b such that the function defined by
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
It is a continuous function.

Answer:

Given that continuous function is
$f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21 , & if\: \: x > 10 \end{matrix}\right.$
The function is continuous so
$\lim\limits_{x\rightarrow 2^-}f(x) = \lim\limits_{x\rightarrow 2^+}f(x)\\ and\\ \lim\limits_{x\rightarrow 10^-}f(x)=\lim\limits_{x\rightarrow 10^+}f(x)$
$\lim\limits_{x\rightarrow 2^-}f(x) = 5\\ \lim\limits_{x\rightarrow 2^+}f(x)=ax+b=2a+b$

$ 2a+b = 5 \ \ \ \ \ \ -(i)$

$\lim\limits_{x\rightarrow 10^-}f(x)=ax+b=10a+b\\ \lim\limits_{x\rightarrow 10^+}f(x)=21$

$ 10a+b=21 \ \ \ \ \ \ \ \ -(ii)$
By solving equation (i) and (ii)
a = 2 and b = 1
Hence, values of a and b such that the function defined by $f (x) = \left\{\begin{matrix} 5 & if\: \: x \leq 2 \\ ax + b & if\: \: 2 < x < 10 \\ 21, & if\: \: x > 10 \end{matrix}\right.$ is a continuous function is 2 and 1 respectively

Question:31.Show that the function defined by $f (x) = \cos (x^2 )$ is a continuous function.

Answer:

Given function is
$f (x) = \cos (x^2 )$
Given function is defined for all real values of x
Let x = k + h
if $x\rightarrow k , \ then \ h \rightarrow 0$
$f(k) = \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = \lim\limits_{x \rightarrow k}\cos x^2 = \lim\limits_{h \rightarrow 0}\cos (k+h)^2 $

$= \cos k^2\\ \lim\limits_{x \rightarrow k}f(x) = f(k)$
Hence, the function $f (x) = \cos (x^2 )$ is a continuous function

Question 32.Show that the function defined by $f (x) = |\cos x |$ is continuous.

Answer:

Given function is
$f (x) = |\cos x |$
Given function is defined for all values of x
f = g o h , $g(x) = \|x\|$ and $h(x) = \cos x$
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) $k < 0$
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when $k < 0$

case (ii) $k > 0$
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) $k = 0$
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$

$ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$

$\lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
$h(x) = \cos x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim\limits_{x\rightarrow c}h(x)$

$= \lim\limits_{x\rightarrow c}\cos x$

$= \lim\limits_{h\rightarrow 0}\cos (c+h)$

$\text{We know that } \cos(a+b) = \cos a \cos b + \sin a\sin b$

$ \lim\limits_{h\rightarrow 0}\cos (c+h) = \lim\limits_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h)$

$= \lim\limits_{h\rightarrow 0}\cos c\cos h + \lim\limits_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
g(x) is continuous , $h(x)$ is continuous
Therefore, $f(x) = g o h$ is also continuous

Question 33. Examine that sin | x| is a continuous function.

Answer:

Given function is
$f(x) = \sin \|x\|$
f(x) = h o g , h(x) =$\sin x$ and g(x) = |x|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) $k < 0$
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) $k > 0$
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) $k = 0$
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$

$\lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$

$ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) =$\sin x$
Let's suppose $x = c + h$
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \sin c$

$ \lim\limits_{x\rightarrow c}h(x) = \lim\limits_{x\rightarrow c}\sin x$

$= \lim\limits_{h\rightarrow 0}\sin (c+h)$

$\text{ We know that } \sin(a+b) = \sin a \cos b + \cos a\sin b$

$ \lim\limits_{h\rightarrow 0}\sin (c+h) = \lim\limits_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h)$

$= \lim\limits_{h\rightarrow 0}\sin c\cos h + \lim\limits_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim\limits_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \sin x$ is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, $f(x) = h o g$ is also continuous

Question:34.Find all the points of discontinuity of f defined by $f (x) = | x| - | x + 1|.$

Answer:

Given function is
$f (x) = | x| - | x + 1|$
Let g(x) = |x| and h(x) = |x+1|
Now,
$g(x)\begin{cases} -x & \text{ if } x<0 \\ 0 & \text{ if } x= 0\\ x& \text{ if } x>0 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < 0
$g(k) = -k\\ \lim\limits_{x\rightarrow k}g(x) = -k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

case (ii) k > 0
$g(k) = k\\ \lim\limits_{x\rightarrow k}g(x) = k\\ \lim\limits_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k > 0

case (iii) k = 0
$g(0) = 0\\ \lim\limits_{x\rightarrow 0^-}g(x) = -x = 0$

$ \lim\limits_{x\rightarrow 0^+}g(x ) = x = 0$

$ \lim\limits_{x\rightarrow 0^-}g(x) = g(0) = \lim\limits_{x\rightarrow 0^+}g(x )$
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x

Now,
$h(x)\begin{cases} -(x+1) & \text{ if } x<-1 \\ 0 & \text{ if } x= -1\\ (x+1)& \text{ if } x>-1 \end{cases}$
g(x) is defined for all real numbers k
case(i) k < -1
$h(k) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = -(k+1)\\ \lim\limits_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k < -1

case (ii) k > -1
$h(k) = k+1\\ \lim\limits_{x\rightarrow k}h(x) = k+1\\ \lim\limits_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k > -1

case (iii) k = -1
$h(-1) = 0\\ \lim\limits_{x\rightarrow -1^-}h(x) = -(x-1)$

$= 0\\ \lim\limits_{x\rightarrow -1^+}h(x ) = x+1 = 0$

$ \lim\limits_{x\rightarrow -1^-}h(x) = h(0) = \lim\limits_{x\rightarrow -1^+}h(x )$
Hence, h(x) is continuous when k = -1
Therefore, h(x) = |x+1| is continuous for all real values of x.
g(x) is continuous and h(x) is continuous
Therefore, $f(x) = g(x) - h(x) = |x| - |x+1|$ is also continuous

NCERT class 12 maths chapter 5 question answer: Exercise: 5.2 (Page no. 122, Total questions- 10)

Question 1. Differentiate the functions with respect to x in

$\sin (x^2 +5 )$

Answer:

Given function is
$f(x)=\sin (x^2 +5 )$
When we differentiate it w.r.t. x.
Let's take $t = x^2+5$. then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (x^2+5)$
$\frac{dt}{dx} = \frac{d(x^2+5 )}{dx} = 2x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (x^2+5).2x$
Therefore, the answer is $2x \cos (x^2+5)$

Question 2.Differentiate the functions with respect to x in

$\cos ( \sin x )$

Answer:

Given function is
$f(x)= \cos ( \sin x )$
Let’s take $t = \sin x$ then,
$f(t) = \cos t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ ( By chain rule)
$\frac{df(t)}{dt} = \frac{d(\cos t)}{dt} = -\sin t = -\sin (\sin x)$
$\frac{dt}{dx} = \frac{d(\sin x)}{dt} = \cos x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = -\sin(\sin x).\cos x$
Therefore, the answer is $-\sin(\sin x).\cos x$

Question 3.Differentiate the functions with respect to x in

$\sin (ax +b )$

Answer:

Given function is
$f(x) = \sin (ax +b )$
When we differentiate it w.r.t. x.
Let's take $t = ax+b$. then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (ax+b)$
$\frac{dt}{dx} = \frac{d(ax+b )}{dx} = a$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (ax+b).a$
Therefore, the answer is $a \cos (ax+b)$

Question 4. Differentiate the functions with respect to x in

$\sec (\tan (\sqrt x) )$

Answer:

Given function is
$f(x)=\sec (\tan (\sqrt x) )$
When we differentiate it w.r.t. x.
Let's take $t = \sqrt x$. then,
$f(t) = \sec (\tan t)$
take $\tan t = k$ . then,
$f(k) = \sec k$
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)$
$(\because k = \tan t \ and \ t = \sqrt x)$
$\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)$
$\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}$
Now,
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$

$=\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}$
Therefore, the answer is $\frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}$

Question 5.Differentiate the functions with respect to x in

$\frac{\sin (ax +b )}{\cos (cx + d)}$

Answer:

Given function is
$f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}$
We know that,
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}$
$g(x) = \sin(ax+b)$ and $h(x) = \cos(cx+d)$
Let's take $u = (ax+b) \ and \ v = (cx+d)$
Then,
$\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c$
$g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx}$ (By chain rule)
$\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)$
$\frac{du}{dx} = \frac{d(ax+b)}{dx} = a$
$g^{'}(x)=a\cos (ax+b)$ -(i)
Similarly,
$h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}$
$\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))$
$\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c$
$h^{'}(x)=-c\sin(cx+d)$ -(ii)
Now, put (i) and (ii) in
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}$
$= \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}$
$= a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$
Therefore, the answer is $a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$

Question 6.Differentiate the functions with respect to x in

$\cos x^3 . \sin ^ 2 ( x ^5 )$

Answer:

Given function is
$f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )$
Differentiation w.r.t. x is
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)$
$g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)$
Lets take $u = x^3 \ and \ v = x^5$
Our functions become,
$\cos x^3 = \cos u$ and $\sin^2(x^5) = \sin^2v$
Now,
$g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx}$ ( By chain rule)
$\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)$
$\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2$
$g^{'}(x) = -\sin x^3.3x^2$ -(i)
Similarly,
$h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}$
$\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5$

$ (\because v = x^5)$
$\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4$
$h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5$ -(ii)
Put (i) and (ii) in
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) $

$= -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5 \cos x^5$
Therefore, the answer is $10x^4\sin x^5 \cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5$

Question 7.Differentiate the functions with respect to x in

$2 \sqrt { \cot ( x^2 )}$

Answer:

Give function is
$f(x)=2 \sqrt { \cot ( x^2 )}$
Let's take $t = x^2$
Now, take $\cot t = k^2$
$f(k) = 2k$
Differentiation w.r.t. x
$\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ -(By chain rule)
$\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2$
$\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}}(\because t = x^2)$
$\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x$
So,
$\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} }$ ( Multiply and divide by $\sqrt 2$ and multiply and divide $\sqrt {\cot x^2}$ by $\sqrt{\sin x^2}$
$(\text{because} \cot x = \frac{\cos x}{\sin x} \ and \csc x = \frac{1}{\sin x })$
$=\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}(\because 2\sin x\cos x=\sin2x)$
There, the answer is $\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}$

Question 8Differentiate the functions with respect to x in

$\cos ( \sqrt x )$

Answer:

Let us assume : $y\ =\ \cos ( \sqrt x )$

Differentiating y with respect to x, we get :

$\frac{dy}{dx}\ =\ \frac{d(\cos ( \sqrt x ))}{dx}$

or $=\ - \sin \sqrt{x}.\frac{d( \sqrt x )}{dx}$

or $=\ \frac{- \sin \sqrt{x}}{2\sqrt{x}}$

Question 9. Prove that the function f given by $f (x) = |x-1 |, x \epsilon R$ is not differentiable at x = 1.

