NCERT Solutions for Exercise 5.1 Class 12 Maths Chapter 5 - Continuity and Differentiability

# NCERT Solutions for Exercise 5.1 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Ramraj Saini | Updated on Dec 03, 2023 05:24 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.1

NCERT Solutions for Exercise 5.1 Class 12 Maths Chapter 5 Continuity and Differentiability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In the previous class, you have already learned about limits, derivatives, limits, and derivatives of trigonometric functions. In this article, you will get NCERT solutions for Class 12 Maths chapter 5 exercise 5.1. NCERT book Class 12 Maths chapter 5 exercise 5.1 consists of questions related to finding whether a function is continuous or not.

Continuity of functions can't be learned without fundamental knowledge of limit which you have learned already. It is a fundamental concept of calculus that you must know to understand more concepts of calculus. Solving exercise 5.1 Class 12 Maths questions are very important to get conceptual clarity about continuity. There are different theorems to check the continuity of different types of functions mentioned in the NCERT syllabus of Class 12 Maths. 12th class Maths exercise 5.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Continuity and Differentiability Exercise: 5.1

Given function is
$f ( x) = 5 x -3$
$f(0) = 5(0)-3 = -3$
$\lim_{x\rightarrow 0} f(x) = 5(0)-3 = -3$
$\lim_{x\rightarrow 0} f(x) =f(0)$
Hence, function is continous at x = 0

$\dpi{100} f(-3)= 5(-3)-3=-15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = 5(-3)-3 = -15-3=-18\\\Rightarrow \lim_{x\rightarrow -3} f(x) = f(-3)$
Hence, function is continous at x = -3

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

Given function is
$f(x) = 2x^2-1$
at x = 3
$f(3) = 2(3)^2-1 = 2\times 9 - 1=18-1=17\\ \lim_{x\rightarrow 3}f(x) = 2(3)^2-1=2\times 9-1=18-1=17$
$\lim_{x\rightarrow 3}f(x) = f(3)$
Hence, function is continous at x = 3

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

Question:3 b) Examine the following functions for continuity.

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

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

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

Given function is
$f(x ) = \frac{x^2-25}{x+5}$
For every real number k , $k \neq -5$
We gwt,
$f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = f(k)$
Hence, function $f(x ) = \frac{x^2-25}{x+5}$ continuous for every real value of x , $x \neq -5$

Given function is
$f (x) = | x - 5|$
for x > 5 , f(x) = x - 5
for x < 5 , f(x) = 5 - x
SO, different cases are their
case(i) x > 5
for every real number k > 5 , f(x) = x - 5 is defined
$f(k) = k - 5\\ \lim_{x\rightarrow k }f(x) = k -5\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, function f(x) = x - 5 is continous for x > 5

case (ii) x < 5
for every real number k < 5 , f(x) = 5 - x is defined
$f(k) = 5-k\\ \lim_{x\rightarrow k }f(x) = 5 -k\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, function f(x) = 5 - x is continous for x < 5

case(iii) x = 5
for x = 5 , f(x) = x - 5 is defined
$f(5) = 5 - 5=0\\ \lim_{x\rightarrow 5 }f(x) = 5 -5=0\\ \lim_{x\rightarrow 5 }f(x) = f(5)$
Hence, function f(x) = x - 5 is continous for x = 5

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

GIven function is
$f (x) = x^n$
the function $f (x) = x^n$ is defined for all positive integer, n
$f(n) = n^n\\ \lim_{x\rightarrow n}f(x) = n^n\\ \lim_{x\rightarrow n}f(x) = f(n)$
Hence, the function $f (x) = x^n$ is continuous at x = n, where n is a positive integer

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

given function is defined for x = 1
Now, for x = 1 Right-hand limit and left-hand limit are not equal
$f(1) = 1\\ \lim_{x\rightarrow 1^-}f(x) = f(x) = 1\\ \lim_{x\rightarrow 1^+}f(x) =f(5) = 5$
R.H.L $\neq$ L.H.L.
Therefore, given function is not continous at x =1
Given function is defined for x = 2 and its value at x = 2 is 5
$f(2) = 2\\ \lim_{x\rightarrow 2}f(x) = f(5) = 5\\\lim_{x\rightarrow 2}f(x) = f(2)$
Hence, given function is continous at x = 2

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

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
$f(k) = 2k-3\\ \lim_{x\rightarrow k}f(x) = 2k-3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
$f(k) = 2k +3\\ \lim_{x\rightarrow k}f(x) = 2k+3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