Answer:

Given function is
$f (x) = |x-1 | , x \epsilon R$
We know that any function is differentiable when both.
$\lim\limits_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim\limits_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
The required condition for the function to be differential at x = 1 is

$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, the Left-hand limit of a function at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h} = \lim\limits_{h\rightarrow 0^-}\frac{|h|-0}{h}$
$= \lim\limits_{h\rightarrow 0^-}\frac{-h}{h} = -1 \ \ \ \ (\because h < 0)$
The right-hand limit of a function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h} = \lim\limits_{h\rightarrow 0^+}\frac{|h|-0}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{h}{h} = 1$
Now, it is clear that.
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1
Therefore, function $f (x) = |x-1 |$ is not differentiable at x = 1

Question 10.Prove that the greatest integer function defined by $f (x) = [x] , 0 < x < 3$ is not differentiable at

x = 1 and x = 2.

Answer:

Given function is
$f (x) = [x] , 0 < x < 3$
We know that any function is differentiable when both.
$\lim\limits_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h}$ and $\lim\limits_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h}$ are finite and equal
The required condition for the function to be differential at x = 1 is

$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}$
Now, the Left-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{[1+h]-[1]}{h} = \lim\limits_{h\rightarrow 0^-}\frac{0-1}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ \ (\because h < 0 \rightarrow 1+h<1, \therefore [1+h] =0)$
The right-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{[1+h]-[1]}{h} = \lim\limits_{h\rightarrow 0^+}\frac{1-1}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \rightarrow 1+h>1, \therefore [1+h] =1)$
Now, it is clear that.
R.H.L. at x= 1 $\neq$ L.H.L. at x= 1 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 1
Similarly, for x = 2
The required condition for the function to be differential at x = 2 is

$\lim\limits_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}$
Now, the Left-hand limit of the function at x = 2 is
$\lim\limits_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^-}\frac{[2+h]-[2]}{h} = \lim\limits_{h\rightarrow 0^-}\frac{1-2}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ (\because h < 0 \rightarrow 2+h<2, \therefore [2+h] =1)$
The right-hand limit of the function at x = 1 is
$\lim\limits_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h } = \lim\limits_{h\rightarrow 0^+}\frac{[2+h]-[2]}{h} = \lim\limits_{h\rightarrow 0^+}\frac{2-2}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \rightarrow 2+h>2, \therefore [2+h] =2)$
Now, it is clear that.
R.H.L. at x= 2 $\neq$ L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 2

NCERT class 12 maths chapter 5 question answer: Exercise: 5.3 (Page no. 125, Total questions- 15)

Question 1.Find $\frac{dy}{dx}$ in the following:

$2 x + 3 y = \sin x$

Answer:

Given function is
$2 x + 3 y = \sin x$
We can rewrite it as
$3y = \sin x - 2x$
Now, differentiation w.r.t. x is
$3\frac{dy}{dx} = \frac{d(\sin x - 2x)}{dx} = \cos x - 2$
$\frac{dy}{dx} = \frac{\cos x-2}{3}$
Therefore, the answer is $\frac{\cos x-2}{3}$

Question 2.Find $\frac{dy}{dx}$ in the following: $2 x + 3y = \sin y$

Answer:

Given function is
$2 x + 3 y = \sin y$
We can rewrite it as
$\sin y - 3y = 2x$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(\sin y - 3y) = \frac{d( 2x)}{dx}$

$(\cos y\frac{dy}{dx} - 3\frac{dy}{dx}) = 2$
$\frac{dy}{dx} = \frac{2}{\cos y -3}$
Therefore, the answer is $\frac{2}{\cos y -3}$

Question 3.Find $\frac{dy}{dx}$ in the following: $ax + by ^2 = \cos y$

Answer:

Given function is
$ax + by ^2 = \cos y$
We can rewrite it as
$by^2-\cos y = -ax$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(2by - (-\sin y)) = \frac{d( -ax)}{dx} = -a$
$\frac{dy}{dx} = \frac{-a}{2b y +\sin y}$
Therefore, the answer is $\frac{-a}{2b y +\sin y}$

Question 4.Find $\frac{dy}{dx}$ in the following:

$xy + y^2 = \tan x + y$

Answer:

Given function is
$xy + y^2 = \tan x + y$
We can rewrite it as
$xy+y^2-y= \tan x$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y-1) = \frac{d( \tan x)}{dx} = \sec^2 x$
$\frac{dy}{dx} = \frac{\sec^2 x- y}{x+2y-1}$
Therefore, the answer is $\frac{\sec^2 x- y}{x+2y-1}$

Question 5.Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$

Answer:

Given function is
$x^2 + xy + y^2 = 100$
We can rewrite it as
$xy + y^2 = 100 - x^2$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y) = \frac{d( 100-x^2)}{dx} = -2x$
$\frac{dy}{dx} = \frac{-2 x- y}{x+2y}$
Therefore, the answer is $\frac{-2 x- y}{x+2y}$

Question 6Find $\frac{dy}{dx}$ in the following:

$x ^3 + x^2 y + xy^2 + y^3 = 81$

Answer:

Given function is
$x ^3 + x^2 y + xy^2 + y^3 = 81$
We can rewrite it as
$x^2 y + xy^2 + y^3 = 81 - x^3$
Now, differentiation w.r.t. x is
$\frac{d(x^2 y + xy^2 + y^3)}{dx} = \frac{d(81 - x^3)}{dx}$
$2xy+y^2+\frac{dy}{dx}(x^2+2xy+3y^2) = -3x^2\\ \frac{dy}{dx}=\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$
Therefore, the answer is $\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$

Question 7. Find $\frac{dy}{dx}$ in the following: $\sin ^ 2 y + \cos xy = k$

Answer:

Given function is
$\sin ^ 2 y + \cos xy = k$
Now, differentiation w.r.t. x is
$\frac{d(\sin^2y+\cos xy)}{dx} = \frac{d(k)}{dx}$
$2\sin y \cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0$

$\frac{dy}{dx}(2\sin y \cos y-x\sin xy)= y\sin xy\\ \frac{dy}{dx}$

$= \frac{y\sin xy}{2\sin y \cos y-x\sin xy}$

$= \frac{y\sin xy}{\sin 2y -x\sin xy} \ \ \ \ \ \ (\because 2\sin x\cos y = \sin 2x)$
Therefore, the answer is $\frac{y\sin xy}{\sin 2y -x\sin xy}$

Question 8.Find $\frac{dy}{dx}$ in the following:

$\sin ^2 x + \cos ^ 2 y = 1$

Answer:

Given function is
$\sin ^2 x + \cos ^ 2 y = 1$
We can rewrite it as
$\cos ^ 2 y = 1-\sin^2x$
Now, differentiation w.r.t. x is
$\frac{d(\cos^2y)}{dx} = \frac{d(1-\sin^2x)}{dx}$
$2\cos y (-\sin y)\frac{dy}{dx} = -2\sin x \cos x\\ \frac{dy}{dx} = \frac{2\sin x\cos x}{2\sin y \cos y} = \frac{\sin 2x }{\sin 2y} \ \ \ \ \ \ (\because2\sin a \cos a = \sin 2a)$
Therefore, the answer is $\frac{\sin 2x}{\sin 2y }$

Question 9Find $\frac{dy}{dx}$ in the following:

$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$

Answer:

Given function is
$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2 x})$
Our equation reduces to
$y = \sin^{-1}(\sin 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

Question 10.Find $\frac{dy}{dx}$ in the following:
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right ) , - \frac{1}{\sqrt 3 } < x < \frac{1}{\sqrt 3 }$

Answer:

Given function is
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{3x-x^3}{1-3x^2} = \frac{3\tan t-\tan^3t }{1-3\tan^2t} = \tan3t \ \ \ \ \ \ (\because \tan 3x = \frac{3\tan x-\tan^3x }{1-3\tan^2x} )$
Our equation reduces to
$y = \tan^{-1}(\tan 3t)$
$y = 3t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(3t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 3.\frac{1}{1+x^2} = \frac{3}{1+x^2}$
Therefore, the answer is $\frac{3}{1+x^2}$

Question 11.Find $\frac{dy}{dx}$ in the following:

$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right ) , 0 < x < 1$

Answer:

Given function is
$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$y = \cos^{-1}(\cos 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

Question 12. Find $\frac{dy}{dx}$ in the following: $y = \sin ^{-1 } \left ( \frac{1- x ^2 }{1+ x^2} \right ) , 0< x < 1$

Answer:

Given function is
$y = \sin ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
We can rewrite it as
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$\sin y = \cos 2t$
Now, differentiation w.r.t. x is
$\frac{d(\sin y)}{dx} = \frac{d(\cos2t)}{dt}.\frac{dt}{dx}$
$\cos y\frac{dy}{dx} = 2(-\sin 2t).\frac{1}{1+x^2} = \frac{-2\sin2t}{1+x^2}$ $= \frac{-2.\frac{2\tan t}{1+\tan^2t}}{1+x^2} =\frac{-2.\frac{2x}{1+x^2}}{1+x^2} =\frac{-4x}{(1+x^2)^2}$
$(\because \sin 2x = \frac{2\tan x}{1+\tan^2x} \ and \ x = \tan t)$
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )\rightarrow \cos y = \frac{2x}{1+x^2}$
$\frac{2x}{1+x^2}\frac{dy}{dx} = \frac{-4x}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question 13.Find $\frac{dy}{dx}$ in the following:

$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right ) , -1 < x < 1$

Answer:

Given function is
$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
We can rewrite it as
$\cos y = \left ( \frac{2x}{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2x} )$
Our equation reduces to
$\cos y = \sin 2t$
Now, differentiation w.r.t. x is
$\frac{d(\cos y)}{dx} = \frac{d(\sin2t)}{dt}.\frac{dt}{dx}$
$(-\sin y)\frac{dy}{dx} = 2(\cos 2t).\frac{1}{1+x^2} = \frac{2\cos2t}{1+x^2}$ $= \frac{2.\frac{1-\tan^2 t}{1+\tan^2t}}{1+x^2} =\frac{2.\frac{1-x^2}{1+x^2}}{1+x^2} =\frac{2(1-x^2)}{(1+x^2)^2}$
$(\because \cos 2x = \frac{1-\tan^2 x}{1+\tan^2x} \ and \ x = \tan t)$
$\cos y = \ \left ( \frac{2 x }{1+ x^2 } \right )\rightarrow \sin y = \frac{1-x^2}{1+x^2}$
$-\frac{1-x^2}{1+x^2}\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question 14. Find $\frac{dy}{dx}$ in the following:

$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} ) , -\frac{1}{\sqrt2} < x \frac{1}{\sqrt 2 }$

Answer:

Given function is
$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} )$
Let's take $x = \sin t$
Then,
$\frac{d(x)}{dx} = \frac{(\sin t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =\cos t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{1}{\cos t } = \frac{1}{\sqrt{1-\sin ^2t}} = \frac{1}{\sqrt{1-x^2}}$
$(\because \cos x = \sqrt{1-\sin^2x} \ and \ x = \sin t )$
And
$2x\sqrt{1-x^2} = 2\sin t \sqrt{1-\sin^2t} = 2\sin t \sqrt{\cos^2 t}$

$= 2\sin t\cos t =\sin 2t$
$(\because \cos x = \sqrt{1-\sin^2x} \text{ and }\ 2\sin x\cos x = \sin2x )$
Now, our equation reduces to
$y = \sin ^ { -1 } ( \sin 2t )$
$y = 2t$
Now, differentiation w.r.t. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{2}{\sqrt{1-x^2}}$

Question 15. Find $\frac{dy}{dx}$ in the following:

$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right ) , 0 < x < 1/ \sqrt 2$

Answer:

Given function is
$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right )$
Let's take $x = \cos t$
Then,
$\frac{d(x)}{dx} = \frac{(\cos t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =-\sin t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{-1}{\sin t } = \frac{-1}{\sqrt{1-\cos ^2t}} = \frac{-1}{\sqrt{1-x^2}}$
$(\because \sin x = \sqrt{1-\cos^2x} \ and \ x = \cos t )$
And
$\frac{1}{2x^2-1} =\frac{1}{2\cos^2 t - 1} = \frac{1}{\cos2t} = \sec2t$
$(\because \cos 2x = \sqrt{2\cos^2x-1} \text{ and }\cos x = \frac{1}{\sec x} )$

Now, our equation reduces to
$y = \sec ^{-1} \left ( \sec 2t \right )$
$y = 2t$
Now, differentiation w.r.t. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{-1}{\sqrt{1-x^2}} = \frac{-2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{-2}{\sqrt{1-x^2}}$

NCERT class 12 maths chapter 5 question answer: Exercise 5.4 (Page no. 130, Total questions- 10)

Question 1.Differentiate the following w.r.t. x:

$\frac{e ^x }{\sin x }$

Answer:

Given function is
$f(x)=\frac{e ^x }{\sin x }$
We differentiate with the help of the Quotient rule.
$f^{'}(x)=\frac{\frac{d(e^x)}{dx}.\sin x-e^x.\frac{(\sin x)}{dx} }{\sin^2 x }$
$=\frac{e^x.\sin x-e^x.\cos }{\sin^2 x } = \frac{e^x(\sin x-\cos x)}{\sin^2x}$
Therefore, the answer is $\frac{e^x(\sin x-\cos x)}{\sin^2x}$

Question 2. Differentiate the following w.r.t. x:

$e ^{\sin ^{-1}x}$

Answer:

Given function is
$f(x)=e ^{\sin ^{-1}x}$
Let $g(x)={\sin ^{-1}x}$
Then,
$f(x)=e^{g(x)}$
Now, differentiation w.r.t. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)
$g(x) = \sin^{-1}x \rightarrow g^{'}(x ) = \frac{1}{\sqrt{1-x^2}}$
Put this value in our equation (i)
$f^{'}(x) = \frac{1}{\sqrt{1-x^2}}.e^{\sin^{-1}x} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$

Question 3. Differentiate the following w.r.t. x:

$e ^{x^3}$

Answer:

Given function is
$f(x)=e ^{x^3}$
Let $g(x)=x^3$
Then,
$f(x)=e^{g(x)}$
Now, differentiation w.r.t. x
$f^{'}(x)=g^{'}(x).e^{g(x)}$ -(i)
$g(x) = x^3 \rightarrow g^{'}(x ) =3x^2$
Put this value in our equation (i)
$f^{'}(x) =3x^2.e^{x^3}$
Therefore, the answer is $3x^2.e^{x^3}$

Question 4.Differentiate the following w.r.t. x:

$\sin ( \tan ^ { -1} e ^{-x })$

Answer:

Given function is
$f(x)=\sin ( \tan ^ { -1} e ^{-x })$
Let's take $g(x ) = \tan^{-1}e^{-x}$
Now, our function reduces to
$f(x) = \sin(g(x))$
Now,
$f^{'}(x) = g^{'}(x)\cos(g(x))$ -(i)
And
$g(x)=\tan^{-1}e^{-x}\\\rightarrow g^{'}(x) = \frac{d(\tan^{-1}e^{-x})}{dx}.\frac{d(e^{-x})}{dx}= \frac{1}{1+(e^{-x})^2}.-e^{-x} = \frac{-e^{-x}}{1+e^{-2x}}$
Put this value in our equation (i)
$f^{'}(x) =\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$
Therefore, the answer is $\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})$

Question 5. Differentiate the following w.r.t. x:

$\log (\cos e ^x )$

Answer:

Given function is
$f(x)=\log (\cos e ^x )$
Let's take $g(x ) = \cos e^{x}$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\cos e^{x}\\\rightarrow g^{'}(x) = \frac{d(\cos e^{x})}{dx}.\frac{d(e^{x})}{dx}= (-\sin e^x).e^{x} = -e^x.\sin e^x$
Put this value in our equation (i)
$f^{'}(x) =-e^x.\sin e^x.\frac{1}{\cos e^x} = -e^x.\tan e^x \ \ \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)$
Therefore, the answer is $-e^x.\tan e^x,\ \ \ e^x\neq (2n+1)\frac{\pi}{2},\ \ n\in N$

Question 6. Differentiate the following w.r.t. x:

$e ^x + e ^{x^2} + .....e ^{x^5}$

Answer:

Given function is
$f(x)= e ^x + e ^{x^2} + .....e ^{x^5}$
Now, differentiation w.r.t. x is
$f^{'}(x)= \frac{d(e^x)}{dx}.\frac{d(x)}{dx}+\frac{d(e^{x^2})}{dx}.\frac{d(x^2)}{dx}+\frac{d(e^{x^3})}{dx}.\frac{d(x^3)}{dx}+\frac{d(e^{x^4})}{dx}.\frac{d(x^4)}{dx}+\frac{d(e^{x^5})}{dx}.\frac{d(x^5)}{dx}$
$=e^x.1+e^{x^2}.2x+e^{x^3}.3x^2+e^{x^4}.4x^3+e^{x^5}.5x^4$
$=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$
Therefore, answer is $e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$

Question 7. Differentiate the following w.r.t. x:

$\sqrt { e ^{ \sqrt x }} , x > 0$

Answer:

Given function is
$f(x)=\sqrt { e ^{ \sqrt x }}$
Let's take $g(x ) = \sqrt x$
Now, our function reduces to
$f(x) = \sqrt {e^{g(x)}}$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.\frac{d({e^{g(x)}})}{dx} = g{'}(x).\frac{1}{2\sqrt{e^{g(x)}}}.{e^{g(x)}} = \frac{g^{'}(x).e^{g(x)}}{2.\sqrt{e^{g(x)}}} = \frac{g^{'}(x).e^{\sqrt x}}{2.\sqrt{e^{\sqrt x}}}$ -(i)
And
$g(x)=\sqrt x\\\rightarrow g^{'}(x) = \frac{(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{e^{\sqrt x}}{2\sqrt x.2.\sqrt{e^{\sqrt x}}} = \frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}$
Therefore, the answer is $\frac{e^{\sqrt x}}{4\sqrt{xe^{\sqrt x}}}.\ \ x>0$

Question 8Differentiate the following w.r.t. x: $\log ( \log x ) , x > 1$

Answer:

Given function is
$f(x)=\log ( \log x )$
Let's take $g(x ) = \log x$
Now, our function reduces to
$f(x) = \log(g(x))$
Now,
$f^{'}(x) = g^{'}(x).\frac{1}{g(x)}$ -(i)
And
$g(x)=\log x\\\rightarrow g^{'}(x) = \frac{1}{x}$
Put this value in our equation (i)
$f^{'}(x) =\frac{1}{x}.\frac{1}{\log x} = \frac{1}{x\log x}$
Therefore, the answer is $\frac{1}{x\log x}, \ \ x>1$

Question 9.Differentiate the following w.r.t. x:

$\frac{\cos x }{\log x} , x > 0$

Answer:

Given function is
$f(x)=\frac{\cos x }{\log x}$
We differentiate with the help of the Quotient rule.
$f^{'}(x)=\frac{\frac{d(\cos x)}{dx}.\log x-\cos x.\frac{(\log x)}{dx} }{(\log x)^2 }$
$=\frac{(-\sin x).\log x-\cos x.\frac{1}{x} }{(\log x)^2 } = \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$
Therefore, the answer is $\frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$

Question 10.Differentiate the following w.r.t. x:

$\cos ( log x + e ^x ) , x > 0$

Answer:

Given function is
$f(x)=\cos ( log x + e ^x )$
Let's take $g(x) = ( log x + e ^x )$
Then, our function reduces to
$f(x) = \cos (g(x))$
Now, differentiation w.r.t. x is
$f^{'}(x) = g^{'}(x)(-\sin) (g(x))$ -(i)
And
$g(x) = ( log x + e ^x )$
$g^{'}(x)= \frac{d(\log x)}{dx}+\frac{d(e^x)}{dx} = \frac{1}{x}+e^x$
Put this value in our equation (i)
$f^{'}(x) = -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x)$
Therefore, the answer is $-\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0$

Class 12 Maths Chapter 5 NCERT solutions: Exercise: 5.5 (Page no. 134, Total questions- 18)

Question 1Differentiate the functions w.r.t. x. $\cos x . \cos 2x .\cos 3x$

Answer:

Given function is
$y=\cos x. \cos 2x .\cos 3x$
Now, take a look at both sides.
$\log y=\log (\cos x . \cos 2x .\cos 3x)$

$\log y = \log \cos x + \log \cos 2x + \log \cos 3x$
Now, differentiation w.r.t. x
$\log y=\log (\cos x . \cos 2x .\cos 3x)$

$\frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}$

$\frac{1}{y}.\frac{dy}{dx} =$

$(-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}$

$\frac{1}{y} \frac{dy}{dx}= (\tan x+ \tan 2x+ \tan 3x )$

$(\because \frac{\sin x }{\cos x} =\tan x)$

$ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)$

$ \frac{dy}{dx}= -\cos x \cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$
There, the answer is $-\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$

Question 2.Differentiate the functions w.r.t. x.