$\lim_{x\rightarrow 2^-}f(x) = 2x+3 = 2\times 2 + 3 = 4 + 3 = 7\\ \lim_{x\rightarrow 2^+}f(x) = 2x-3 = 2\times 2-3 = 4-3 = 1$
Right hand limit at x= 2 $\neq$ Left hand limit at x = 2
Therefore, x = 2 is the point of discontinuity

$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3

Given function is
$f (x) = \left\{\begin{matrix} |x|+3 & if \: \: x \leq -3 & \\ -2x & if \: \: -3
GIven function is defined for every real number k
Different cases are their
case (i) k < -3
$f(k) = -k + 3\\ \lim_{x\rightarrow k}f(x) = -k + 3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for every value of k < -3

case(ii) k = -3
$f(-3) = -(-3) + 3 = 6\\ \lim_{x\rightarrow -3^-}f(x) = -k + 3=-(-3)+3 = 6\\ \lim_{x\rightarrow -3^+}f(x) = -2x = -2(-3) = 6\\ R.H.L. = L.H.L. = f(-3)$
Hence, given function is continous for x = -3

case(iii) -3 < k < 3
$f(k) = -2k \\ \lim_{x\rightarrow k}f(x) = -2k\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in -3 < k < 3 given function is continous

case(iv) k = 3
$f(3) = 6x+2 = 6\times3+2 =18+2=20\\ \lim_{x\rightarrow 3^-}f(x) = -2x = -2(3) = -6\\ \lim_{x\rightarrow 3^+}f(x) = 6x+2 = 6\times3+2 = 20\\ R.H.L. = f(3) \neq L.H.L.$
Hence. x = 3 is the point of discontinuity

case(v) k > 3
$f(k) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = 6k+2 \\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each and every value of k > 3

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

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
$f(k) = -1\\ \lim_{x\rightarrow k }f(x) = -1\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, given function is continuous for every value of k < 0
case(ii) k > 0
$f(k) = 1\\ \lim_{x\rightarrow k }f(x) = 1\\ \lim_{x\rightarrow k }f(x) = f(k)$
Hence, given function is continuous for every value of k > 0
case(iii) x = 0
$f(0) = 0\\ \lim_{x\rightarrow 0^- }f(x) = -1\\ \lim_{x\rightarrow 0^+}f(x) = 1\\ f(0) \neq R.H.L. \neq L.H.L.$
Hence, 0 is the only point of discontinuity

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

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 each and every value of x
Hence, no point of discontinuity

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

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

case(ii) k < 1
$f(k) = k^2 ++1\\ \lim_{x\rightarrow k}f(x) = k^2+1\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

$\lim_{x\rightarrow 1^-}f(x) = x^2+1 = 1^2 + 1 = 1 + 1 = 2\\ \lim_{x\rightarrow 1^+}f(x) = x+1 = 1+1 = 2\\ f(1) = 1^2+1 = 2 \\ R.H.L. = L.H.L. = f(1)$

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

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

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
$f(k) = k^2+1\\ \lim_{x\rightarrow k}f(x) = k^2+1\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
$f(k) = k^3 -3\\ \lim_{x\rightarrow k}f(x) = k^3-3\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

$\lim_{x\rightarrow 2^-}f(x) = x^3-3 = 2^3- 3 = 8- 3 = 5\\ \lim_{x\rightarrow 2^+}f(x) = x^2+1= 2^2+1 = 4+1 = 5\\ f(2) = 2^3-3 = 8 - 3 = 5\\ f(2)=R.H.L.=L.H.L.$
Hence, given function is continuous at x = 2
There, no point of discontinuity

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

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
$f(k) = k^2\\ \lim_{x\rightarrow k}f(x) = k^2\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
$f(k) = k^{10} -1\\ \lim_{x\rightarrow k}f(x) = k^{10}-1\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

$\lim_{x\rightarrow 1^-}f(x) = x^{10}-1 = 1^{10} - 1 = 1 - 1 = 0\\ \lim_{x\rightarrow 1^+}f(x) = x^2 = 1^2 = 1\\ f(1) = x^{10}-1 = 0\ f(1) = L.H.L. \neq R.H.L.$

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?