$\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$

Answer:

Given function is
$y=\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$
Take logs on both sides.
$\log y=\frac{1}{2}\log\left ( \frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)} \right )$

$\log y =$

$\frac{1}{2} (\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)$

$-\log(x-5))$
Now, differentiation w.r.t. x is
$\frac{d(\log y)}{dx} = \frac{1}{2} (\frac{d(\log(x-1))}{dx}+\frac{d(\log(x-2))}{dx}-\frac{d(\log(x-3))}{dx}$

$-\frac{d(\log(x-4))}{dx}- \frac{d(\log(x-5))}{dx})$
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx}$

$=y\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

$ \frac{dy}{dx} = \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}$

$(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$
Therefore, the answer is $\frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$

Question 3Differentiate the functions w.r.t. x. $(\log x ) ^{\cos x}$

Answer:

Given function is
$y=(\log x ) ^{\cos x}$
Take logs on both sides.
$\log y=\cos x\log (\log x )$
Now, differentiation w.r.t x is

$\frac{d(\log y)}{dx}= \frac{d(\cos x\log(\log x))}{dx}$

$\frac{1}{y} \frac{dy}{dx}= (-\sin x)(\log(\log x))+\cos x (\frac{1}{\log x} \cdot \frac{1}{x})$

$\frac{dy}{dx}= y( \cos x.\frac{1}{\log x}.\frac{1}{x}-\sin x\log(\log x) )\\$

$\frac{dy}{dx} = (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$
Therefore, the answer is $(\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )$

Question 4Differentiate the functions w.r.t. x. $x ^x - 2 ^{ \sin x }$

Answer:

Given function is
$y = x ^x - 2 ^{ \sin x }$
Let's take $t = x^x$
Take logs on both sides.
$\log t=x\log x\\$
Now, differentiation w.r.t x is
$\log t=x\log x\\ \frac{d(\log t)}{dt}.\frac{dt}{dx} = \frac{d(x\log x)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)$

$\frac{1}{t}.\frac{dt}{dx} = \log x +1$

$\frac{dt}{dx} = t(\log x+1)$

$\frac{dt}{dx}= x^x(\log x+1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because t = x^x )$
Similarly, take $k = 2^{\sin x}$
Now, take log on both sides and differentiate w.r.t. x
$\log k=\sin x\log 2\\ \frac{d(\log k)}{dk}.\frac{dk}{dx} = \frac{d(\sin x\log 2)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)$

$\frac{1}{k}.\frac{dk}{dx} = \cos x \log 2$

$\frac{dk}{dx} = k(\cos x \log 2)$

$\frac{dk}{dx}= 2^{\sin x}(\cos x\log 2)$

$(\because k = 2^{\sin x} )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}-\frac{dk}{dx}$

$\frac{dy}{dx} = x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)$
Therefore, the answer is $x^x(\log x+1 )- 2^{\sin x}(\cos x \log 2)$

Question 5Differentiate the functions w.r.t. x. $( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$

Answer:

Given function is
$y=( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4$
Take logs on both sides.
$\log y=\log [( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4]$

$ \log y = 2\log(x+3)+3\log(x+4)+4\log(x+5)$
Now, differentiate w.r.t. x we get,
$\frac{1}{y}.\frac{dy}{dx} = 2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}$

$\frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right ) $

$\frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )$

$\frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.$

$\left ( \frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right )$

$ \frac{dy}{dx} = (x + 3) (x + 4)^2 (x + 5)^3$

$(9x^2 + 70x + 133)$
Therefore, the answer is $(x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)$

Question 6Differentiate the functions w.r.t. x. $( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$

Answer:

Given function is
$y = ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }$
Let's take $t = ( x+ \frac{1}{x} ) ^ x$
Now, take a look at both sides.
$\log t =x \log ( x+ \frac{1}{x} )$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}.\frac{dt}{dx}=\log \left ( x+\frac{1}{x} \right )+x(1-\frac{1}{x^2}).\frac{1}{\left ( x+\frac{1}{x} \right )}$

$= \frac{x^2-1}{x^2+1}+\log \left ( x+\frac{1}{x} \right )$

$ \frac{dt}{dx} = t(\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))$

$ \frac{dt}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))$
Similarly, take $k = x^{1+\frac{1}{x}}$
Now, take a look at both sides.
$\log k = ({1+\frac{1}{x}})\log x$
Now, differentiate w.r.t. x
We get,
$\frac{1}{k}.\frac{dk}{dx}=\frac{1}{x} \left ( 1+\frac{1}{x} \right )+(-\frac{1}{x^2}).\log x$

$= \frac{x^2+1}{x^2}+\frac{-1}{x^2}.\log x$

$ \frac{dk}{dx} = t(\frac{x^2+1}{x^2}+\left (\frac{-1}{x^2} \right )\log x)$

$\frac{dk}{dx} = x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+$

$\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$
Therefore, the answer is $\left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )$

Question 7Differentiate the functions w.r.t. x. $(\log x )^x + x ^{\log x }$

Answer:

Given function is
$y = (\log x )^x + x ^{\log x }$
Let's take $t = (\log x)^x$
Now, take a look at both sides.
$\log t = x \log(\log x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \log (\log x) + x.\frac{1}{x}.\frac{1}{\log x}$

$= \log (\log x)+\frac{1}{\log x}\\$

$ \frac{dt}{dx}= t.(\log (\log x)+\frac{1}{\log x})\\$

$ \frac{dt}{dx} =(\log x)^x(\log (\log x)) + (\log x)^x.\frac{1}{\log x}$

$=(\log x)^x(\log (\log x))+ (\log x )^{x-1}$
Similarly, take $k = x^{\log x}$
Now, take a look at both sides.
$\log k = \log x \log x = (\log x)^2$
Now, differentiate w.r.t. x
We get,
$\frac{1}{k}\frac{dk}{dx} =2 (\log x).\frac{1}{x} \\$

$ \frac{dt}{dx}= k.\left ( 2 (\log x).\frac{1}{x} \right )\\$

$ \frac{dt}{dx} = x^{\log x}.\left (2 (\log x).\frac{1}{x} \right ) = 2x^{\log x-1}.\log x$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$
Therefore, the answer is $(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x$

Question 8: Differentiate the functions w.r.t. x. $(\sin x )^x + \sin ^{-1} \sqrt x$

Answer:

Given function is
$(\sin x )^x + \sin ^{-1} \sqrt x$
Let's take $t = (\sin x)^x$
Now, take a look at both sides.
$\log t = x \log(\sin x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \log (\sin x) + x.\cos x.\frac{1}{\sin x}$

$= \log (\sin x)+x.\cot x$

$ \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)\\$

$ \frac{dt}{dx}= t.(\log (\sin x)+x.\cot x)\\$

$ \frac{dt}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)$
Similarly, take $k = \sin^{-1}\sqrt x$
Now, differentiate w.r.t. x
We get,
$\frac{dk}{dt} = \frac{1}{\sqrt{1-(\sqrt x)^2}}.\frac{1}{2\sqrt x}$

$= \frac{1}{2\sqrt{x-x^2}}\\$

$ \frac{dk}{dt}=\frac{1}{2\sqrt{x-x^2}}\\$
Now,
$\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$
Therefore, the answer is $(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}$

Question 9Differentiate the functions w.r.t. x $y=x^{\sin x}+(\sin x)^{\cos x}$

Answer:

Given function is
$y=x^{\sin x}+(\sin x)^{\cos x}$

Now, take $t = x^{\sin x}$
Now, take a look at both sides.
$\log t = \sin x \log x$
Now, differentiate it w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\$

$\frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\$

$\frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )$
Similarly, take $k = (\sin x)^{\cos x}$
Now, take a look at both sides.
$\log k = \cos x \log (\sin x)$
Now, differentiate it w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x$

$=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}$

$= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ $

$\frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Now,
$\frac{dy}{dx} = x^{\sin x}(\cos x \log x+\frac{1}{x}.\sin x)+$

$(\sin x)^{\cos x} ( -\sin x\log(\sin x)+\cot x.\cos x)$
Therefore, the answer is $x^{\sin x}\left (\cos x\log x+\frac{1}{x}.\sin x \right )+$

$(\sin x)^{\cos x}\left (-\sin x\log(\sin x)+\cot x.\cos x\right )$

Question 10: Differentiate the functions w.r.t. x. $x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }$

Answer:

Given function is
$y=x^{\sin x}+(\sin x)^{\cos x}$

Now, take $t = x^{\sin x}$
Now, take a look at both sides.
$\log t = \sin x \log x$
Now, differentiate it w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\$

$\frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\$

$\frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )$
Similarly, take $k = (\sin x)^{\cos x}$
Now, take a look at both sides.
$\log k = \cos x \log (\sin x)$
Now, differentiate it w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x$

$=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}$

$= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ $

$\frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )$
Now,
$\frac{dy}{dx} = x^{\sin x}(\cos x \log x+\frac{1}{x}.\sin x)+$

$(\sin x)^{\cos x} ( -\sin x\log(\sin x)+\cot x.\cos x)$
Therefore, the answer is $x^{\sin x}\left (\cos x\log x+\frac{1}{x}.\sin x \right )+$

$(\sin x)^{\cos x}\left (-\sin x\log(\sin x)+\cot x.\cos x\right )$

Question 11: Differentiate the functions w.r.t. x. $( x \cos x )^ x + ( x \sin x )^{1/ x}$

Answer:

Given function is
$f(x)=( x \cos x)^ x + ( x \sin x )^{1/ x}$
Let's take $t = (x\cos x)^x$
Now, take a look at both sides.
$\log t =x\log (x\cos x) = x(\log x+\log \cos x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{t}\frac{dt}{dx} =(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}.(-\sin x))\\$

$\frac{dt}{dx} = t(\log x + \log \cos x+1-x\tan x) $

$\ \ \ \ \ \ \ \ \ (\because \frac{\sin x}{\cos x}= \tan x)\\$

$\frac{dt}{dx}= (x\cos x)^x(\log x + \log \cos x+1-x\tan x)\\$

$\frac{dt}{dx}=(x\cos x)^x(+1-x\tan x+\log (x\cos x))$
Similarly, take $k = (x\sin x)^{\frac{1}{x}}$
Now, take a look at both sides.
$\log k = \frac{1}{x}(\log x+\log \sin x)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} =(\frac{-1}{x^2})(\log x+\log \sin x)+$

$\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.(\cos x))\\$

$\frac{dk}{dx} = \frac{k}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})$

$\ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x}= \cot x)\\$

$\frac{dk}{dx}=\frac{(x\sin x)^{\frac{1}{x}}}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})\\$

$\frac{dk}{dx}=(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Now,
$\frac{dy}{dx}= \frac{dt}{dx}+\frac{dk}{dx}$
$\frac{dy}{dx}= (x\cos x)^x(+1-x\tan x+\log (x\cos x))+$

$(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$
Therefore, the answer is $(x\cos x)^x(1-x\tan x+\log (x\cos x))+$

$(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}$

Question 12Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15

$x ^ y + y ^ x = 1$ .

Answer:

Given function is
$f(x)=x ^ y + y ^ x = 1$
Now, take $t = x^y$
Take logs on both sides.
$\log t = y\log x$
Now, differentiate w.r.t x
we get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\$

$ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\$

$ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take a look at both sides.
$\log k = x\log y$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\$

$ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\$

$ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)= \frac{dt}{dx}+\frac{dk}{dx}= 0$

$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) + (y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) = 0 $

$\frac{dy}{dx}(x^y(\log x)+xy^{x-1}) = -(yx^{y-1}+y^x(\log y))\\ $

$\frac{dy}{dx}= \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$

Therefore, the answer is $\frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}$

Question 13 Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15.

$y^x = x ^y$

Answer:

Given function is
$f(x)\rightarrow x ^ y = y ^ x$
Now, take $t = x^y$
Take logs on both sides.
$\log t = y\log x$
Now, differentiate w.r.t x
we get,
$\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\$

$ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})$

$ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})$
Similarly, take $k = y^x$
Now, take a look at both sides.
$\log k = x\log y$
Now, differentiate w.r.t. x
we get,
$\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\$

$ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ $

$\frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})$
Now,
$f^{'}(x)\rightarrow \frac{dt}{dx}= \frac{dk}{dx}$

$( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) = (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})\\$

$ \frac{dy}{dx}(x^y(\log x)-xy^{x-1}) = (y^x(\log y)-yx^{y-1})\\$

$ \frac{dy}{dx}= \frac{ y^x(\log y)-yx^{y-1}}{(x^y(\log x)-xy^{x-1})} $

$= \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$

Therefore, the answer is $\frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )$

Question 14Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15. $(\cos x )^y = ( \cos y )^x$

Answer:

Given function is
$f(x)\rightarrow (\cos x) ^ y = (\cos y) ^ x$
Now, take log on both sides.
$y\log \cos x = x \log \cos y$
Now, differentiate w.r.t x
$\frac{dy}{dx}(\log \cos x)-y\tan x = \log \cos y-x\tan y\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{dy}{dx}(\log \cos x)-y\tan x)$

$= (\log \cos y-x\tan y\frac{dy}{dx})\\ \frac{dy}{dx} \left (\log \cos x+(\cos y)^x.x\tan y) \right )$

$= \left ( \log \cos y+(\cos x)^y.y\tan x \right )\\ \frac{dy}{dx}$

$= \frac{\left (\log \cos y+y\tan x \right )}{\left ( \log \cos x+x\tan y) \right )}$

$= \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

Therefore, the answer is $\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$

Question 15Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15. $xy = e ^{x-y}$

Answer:

Given function is
$f(x)\rightarrow xy = e ^{x-y}$
Now, take a look at both sides.
$\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)$
Now, differentiate w.r.t x
$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
By taking similar terms on the same side
We get,
$(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}$
Therefore, the answer is $\frac{y}{x}.\frac{x-1}{y+1}$

Question 16Find the derivative of the function given by $f (x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find

f ' (1)

Answer:

Given function is
$y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$
Take logs on both sides.
$\log y =\log (1 + x) + \log (1 + x^2) +\log (1 + x^4) +\log (1 + x^8)$
NOW, differentiate w.r.t. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}\\$

$ \frac{dy}{dx}=y.\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )\\ $

$\frac{dy}{dx}= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Therefore, $f^{'}(x)=$

$ (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )$
Now, the value of $f^{'}(1)$ is
$f^{'}(1)= (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8).\left ( \frac{1}{1+1}+ \frac{2(1)}{1+1^2}+\frac{4(1)^3}{1+1^4}+ \frac{8(1)^7}{1+1^8} \right )\\$

$ f^{'}(1)=16.\frac{15}{2} = 120$

Question 17 (1)Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(i) By using product rule

Answer:

Given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, we need to differentiate using the product rule.
$f^{'}(x)=\frac{d((x^2 - 5x + 8))}{dx}. (x^3 + 7x + 9)+(x^2 - 5x + 8).\frac{d( (x^3 + 7x + 9))}{dx}\\$
$= (2x-5).(x^3+7x+9)+(x^2-5x+8)(3x^2+7)\\$

$ =2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56\\$

$ = 5x^4 -20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4 -20x^3+45x^2-52x+11$

Question 17 (2)Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(ii) by expanding the product to obtain a single polynomial.

Answer:

Given function is
$f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Multiply both to obtain a single higher-degree polynomial.
$f(x) = x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)$
$= x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72$
$= x^5-5x^4+15x^3-26x^2+11x+72$
Now, differentiate w.r.t. x
we get,
$f^{'}(x)=5x^4-20x^3+45x^2-52x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-52x+11$

Question 17 (3)Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
(iii) by logarithmic differentiation.
Do they all give the same answer?

Answer:

Given function is
$y=(x^2 - 5x + 8) (x^3 + 7x + 9)$
Now, take a look at both sides.
$\log y = \log (x^2-5x+8)+\log (x^3+7x+9)$
Now, differentiate w.r.t. x
we get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2-5x+8}.(2x-5) + \frac{1}{x^3+7x+9}.(3x^2+7)\\$

$ \frac{dy}{dx}= y.\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\$

$ \frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9).\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\$

$ \frac{dy}{dx} = (2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)\\ \frac{dy}{dx} = 5x^4-20x^3+45x^2-56x+11$
Therefore, the answer is $5x^4-20x^3+45x^2-56x+11$
And yes, they all give the same answer.

Question 18 If u, v and w are functions of x, then show that $\frac{d}{dx} ( u,v,w) = \frac{du}{dx} v. w +u. \frac{dv }{dx } v. w+ u . \frac{dv}{dx } . w+u.v \frac{dw}{dx}$ in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Answer:

It is given that u, v and w are the functions of x
Let $y = u.v.w$
Now, we differentiate using the product rule w.r.t x
First, take $y = u.(vw)$
Now,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{d(v.w)}{dx}.u$ -(i)
Now, again, by the product rule
$\frac{d(v.w)}{dx}= \frac{dv}{dx}.w + \frac{dw}{dx}.v$
Put this in equation (i)
we get,
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, by the product rule, we proved it.

Now, by taking the log
Again take $y = u.v.w$
Now, take a look at both sides.
$\log y = \log u + \log v + \log w$
Now, differentiate w.r.t. x
we get,
$\frac{1}{y}.\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\ \frac{dy}{dx}= y. \left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\ \frac{dy}{dx} $

$= (u.v.w)\left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\$
$\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)$
Hence, we proved it by taking the log.

Class 12 Maths Chapter 5 NCERT solutions: Exercise 5.6 (Page no. 137, Total questions- 11)

Question 1: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$.

$x = 2at^2, y = at^4$

Answer:

Given equations are
$x = 2at^2, y = at^4$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(2at^2)}{dt}= 4at$
Similarly,
$\frac{dy}{dt}=\frac{d(at^4)}{dt}= 4at^3$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{4at^3}{4at} = t^2$
Therefore, the answer is $\frac{dy}{dx}= t^2$

Question 2: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$.

Answer:

Given equations are
$x= a \cos \theta , y = b \cos \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\cos \theta)}{d\theta}= -a\sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\cos \theta)}{d\theta}= -b\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-b\sin \theta}{-a\sin \theta} = \frac{b}{a}$
Therefore, answer is $\frac{dy}{dx}= \frac{b}{a}$

Question 3: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$. $x = \sin t , y = \cos 2 t$

Answer:

Given equations are
$x = \sin t , y = \cos 2 t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t$
Similarly,
$\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t $

$\ \ \ \ \ (\because \sin 2x = \sin x\cos x)$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t$
Therefore, the answer is $\frac{dy}{dx} = -4\sin t$

Question 4If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\ \frac {dy}{dx}$

$x = 4t , y = 4/t$

Answer:

Given equations are
$x = 4t , y = 4/t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(4 t)}{dt}= 4$
Similarly,
$\frac{dy}{dt}=\frac{d(\frac{4}{t})}{dt}= \frac{-4}{t^2}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{ \frac{-4}{t^2} }{4} = \frac{-1}{t^2}$
Therefore, the answer is $\frac{dy}{dx} = \frac{-1}{t^2}$

Question 5: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2 \theta$

Answer:

Given equations are
$x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$
Therefore, answer is $\frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$

Question 6If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = a ( \theta - \sin \theta ), y = a ( 1+ \cos \theta )$

Answer:

Given equations are
$x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(1+\cos \theta))}{d\theta}=-a\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-a\sin \theta}{a(1-\cos \theta)} = \frac{-\sin }{1-\cos \theta} =- \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})$
Therefore, the answer is $\frac{dy}{dx}=-\cot \frac{\theta}{2}$

Question 7If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = \frac{\sin ^3 t }{\sqrt {\cos 2t }}, y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$

Answer:

Given equations are
$x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$
Now, differentiate both w.r.t
We get,
$\frac{dx}{dt}=\frac{d(\frac{\sin ^3 t }{\sqrt {\cos 2t }})}{dt}$

$=\frac{\sqrt{\cos 2t}.\frac{d(\sin^3t)}{dt}-\sin^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2}$

$=\frac{3\sin^2 t\cos t.\sqrt{\cos 2t}-\sin^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{{\cos 2t}}$
$=\frac{3\sin^2t\cos t . \cos 2t+sin^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}} \ \ \ \ \ (\because \frac{\cos }{\sin x}=\cot x)$
Similarly,
$\frac{dy}{dt}=\frac{d( \frac{\cos ^3 t }{\sqrt {\cos 2t }})}{dt}$

$=\frac{\sqrt{\cos 2t}.\frac{d(\cos^3t)}{dt}-\cos^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2}$

$=\frac{3\cos^2 t(-\sin t).\sqrt{\cos 2t}-\cos^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{(\sqrt{\cos 2t})^2}$
$=\frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$= \frac{\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}} }{\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}}}$

$= \frac{\cot^3t(1-3\tan t \cot 2t)}{(3\cot t \cot 2t+1)}$
$= \frac{\cos^3t(1-3.\frac{\sin t}{\cos t}.\frac{\cos2t}{\sin 2t})}{\sin^3t(3.\frac{\cos t}{\sin t}.\frac{\cos 2t}{\sin 2t}+1)} $

$= \frac{\cos^2t(\cos t\sin2t -3\sin t \cos 2t)}{\sin^2t(3\cos t \cos2t+\sin t \sin 2t)}$
$=\frac{\cos^2t(\cos t .2\sin t \cos t - 3\sin t (2\cos^2t-1))}{\sin^2t(3\cos t(1-2\sin^2 2t)+\sin t.2\sin t \cos t)}$
$(\because \sin 2x = 2\sin x\cos x \ and \ \cos 2x = 2\cos^2x-1 \ and \ \cos 2x = 1-2\sin^2x)$
$=\frac{\cos^2t(2\sin t\cos^2 t-6\sin t\cos^2t+3\sin t)}{\sin^2t(3\cos t-6\cos t \sin^2t+2\sin^2\cos t)}\\=\frac{sint cost(-4cos^3t+3cost)}{sintcost(3sint-4sin^3t)}$