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
$f(k) = k-5\\ \lim_{x\rightarrow k}f(x) = k-5\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for each value of k > 1

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

case(iii) x = 1

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

Hence, x = 1 is the point of discontinuity

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

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

case(ii) k = 1
$f(1) = 3 \\ \lim_{x\rightarrow 1^-}f(x) = 3\\ \lim_{x\rightarrow 1^+}f(x) = 4\\ R.H.L. \neq L.H.L. = f(1)$
Hence, given function is discontinous at x = 1
Therefore, x = 1 is he point od discontinuity

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

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

case(v) k > 3
$f(k) = 5 \\ \lim_{x\rightarrow k}f(x) = 5 \\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for each and every value of k > 3
case(vi) when k < 3

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

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

Given function is
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$
Given function is satisfies for the all real values of x
case (i) k < 0
$f(k) = 2k$
$\lim_{k \rightarrow 0^-}f(x)= 2k = f(k)$
Hence, function is continuous for all values of x < 0

case (ii) x = 0
$f(0 )= 0$
L.H.L at x= 0
$\lim_{x\rightarrow 0^-}f(x)= 2(0)= 0$
R.H.L. at x = 0
$\lim_{x\rightarrow 0^+}f(x)= 0$
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii) k > 0
$f(k)=0$
$\lim_{k\rightarrow 0^+}f(x)= 0= f(k)$
Hence , function is continuous for all values of x > 0

case (iv) k < 1
$f(k) = 0$
$\lim_{x\rightarrow 1^-}f(x)= 0 = f(k)$
Hence , function is continuous for all values of x < 1

case (v) k > 1
$f(k) = 4k$
$\lim_{x\rightarrow 1^+}f(x)= 4k = f(k)$
Hence , function is continuous for all values of x > 1

case (vi) x = 1
$f(1)= 0$
$\lim_{x\rightarrow 1^-}f(1)= 0$
$\lim_{x\rightarrow 1^+}f(1)= 4(1) = 4$
Hence, function is not continuous at x = 1

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

Given function is
$f ( x ) = \left\{\begin{matrix} -2 & if x \leq -1 \\ 2x & if -1< x \leq 1 \\ 2 & if x > 1 \end{matrix}\right.$
GIven function is defined for every real number k
Different cases are their
case (i) k < -1
$f(k) = -2\\ \lim_{x\rightarrow k}f(x) = -2\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continuous for every value of k < -1

case(ii) k = -1
$f(-1) = -2 \\ \lim_{x\rightarrow -1^-}f(x) = -2\\ \lim_{x\rightarrow -1^+}f(x) = 2x = 2(-1) = -2\\ R.H.L. =L.H.L. = f(-1)$
Hence, given function is continous at x = -1

case(iii) k > -1
$f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for all values of x > -1

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

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

case(vi) k > 1
$f(k) = 2 \\ \lim_{x\rightarrow k}f(x) = 2 \\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, given function is continous for each and every value of k > 1
case(vii) when k < 1

$f(k) = 2k \\ \lim_{x\rightarrow k}f(x) = 2k\\ \lim_{x\rightarrow k}f(x) = f(k)$
Hence, for every value of k in k < 1 given function is continuous

Therefore, continuous at all points

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.
$\lim_{x\rightarrow 3^-}f(x)= ax + 1 = 3a+1\\ \lim_{x\rightarrow 3^+}f(x) = bx+3=3b+3$
For the function to be continuous
$\lim_{x\rightarrow 3^-}f(x) = \lim_{x\rightarrow 3^+}f(x) \\ 3a + 1= 3b+3\\ 3(a-b)=2\\ a-b = \frac{2}{3}\\ a = b+\frac{2}{3}$

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.
$\lim_{x\rightarrow 0^-}f(x) = \lambda (x^2-2x) = 0\\ \lim_{x\rightarrow 0^+}f(x) = 4x+1=1$
For the function to be continuous
$\lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0^+}f(x) \\ 0\neq 1$
Hence, for no value of function is continuous at x = 0

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

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

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

Given function is
$f (x) = \sin x + \cos x$
Given function is defined for all real number
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) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) + h(x) = sin x + cos x is also a continuous function

Given function is
$f (x) = \sin x - \cos x$
Given function is defined for all real number
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) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

Given function is
$f (x) = \sin x . \cos x$
Given function is defined for all real number
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) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x).h(x) = sin x .cos x is also a continuous function

We, know that if two function g(x) and h(x) are continuous then
$\frac{g(x)}{h(x)} , h(x) \neq0\ is \ continuous\\ \frac{1}{h(x)} , h(x) \neq 0\ is \ continuous\\ \frac{1}{g(x)} , g(x) \neq0\ is \ continuous\\$
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, the function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is a continous function
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

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

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

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
$f(0) = 0\\ \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\left ( x^2\sin\frac{1}{x} \right )=\lim_{x\rightarrow 0}\left ( \frac{x.\sin\frac{1}{x}}{\frac{1}{x}} \right ) = 0(1)=0 \ \ \ \ \ \ (\because\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1)$
$\lim_{x\rightarrow 0}f(x) = f(0)$
Hence, function is continuous at x = 0
when $x \neq 0$
$f(k) = k^2\sin \frac{1}{k}\\ \lim_{x\rightarrow k}f(x)=\lim_{x\rightarrow k}\left ( x^2\sin\frac{1}{x} \right )=k^2\sin \frac{1}{k}\\ \lim_{x\rightarrow k} = f(k)$
Hence, the given function is continuous for all points