$\frac{dy}{dx} = \frac{-4\cos^3t+3\cos t}{3\sin t -4\sin^3 t}= \frac{-\cos 3t}{\sin 3t} = -\cot 3t$ $\left ( \because \sin3t = 3\sin t-4\sin^3t \\ \ and \ \cos3t = 4\cos^3t - 3\cos t \right )$

Therefore, the answer is $\frac{dy}{dx} = -\cot 3t$

Question 8: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = a ( \cos t + \log \tan t/2 ),y = a \sin t$

Answer:

Given equations are
$x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})$
$= a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})$
$=a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})$
Similarly,
$\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t$
Therefore, the answer is $\frac{dy}{dx} = \tan t$

Question 9: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find $\frac{dy}{dx}$ $x = a \sec \theta, y = b \ tan \theta$

Answer:

Given equations are
$x = a \sec \theta , y = b \ tan \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}$
Therefore, the answer is $\frac{dy}{dx} = \frac{b cosec \theta}{a}$

Question 10If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{dy}{dx}$ $x = a ( \cos \theta + \theta \sin \theta ), y = a ( \sin \theta - \theta \cos \theta )$

Answer:

Given equations are
$x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta$
Therefore, the answer is $\frac{dy}{dx}= \tan \theta$

Question 11If $x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$ , show that $\frac{dy}{dx}$ = - y /x$

Answer:

Given equations are
$x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$

$xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\sin ^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c$

Differentiating with respect to x

$x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x}$

Class 12 Maths Chapter 5 NCERT solutions: Exercise: 5.7 (Page no. 139, Total questions- 17)

Question 1Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x^2 + 3x+ 2$

Answer:

Given function is
$y=x^2 + 3x+ 2$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= 2x+3$
Now, second-order derivative
$\frac{d^2y}{dx^2}= 2$
Therefore, the second order derivative is $\frac{d^2y}{dx^2}= 2$

Question 2Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x ^{20}$

Answer:

Given function is
$y=x ^{20}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= 20x^{19}$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}$
Therefore, second-order derivative is $\frac{d^2y}{dx^2}= 380x^{18}$

Question 3Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x \cos x$

Answer:

Given function is
$y = x \cos x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2}= -2\sin x - x\sin x$

Question 4Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\log x$

Answer:

Given function is
$y=\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{1}{x}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2}= \frac{-1}{x^2}$

Question 5Find the second-order derivatives of the functions given in Exercises 1 to 10.

$x ^3 \log x$

Answer:

Given function is
$y=x^3\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2} = x(6.\log x+5)$

Question 6Find the second-order derivatives of the functions given in Exercises 1 to 10.

$e ^x \sin5 x$

Answer:

Given function is
$y= e^x\sin 5x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))$
$= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)$
$=2e^x(5\cos5x-12\sin5x)$
Therefore, second order derivative is $\frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)$

Question 7Find the second-order derivatives of the functions given in Exercises 1 to 10.

$e ^{6x}\cos 3x$

Answer:

Given function is
$y= e^{6x}\cos 3x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)$
$= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)$
$e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)$
Therefore, second order derivative is $\frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)$

Question 8Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\tan ^{-1} x$

Answer:

Given function is
$y = \tan^{-1}x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}$

Question 9Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\log (\log x )$

Answer:

Given function is
$y = \log(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}$

Question 10Find the second-order derivatives of the functions given in Exercises 1 to 10.

$\sin (\log x )$

Answer:

Given function is
$y = \sin(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}$
Now, the second order derivative is
Using Quotient rule
$\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$

Question 11If $y = 5 \cos x - 3 \sin x$ prove that $\frac{d^2y}{dx^2}+y = 0$

Answer:

Given function is
$y = 5 \cos x - 3 \sin x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}$

$=-5\sin x-3\cos x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}$

$=-5\cos x+3\sin x$
Now,
$\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0$
Hence proved

Question 12If $y = \cos ^{-1} x$ Find $\frac{d ^2 y }{dx^2 }$ in terms of y alone.

Answer:

Given function is
$y = \cos ^{-1} x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}$
Now, the second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2}$ -(i)
Now, we want $\frac{d^2y}{dx^2}$ in terms of y
$y = \cos ^{-1} x$
$x = \cos y$
Now, put the value of x in (i)
$\frac{d^2y}{dx^2} = \frac{-2\cos y }{1-\cos^2 y } = \frac{-2\cos y}{\sin ^2 y}= -2\cot y cosec y$
$(\because 1-\cos^2x =\sin^2 x\ and \ \frac{\cos x}{\sin x} = \cot x \ and \ \frac{1}{\sin x}= cosec x)$
Therefore, answer is $\frac{d^2y}{dx^2} = -2\cot y cosec y$

Question 13 If $y = 3 \cos (\log x) + 4 \sin (\log x)$ , show that $x^2 y_2 + xy_1 + y = 0$

Answer:

Given function is
$y = 3 \cos (\log x) + 4 \sin (\log x)$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}$
$=\frac{4\cos (\log x)-3\sin(\log x)}{x}$ -(i)
Now, the second order derivative is
By using the Quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{4\cos (\log x)-3\sin(\log x)}{x})}{dx^2}= \frac{(-4\sin(\log x).\frac{1}{x}-3\cos(\log x).\frac{1}{x}).x-1.(4\cos (\log x)-3\sin(\log x))}{x^2}$
$=\frac{-\sin(\log x)+7\cos (\log x)}{x^2}$ -(ii)
Now, from equation (i) and (ii), we will get $y_1 \ and \ y_2$
Now, we need to show.
$x^2 y_2 + xy_1 + y = 0$
Put the value of $y_1 \ and \ y_2$ from equation (i) and (ii)
$x^2\left ( \frac{-\sin(\log x)+7\cos (\log x)}{x^2} \right )+x\left ( \frac{4\cos (\log x)-3\sin(\log x)}{x} \right )+ 3 \cos (\log x)$ $+4\sin(\log x)$
$-\sin(\log x)-7\cos(\log x)+4\cos(\log x)-3\sin(\log x)+3\cos (\log x)$ $+4\sin(\log x)$
$=0$
Hence proved

Question 14If $y = A e ^{mx} + Be ^{nx}$ , show that $\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + many = 0$

Answer:

Given function is
$y = A e ^{mx} + Be ^{nx}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx}$ -(i)
Now, the second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx}$ -(ii)
Now, we need to show.
$\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0$
Put the value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})$
$m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}$ $+mnBe^{nx}$
$=0$
Hence proved

Question 15If $y = 500 e ^{7x} + 600 e ^{- 7x }$ , show that $\frac{d^2 y}{dx ^2} = 49 y$
Answer:

Given function is
$y = 500 e ^{7x} + 600 e ^{- 7x }$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x}$ -(i)
Now, the second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}$
$= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x}$ -(ii)
Now, we need to show.
$\frac{d^2 y}{dx ^2} = 49 y$
Put the value of $\frac{d^2y}{dx^2}$ from equation (ii)
$24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})$
$= 24500e^{7x}+29400e^{-7x}$
Hence, L.H.S. = R.H.S.
Hence proved

Question 16If $e ^y (x+1) = 1$ show that $\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$

Answer:

Given function is
$e ^y (x+1) = 1$
We can rewrite it as
$e^y = \frac{1}{x+1}$
Now, differentiation w.r.t. x
$\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1}$ -(i)
Now, the second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2}$ -(ii)
Now, we need to show.
$\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$
Put value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2$
$=\frac{1}{(x+1)^2}$
Hence, L.H.S. = R.H.S.
Hence proved

Question 17If $y = (\tan^{-1} x)^2$ show that $(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$

Answer:

Given function is
$y = (\tan^{-1} x)^2$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2}$ -(i)
Now, the second-order derivative is
By using the quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2}$ -(ii)
Now, we need to show.
$(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$
Put the value from equation (i) and (ii)
$(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2$
Hence, L.H.S. = R.H.S.
Hence proved

NCERT Class 12 continuity and differentiability NCERT solutions Miscellaneous Exercise (Page no. 144, Total questions- 22)

Question 1Differentiate w.r.t. x the function in Exercises 1 to 11.

$( 3x^2 - 9x + 5 )^9$

Answer:

Given function is
$f(x)=( 3x^2 - 9x + 5 )^9$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d((3x^2-9x+5)^9)}{dx}= 9(3x^2-9x+5)^8.(6x-9)$
$= 27(2x-3)(3x^2-9x+5)^8$
Therefore, differentiation w.r.t. x is $27(3x^2-9x+5)^8(2x-3)$

Question 2Differentiate w.r.t. x the function in Exercises 1 to 11.

$\sin ^3 x + \cos ^6 x$

Answer:

Given function is
$f(x)= \sin ^3 x + \cos ^6 x$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^3x +\cos^6x)}{dx}=3\sin^2x.\frac{d(\sin x)}{dx}+6\cos^5x.\frac{d(\cos x)}{dx}$
$=3\sin^2x.\cos x+6\cos^5x.(-\sin x)$
$=3\sin^2x\cos x- 6\cos^5x\sin x = 3\sin x\cos x(\sin x- 2\cos ^4x)$

Therefore, differentiation w.r.t. x is $3\sin x\cos x(\sin x- 2\cos ^4x)$

QuestionDifferentiate w.r.t. x the function in Exercises 1 to 11.

$( 5 x) ^{ 3 \cos 2x }$

Answer:

Given function is
$y=( 5 x) ^{ 3 \cos 2x }$
Take a log on both sides.
$\log y = 3\cos 2x\log 5x$
Now, differentiation w.r.t. x is
By using the product rule
$\frac{1}{y}.\frac{dy}{dx} = 3.(-2\sin 2x)\log 5x + 3\cos 2x.\frac{1}{5x}.5$

$= -6\sin2x\log 5x +\frac{3\cos 2x}{x}\\ \frac{dy}{dx} $

$= y.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )\\$

$ \frac{dy}{dx} = (5x)^{3\cos 2x}.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )$

Therefore, differentiation w.r.t. x is $(5x)^{3\cos 2x}.\left ( \frac{3\cos 2x}{x}-6\sin2x\log 5x \right )$

Question 4Differentiate w.r.t. x the function in Exercises 1 to 11.

$\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$

Answer:

Given function is
$f(x)=\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^{-1}x\sqrt x)}{dx}=\frac{1}{\sqrt{1-(x\sqrt x)^2}}.\frac{d(x\sqrt x)}{dx}$
$=\frac{1}{\sqrt{1-x^3}}.\left ( 1.\sqrt x+x\frac{1}{2\sqrt x} \right )$
$=\frac{1}{\sqrt{1-x^3}}.\left ( \frac{3\sqrt x}{2} \right )$
$=\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$

Therefore, differentiation w.r.t. x is $\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$

Question 5Differentiate w.r.t. x the function in Exercises 1 to 11.

$\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$

Answer:

Given function is
$f(x)=\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$
Now, differentiation w.r.t. x is
By using the Quotient rule
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}})}{dx}$

$=\frac{\frac{d(\cos^{-1}\frac{x}{2})}{dx}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{d(\sqrt{2x+7})}{dx}}{(\sqrt{2x+7})^2}\\$

$ f^{'}(x) = \frac{\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{1}{2}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{1}{2.\sqrt{2x+7}}.2}{2x+7}\\$

$ f^{'}(x)= -\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$
Therefore, differentiation w.r.t. x is $-\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$

Question 6Differentiate w.r.t. x the function in Exercises 1 to 11.

$\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$

Answer:

Given function is
$f(x)=\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$
Now, rationalize the part.
$\left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ]$

$= \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} .\frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}\right ]$
$=\frac{(\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x })^2}{(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2}$

$ \ \ \ \ \ \ (Using \ (a-b)(a+b)=a^2-b^2)$
$=\frac{((\sqrt { 1+ \sin x })^2+ (\sqrt { 1- \sin x })^2+2(\sqrt { 1+ \sin x })(\sqrt { 1- \sin x }))}{1+\sin x-1+\sin x}$
$(Using \ (a+b)^2=a^2+b^2+2ab)$
$=\frac{1+\sin x+1-\sin x+2\sqrt{1-\sin^2x} }{2\sin x}$
$=\frac{2(1+\cos x)}{2\sin x} = \frac{1+\cos x}{\sin x}$
$=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \ \ \ \ \ (\because 2\cos^2= 1+\cos2x \ and \ \sin2x = 2\sin x\cos x)$
$=\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot \frac{x}{2}$
Given function reduces to
$f(x) = \cot^{-1}(\cot \frac{x}{2})\\ f(x) = \frac{x}{2}$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}$

$=\frac{d(\frac{x}{2})}{dx} = \frac{1}{2}$
Therefore, differentiation w.r.t. x is $\frac{1}{2}$

Question 7Differentiate w.r.t. x the function in Exercises 1 to 11. $( \log x )^{ \log x } , x > 1$

Answer:

Given function is
$y=( \log x )^{ \log x } , x > 1$
Take logs on both sides.
$\log y=\log x\log( \log x )$
Now, differentiate w.r.t.
$\frac{1}{y}.\frac{dy}{dx}= \frac{1}{x}.\log (\log x)+\log x.\frac{1}{\log x}.\frac{1}{x} = \frac{\log x+1}{x}$
$\frac{dy}{dx} = y.\left ( \frac{\log x+1}{x} \right )\\$
$\frac{dy}{dx} = (\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$
Therefore, differentiation w.r.t x is $(\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$

Question 8 $\cos ( a \cos x + b \sin x )$, for some constant a and b.

Answer:

Given function is
$f(x)=\cos ( a \cos x + b \sin x )$
Now, differentiation w.r.t x
$f^{'}(x)= \frac{d(f(x))}{dx}= \frac{d(\cos(a\cos x+ b \sin x))}{dx}$
$= -\sin(a\cos x+b\sin x).\frac{d(a\cos x+b\sin x)}{dx}$
$= -\sin(a\cos x+b\sin x).(-a\sin x+b\cos x)$
$= (a\sin x-b\cos x)\sin(a\cos x+b\sin x).$
Therefore, differentiation w.r.t x $(a\sin x-b\cos x)\sin(a\cos x+b\sin x)$

Question 9 $(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$

Answer:

Given function is
$y=(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$
Take logs on both sides.
$\log y=(\sin x - \cos x)\log (\sin x - \cos x)$
Now, differentiate w.r.t. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{d(\sin x-\cos x)}{dx}.\log(\sin x- \cos x)+(\sin x- \cos x).\frac{d(\log(\sin x- \cos x))}{dx}$
$\frac{1}{y}.\frac{dy}{dx} =(\cos x -(-\sin x)).\log(\sin x-\cos x)+(\sin x- \cos x).\frac{(\cos x -(-\sin x))}{(\sin x- \cos x)}$
$\frac{dy}{dx} =y.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
$\frac{dy}{dx} =(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
Therefore, differentiation w.r.t x is $(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right ), sinx>cosx$

Question 10$x ^x + x ^a + a ^x + a ^a$ , for some fixed a > 0 and x > 0

Answer:

Given function is
$f(x)=x ^x + x ^a + a ^x + a ^a$
Let's take
$u = x^x$
Now, take a look at both sides.
$\log u = x \log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{dx}{dx}.\log x+x.\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx}= 1.\log x+x.\frac{1}{x}\\ \\ \frac{du}{dx}= y.(\log x+1)\\ \\ \frac{du}{dx}= x^x.(\log x+1)$ -(i)
Similarly, take $v = x^a$
Take logs on both sides.
$\log v = a\log x$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= a.\frac{d(\log x)}{dx}=a.\frac{1}{x}= \frac{a}{x}\\ \\ \frac{dv}{dx}= v.\frac{a}{x}\\ \\ \frac{dv}{dx}= x^a.\frac{a}{x}$ -(ii)

Similarly, take $z = a^x$
Take logs on both sides.
$\log z = x\log a$
Now, differentiate w.r.t x
$\frac{1}{z}.\frac{dz}{dx}=\log a.\frac{d(x)}{dx}=\log a.1= \log a\\ \\ \frac{dz}{dx}= z.\log a\\ \\ \frac{dz}{dx}= a^x.\log a$ -(iii)

Similarly, take $w = a^a$
Take logs on both sides.
$\log w = a\log a= \ constant$
Now, differentiate w.r.t x
$\frac{1}{w}.\frac{dw}{dx}= a.\frac{d(a\log a)}{dx}= 0\\ \\ \frac{dw}{dx} = 0$ -(iv)
Now,
$f(x)=u+v+z+w$
$f^{'}(x) = \frac{du}{dx}+\frac{dv}{dx}+\frac{dz}{dx}+\frac{dw}{dx}$
Put values from equation (i), (ii),(iii) and (iv)
$f^{'}(x)= x^x(\log x+1)+ax^{a-1}+a^x\log a$
Therefore, differentiation w.r.t. x is $x^x(\log x+1)+ax^{a-1}+a^x\log a$

Question 11$x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$

Answer:

Given function is
$f(x)=x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$
take $u=x ^{x^2 -3}$
Now, take a look at both sides.
$\log u=(x^2-3)\log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{d(x^2-3)}{dx}.\log x+(x^2-3).\frac{d(\log x)}{dx}\\$

$\frac{1}{u}.\frac{du}{dx} = 2x.\log x+(x^2-3).\frac{1}{x}\\$

$ \frac{1}{u}.\frac{du}{dx} = \frac{2x^2\log x+x^2-3}{x}\\$

$ \frac{du}{dx}= u.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\$

$ \frac{du}{dx}= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\$ -(i)
Similarly,
take $v=(x-3)^{x^2}\\$
Now, take a look at both sides.
$\log v=x^2\log (x-3)$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= \frac{d(x^2)}{dx}.\log (x-3)+x^2.\frac{d(\log (x-3))}{dx}\\$

$ \frac{1}{v}.\frac{dv}{dx} = 2x.\log (x-3)+x^2.\frac{1}{(x-3)}\\$

$ \frac{1}{v}.\frac{dv}{dx} = 2x\log(x-3)+\frac{x^2}{x-3}\\$

$ \frac{dv}{dx}= v.\left ( 2x\log(x-3)+\frac{x^2}{x-3} \right )\\$

$ \frac{dv}{dx}= (x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )\\ \\$ -(ii)
Now
$f(x)= u + v$
$f^{'}(x)= \frac{du}{dx}+\frac{dv}{dx}$
Put the value from equation (i) and (ii)
$f^{'}(x)= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$
Therefore, differentiation w.r.t x is $x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$

Question 12Find $\frac{dy}{dx}$ if $y = 12 (1 - \cos t), x = 10 (t - \sin t),$ $-\frac{\pi }{2} <t< \frac{\pi }{2}$

Answer:

Given equations are
$y = 12 (1 - \cos t), x = 10 (t - \sin t),$
Now, differentiate both y and x w.r.t t independently.
$\frac{dy}{dt}=\frac{d(12(1-\cos t))}{dt}= -12(-\sin t)=12\sin t$
And
$\frac{dx}{dt}=\frac{d(10(t-\sin t))}{dt}= 10-10\cos t$
Now
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{12\sin t}{10(1-\cos t)} = \frac{6}{5}.\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2\sin^2\frac{t}{2}} = \frac{6}{5}.\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}\\ \\$
$(\because \sin 2x = 2\sin x\cos x \ and \ 1-\cos 2x = 2\sin^2x)$
$\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}$
Therefore, differentiation w.r.t x is $\frac{6}{5}.\cot \frac{t}{2}$

Question 13Find $\frac{dy}{dx}$ if $y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$

Answer:

Given function is
$y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$
Now, differentiate w.r.t. x
$\frac{dy}{dx}= \frac{d(sin ^{-1} x + sin^{-1} \sqrt{1- x^2})}{dx} = \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}}.\frac{d(\sqrt{1-x^2})}{dx}\\$

$ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-1+x^2}}.\frac{1}{2\sqrt{1-x^2}}.(-2x)\\$

$ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\\$

$ \frac{dy}{dx}= 0$
Therefore, differentiate w.r.t. x is 0

Question 14If $x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0 \: \: for \: \: , -1 < x < 1 \: \:prove \: \: that \: \frac{dy}{dx} = -\frac{1}{(1+x)^2}$

Answer:

Given function is
$x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0$
$x\sqrt{1+y} = - y\sqrt{1+x}$
Now, squaring both sides.
$(x\sqrt{1+y})^2 = (- y\sqrt{1+x})^2\\ x^2(1+y)=y^2(1+x)\\$

$ x^2+x^2y=y^2x+y^2\\$

$ x^2-y^2=y^2x-x^2y\\ (x-y)(x+y) = -xy(x-y) \\$

$ x+y =-xy\\ y = \frac{-x}{1+x}$
Now, differentiate w.r.t. x is
$\frac{dy}{dx} = \frac{d(\frac{-x}{1+x})}{dx}= \frac{-1.(1+x)-(-x).(1)}{(1+x)^2}= \frac{-1}{(1+x)^2}$
Hence proved

Question 15If $(x - a)^2 + (y - b)^2 = c^2$ , for some c > 0, prove that $\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}}\:$ is a constant independent of a and b.