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

Given function is
$f (x) = \sin x - \cos x$
Given function is defined for all real number
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) allare continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$g(c) = \sin c\\ \lim_{x\rightarrow c}g(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}g(x) = g(c)$
Hence, function $g(x) = \sin x$ is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
Hence, function $h(x) = \cos x$ is a continuous function
We proved independently that sin x and cos x is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

When x = 0
$f (0) = -1\\ \lim_{x\rightarrow 0^-}f(x) = \sin 0 - \cos 0 = -1\\ \lim_{x\rightarrow 0^+}f(x) = \sin 0 - \cos 0 = -1 \\ R.H.L. = L.H.L. = f(0)$
Hence, function is also continuous at x = 0

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

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 $x = \frac{\pi}{2}$
$f(\frac{\pi}{2}) = 3\\let\ x=\pi +h\\ \lim_{x\rightarrow \frac{\pi}{2}}f(x)= \lim_{h\rightarrow 0}\frac{k\cos\left ( \frac{\pi}{2}+h \right )}{\pi-2\left ( \frac{\pi}{2}+h \right )} = k. \lim_{h\rightarrow 0}\frac{-\sin h}{-2h} = \frac{k}{2}\\$
For the function to be continuous
$\lim_{x\rightarrow \frac{\pi}{2}}f(x)= f(\frac{\pi}{2})\\ \frac{k}{2} = 3\\ k = 6$
Therefore, the values of k so that the function f is continuous is 6

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

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.
$f(2) = 4k\\ \lim_{x\rightarrow 2^-}f(x)= 4k\\ \lim_{x\rightarrow 2^+}f(x) = 3\\ f(2) = \lim_{x\rightarrow 2^-}f(x) = \lim_{x\rightarrow 2^+}f(x)\\ 4k = 3\\ k = \frac{3}{4}$
Hence, the values of k so that the function f is continuous at x= 2 is $\frac{3}{4}$

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

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.
$f(\pi) = k\pi+1\\ \lim_{x\rightarrow \pi^-}f(x)= k\pi+1\\ \lim_{x\rightarrow \pi^+}f(x) = \cos \pi = -1\\ f(\pi) = \lim_{x\rightarrow \pi^-}f(x) = \lim_{x\rightarrow \pi^+}f(x)\\ k\pi+1 = -1\\ k = \frac{-2}{\pi}$
Hence, the values of k so that the function f is continuous at x= $\pi$ is $\frac{-2}{\pi}$

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

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.
$f(5) = 5k+1\\ \lim_{x\rightarrow 5^-}f(x)= 5k+1\\ \lim_{x\rightarrow 5^+}f(x) = 3(5)-5 = 15-5=10\\ f(5) = \lim_{x\rightarrow 5^-}f(x) = \lim_{x\rightarrow 5^+}f(x)\\ 5k+1 = 10\\ k = \frac{9}{5}$
Hence, the values of k so that the function f is continuous at x= 5 is $\frac{9}{5}$

Given 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
$\lim_{x\rightarrow 2^-}f(x) = \lim_{x\rightarrow 2^+}f(x)\\ and\\ \lim_{x\rightarrow 10^-}f(x)=\lim_{x\rightarrow 10^+}f(x)$
$\lim_{x\rightarrow 2^-}f(x) = 5\\ \lim_{x\rightarrow 2^+}f(x)=ax+b=2a+b\\ 2a+b = 5 \ \ \ \ \ \ \ \ \ \ \ -(i)\\ and\\ \lim_{x\rightarrow 10^-}f(x)=ax+b=10a+b\\ \lim_{x\rightarrow 10^+}f(x)=21\\ 10a+b=21 \ \ \ \ \ \ \ \ -(ii)$
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

Given function is
$f (x) = \cos (x^2 )$
given function is defined for all real values of x
Let x = k + h
if $x\rightarrow k , \ then \ h \rightarrow 0$
$f(k) = \cos k^2\\ \lim_{x \rightarrow k}f(x) = \lim_{x \rightarrow k}\cos x^2 = \lim_{h \rightarrow 0}\cos (k+h)^2 = \cos k^2\\ \lim_{x \rightarrow k}f(x) = f(k)$
Hence, the function $f (x) = \cos (x^2 )$ is a continuous function

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
$g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

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

case (iii) k = 0
$g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )$
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 suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \cos c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\cos x = \lim_{h\rightarrow 0}\cos (c+h)\\ We \ know \ that\\ \cos(a+b) = \cos a \cos b + \sin a\sin b\\ \lim_{h\rightarrow 0}\cos (c+h) = \lim_{h\rightarrow 0}(\cos c\cos h + \sin c \sin h) = \lim_{h\rightarrow 0}\cos c\cos h + \lim_{h\rightarrow 0}\sin c \sin h$
$=\cos c\cos 0 + \sin c \sin 0 = \cos c$
$\lim_{x\rightarrow c}h(x) = h(c)$
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.