Answer:

Given function is
$(x - a)^2 + (y - b)^2 = c^2$
$(y - b)^2 = c^2-(x - a)^2$ - (i)
Now, differentiate w.r.t. x
$\frac{d((x-a)^2)}{dx}+\frac{((y-b)^2)}{dx}=\frac{d(c^2)}{dx}\\ \\ 2(x-a)+2(y-b).\frac{dy}{dx}=0\\ \\ \frac{dy}{dx} = \frac{a-x}{y-b}$ -(ii)
Now, the second derivative
$\frac{d^2y}{dx^2} = \frac{\frac{d(a-x)}{dx}.(y-b) -(a-x).\frac{d(y-b)}{dx}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} =\frac{ (-1).(y-b)-(a-x).\frac{dy}{dx}}{(y-b)^2}\\ \\$
Now, put values from equation (i) and (ii)
$\frac{d^2y}{dx^2} =\frac{-(y-b)-(a-x).\frac{a-x}{y-b}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} = \frac{-((y-b)^2+(a-x)^2)}{(y-b)^\frac{3}{2}} = \frac{-c^2}{(y-b)^\frac{3}{2}}$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Now,
$\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}} = \frac{\left ( 1+\left ( \frac{x-a}{y-b} \right )^2 \right )^\frac{3}{2}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{\frac{\left ( (y-b)^2 +(x-a)^2\right )^\frac{3}{2}}{(y-b)^\frac{3}{2}}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{(c^2)^\frac{3}{2}}{-c^2}= \frac{c^3}{-c^2}= c$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Which is independent of a and b
Hence proved

Question 16If $\cos y = x \cos (a + y)$ , with $\cos a \neq \pm 1$ , prove that $\frac{dy}{dx} = \frac{\cos ^2 (a+y )}{\sin a }$

Answer:

Given function is
$\cos y = x \cos (a + y)$
Now, differentiate w.r.t x
$\frac{d(\cos y)}{dx} = \frac{dx}{dx}.\cos(a+y)+x.\frac{d(\cos (a+y))}{dx}\\$

$-\sin y \frac{dy}{dx} = 1.\cos (a+y)+x.(-\sin(a+y)).\frac{dy}{dx}\\$

$ \frac{dy}{dx}.(x\sin(a+y)-\sin y)= \cos(a+y)\\$

$ \frac{dy}{dx}.(\frac{\cos y}{\cos (a+b)}.\sin(a+y)-\sin y)= \cos(a+b) $

$\ \ \ \ \ (\because x = \frac{\cos y}{\cos (a+b)})\\$
$ \frac{dy}{dx}.(\cos y\sin(a+y)-\sin y\cos(a+y))=\cos^2(a+b)\\$

$ \frac{dy}{dx}.(\sin(a+y-y))=\cos^2(a+b)$

$ \ \ \ \ \ \ \ (\because \cos A\sin B-\sin A\cos B = \sin(A-B))\\$

$ \frac{dy}{dx}= \frac{\cos^2(a+y)}{\sin a}$
Hence proved

Question 17If $x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t),$ find $\frac{d^2 y }{dx^2 }$

Answer:

Given functions are
$x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t)$
Now, differentiate both the functions w.r.t. t independently.
We get
$\frac{dx}{dt} = \frac{d(a(\cos t +t\sin t))}{dt}= a(-\sin t)+a(\sin t+t\cos t)$
$=-a\sin t+a\sin t+at\cos t = at\cos t$
Similarly,
$\frac{dy}{dt} = \frac{d(a(\sin t - t\cos t))}{dt}= a\cos t -a(\cos t+t(-\sin t))$
$= a\cos t -a\cos t+at\sin t =at\sin t$
Now,
$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at \cos t} = \tan t$
Now, the second derivative
$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \sec^2 t.\frac{dt}{dx}=\frac{\sec^2t.\sec t }{at}=\frac{\sec^3t}{at}$
$(\because \frac{dx}{dt} = at\cos t \rightarrow \frac{dt}{dx}= \frac{1}{at\cos t}=\frac{\sec t}{at})$
Therefore, $\frac{d^2y}{dx^2}=\frac{\sec^3t}{at}$

Question 18If $f (x) = |x|^3$, show that f ''(x) exists for all real x and find it.

Answer:

Given function is
$f (x) = |x|^3$
$f(x)\left\{\begin{matrix} -x^3 & x<0\\ x^3 & x>0 \end{matrix}\right.$
Now, differentiate in both cases.
$f(x)= x^3\\ f^{'}(x)=3x^2\\ f^{''}(x)= 6x$
And
$f(x)= -x^3\\ f^{'}(x)=-3x^2\\ f^{''}(x)= -6x$
In both cases, f ''(x) exists.
Hence, we can say that f ''(x) exists for all real x
And values are
$f^{''}(x)\left\{\begin{matrix} -6x &x<0 \\ 6x& x>0 \end{matrix}\right.$

Question 19Using the fact that $\sin (A + B) = \sin A \cos B + \cos A \sin B$ and the differentiation,
Obtain the sum formula for cosines.

Answer:

Given function is
$\sin (A + B) = \sin A \cos B + \cos A \sin B$
Now, differentiate w.r.t. x
$\frac{d(\sin(A+B))}{dx} = \frac{d\sin A}{dx}.\cos B+\sin A.\frac{d\cos B}{dx}+\frac{d\cos A}{dx}.\sin B+\cos A.\frac{d\sin B}{dx}$
$\cos (A+b)\frac{d(A+B)}{dx}$ $=\frac{dA}{dx}(\cos A\cos B-\sin A\cos B)+\frac{dB}{dx}(\cos A \sin B-\sin A\sin B)$
$=(\cos A \sin B-\sin A\sin B).\frac{d(A+B)}{dx}$
$\cos(A+B)= \cos A\sin B-\sin A\cos B$
Hence, we get the formula by differentiation of sin(A + B)

Question 20Does there exist a function which is continuous everywhere but not differentiable
At exactly two points? Justify your answer.

Answer:

Consider f(x) = |x| + |x +1|
We know that modulus functions are continuous everywhere, and the sum of two continuous functions is also a continuous function.
Therefore, our function f(x) is continuous.
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
$\lim\limits_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{f(h)-f(0)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{|h|+|h+1|-|1|}{h}$
$=\lim\limits_{h\rightarrow 0^-}\frac{-h-(h+1)-1}{h}= 0$ $(|h| = - h \ because\ h < 0)$
R.H.L. at x = 0
$\lim\limits_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{|h|+|h+1|-|1|}{h}$
$=\lim\limits_{h\rightarrow 0^+}\frac{h+h+1-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{2h}{h}= 2$ $(|h| = h \ because \ h > 0)$
R.H.L. is not equal to L.H.L.
Hence. At x = 0, the function is not differentiable.
Now, Similarly
R.H.L. at x = -1
$\lim\limits_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{f(-1+h)-f(-1)}{h}= \lim\limits_{h\rightarrow 0^+}\frac{|-1+h|+|h|-|-1|}{h}$
$=\lim\limits_{h\rightarrow 0^+}\frac{1-h+h-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{0}{h}= 0$ $(|h| = h \ because \ h > 0)$
L.H.L. at x = -1
$\lim\limits_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h}= \lim\limits_{h\rightarrow 0^-}\frac{|-1+h|+|h|-|1|}{h}$
$=\lim\limits_{h\rightarrow 1^+}\frac{1-h-h-1}{h}= \lim\limits_{h\rightarrow 0^+}\frac{-2h}{h}= -2$ $(|h| = - h \ because\ h < 0)$
L.H.L. is not equal to R.H.L, so not differentiable at x=-1

Hence, exactly two points where it is not differentiable

Question 21If $y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$ , prove that $\frac{dy}{dx}$ = \begin{vmatrix} f '(x) & g'(x) & h' (x) \\ l& m &n \\ a& b &c \end{vmatrix}$

Answer:

Given that
$y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$
We can rewrite it as
$y = f(x)(mc-bn)-g(x)(lc-an)+h(x)(lb-am)$
Now, differentiate w.r.t x
We will get
$\frac{dy}{dx} = f^{'}(x)(mc-bn)-g^{'}(x)(lc-an)+h^{'}(x)(lb-am) \rightarrow \begin{bmatrix} f^{'}(x) &g^{'}(x) &h^{'}(x) \\ l&m &n \\ a& b &c \end{bmatrix}$
Hence proved

Question 22If $y=e^{a \cos ^{-1} x},-1 \leq 1$, show that $( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$

Answer:

Given function is
$y=e^{a \cos ^{-1} x},-1 \leq 1$

Now, differentiate w.r.t x we will get.
$\frac{dy}{dx}= \frac{d(e^{a\cos^{-1}x})}{dx}.\frac{d(a\cos^{-1}x)}{dx} = e^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}$ -(i)
Now, again differentiate w.r.t x
$\frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx}= \frac{-ae^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}.\sqrt{1-x^2}+ae^{a\cos^{-1}x}.\frac{1.(-2x)}{2\sqrt{1-x^2}}}{(\sqrt{1-x^2})^2}$
$=\ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2}$ -(ii)
Now, we need to show that.
$( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$
Put the values from equation (i) and (ii)
$(1-x^2).\left ( \ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2} \right )-x.\left ( \frac{-ae^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x}$
$a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}+\left ( \frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x} = 0$
Hence proved

If you are looking for continuity and differentiability class 12 NCERT solutions of exercises, then these are listed below.

• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.1
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.2
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.3
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.4
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.5
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.6
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.7
• Continuity And Differentiability Class 12 NCERT Solutions Exercise 5.8
• Continuity And Differentiability Class 12 NCERT Solutions Miscellaneous Exercise

Also read,

For the solution of other examples, you can refer to these

• NCERT Exemplar Class 12 Chemistry Solutions
• NCERT Exemplar Class 12 Mathematics Solutions
• NCERT Exemplar Class 12 Biology Solutions
• NCERT Exemplar Class 12 Physics Solutions

NCERT solutions for class 12 maths - Chapter-wise

For subject-wise solutions, you can refer here

• NCERT Solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

NCERT Solutions class-wise

For the solution of other classes, you can refer here

• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

Here, you can refer to the latest syllabus and NCERT Books

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12

Importance of solving NCERT Class 12 Maths Chapter 5

NCERT solutions for class 12 maths chapter 5 continuity and differentiability are very helpful in the preparation of this chapter. But here are some tips to get command of this chapter.

• You should make sure that concepts related to 'limit' are clear to you as they form the base for continuity.
• First, go for the theorem and solve examples of continuity given in the NCERT textbook then try to solve exercise questions. You may find some difficulties in solving them. Go through the NCERT solutions for class 12 maths chapter 5 continuity and differentiability, it will help you to understand the concepts in a much easier way.
• This chapter seems very easy, but at the same time, the chances of silly mistakes are also high. So, it is advised to understand the theory and concepts properly before practising questions of NCERT.
• Once you are good in continuity, then go for the differentiability. Practice more and more questions to get command of it.
• Differentiation is mostly formula-based, so practice NCERT questions; it won't take much effort to remember the formulas.