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
$g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

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

case (iii) k = 0
$g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )$
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 suppose x = c + h
if $x \rightarrow c , \ then \ h \rightarrow 0$
$h(c) = \sin c\\ \lim_{x\rightarrow c}h(x) = \lim_{x\rightarrow c}\sin x = \lim_{h\rightarrow 0}\sin (c+h)\\ We \ know \ that\\ \sin(a+b) = \sin a \cos b + \cos a\sin b\\ \lim_{h\rightarrow 0}\sin (c+h) = \lim_{h\rightarrow 0}(\sin c\cos h + \cos c \sin h) = \lim_{h\rightarrow 0}\sin c\cos h + \lim_{h\rightarrow 0}\cos c \sin h$
$=\sin c\cos 0 + \cos c \sin 0 = \sin c$
$\lim_{x\rightarrow c}h(x) = h(c)$
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

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
$g(k) = -k\\ \lim_{x\rightarrow k}g(x) = -k\\ \lim_{x\rightarrow k}g(x) = g(k)$
Hence, g(x) is continuous when k < 0

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

case (iii) k = 0
$g(0) = 0\\ \lim_{x\rightarrow 0^-}g(x) = -x = 0\\ \lim_{x\rightarrow 0^+}g(x ) = x = 0\\ \lim_{x\rightarrow 0^-}g(x) = g(0) = \lim_{x\rightarrow 0^+}g(x )$
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
$h(k) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = -(k+1)\\ \lim_{x\rightarrow k}h(x) = h(k)$
Hence, h(x) is continuous when k < -1

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

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

## More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

In Class 12th Maths chapter 5 exercise 5.1 there are 34 long answer types questions only checking your knowledge of continuity. Many questions in Class 12 Maths ch 5 ex 5.1 are related to checking the continuity of trigonometric functions. There are 20 examples and some important theorems given before this exercise in the NCERT textbook. Solving these examples is a must to do before going to the Class 12th Maths chapter 5 exercise 5.1 questions because it will help you in solving NCERT problems.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

• Class 12 Maths chapter 5 exercise 5.1 solutions are described in a detailed manner, so you will get the concept of continuity easily.
• In Class 12 Maths chapter 5 exercise 5.1 solutions , you will get different ways to approach the problem.
• There are some properties of continuous functions which make it easy to check the continuity of the given functions.
• In Class 12 Maths chapter 5 exercise 5.1 solutions, you will learn about the algebra of continuous functions.
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## Key Features Of NCERT Solutions for Exercise 5.1 Class 12 Maths Chapter 5

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.1 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 5.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 5.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 5.1 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 5.1 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 5.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Check whether function f(x) = x + 3 is continuous at x = 1 ?

f(1^+)= 1+3 = 4

f(1^-)= 1+3 = 4 = f(1)

Hence f(x) is continuous at x=1.

2. If f(x), g(x) are continuous functions then what about continuity of f(x) + g(x) ?

If f(x), g(x) are continuous functions then f(x) + g(x) is also a continuous function.

3. If f(x), g(x) are continuous functions then what about continuity of f(x) - g(x) ?

If f(x), g(x) are continuous functions then f(x) - g(x) is also a continuous function.

4. If two functions are continuous then check continuity of product of the given two functions.

If two functions are continuous then check the product of the given two functions is also a continuous function.

5. Do every continuous function is a differential function ?

No, every continuous function need not to be a differential function.

6. Do every differential function is a continuous function ?

Yes, every differential function is a continuous function.

7. How many questions are there in the exercise 5.1 Class 12 Maths ?

NCERT book Exercise 5.1 Class 12 Maths is consists of 34 long answer questions related to checking the continuity of the functions. For more questions students can refer to NCERT exemplar problems.

8. How many exercises are in there in the chapter 5 Class 12 Maths ?

There are eight main exercises and one miscellaneous exercise given in the chapter 5 Class 12 Maths.

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#### National Institute of Open Schooling 10th examination

Exam Date:20 September,2024 - 07 October,2024

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9