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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

Edited By Komal Miglani | Updated on Mar 31, 2025 08:07 AM IST | #CBSE Class 12th
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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.

This Story also Contains
  1. NCERT Continuity And Differentiability Class 12 Questions And Answers PDF Free Download
  2. Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae
  3. NCERT Continuity and Differentiability Class 12 Questions And Answers (Exercise)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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 Continuity And Differentiability Class 12 Questions And Answers 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).
  • limxaf(x) exists.
  • limxaf(x)=f(a).
  • limx0f(x)=f(0)
  • limxaf(x) exists
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JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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=vou, where t=u(x), and if both dtdx and dvdx exist, then: dfdx=dvdtdtdx.

Derivatives of Some Standard Functions:

  • ddt(xn)=nxn1
  • ddt(sinx)=cosx
  • ddt(cosx)=sinx
  • ddt(tanx)=sec2x
  • ddt(cotx)=csc2x
  • ddt(secx)=secxtanx
  • ddt(cscx)=cscxcotx
  • ddt(ax)=axln(a)
  • ddt(ex)=ex
  • ddt(lnx)=1x
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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)=(f(b)f(a))(ba).

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)=(f(b)f(a))(ba).

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

NCERT Continuity and Differentiability Class 12 Solutions: Exercise - 5.1
Page no. - 116
Total questions - 34

Question 1. Prove that the function f(x)=5x3 is continuous at x=0,atx=3 and at x=5

Answer:

Given function is
f(x)=5x3
f(0)=5(0)3=3
limx0f(x)=5(0)3=3
limx0f(x)=f(0)
Hence, the function is continuous at x = 0
f(3)=5(3)3=153=18limx3f(x)=5(3)3=153=18

limx3f(x)=f(3)
Hence, the function is continuous at x=3
f(5)=5(5)3=253=22limx5f(x)=5(5)3=253=22

limx5f(x)=f(5)
Hence, the function is continuous at x=5

Question 2. Examine the continuity of the function f(x)=2x21atx=3.

Answer:

Given function is
f(x)=2x21
at x=3
f(3)=2(3)21=2×91=181=17

limx3f(x)=2(3)21=2×91=181=17
limx3f(x)=f(3)
Hence, the function is continuous at x=3

Question 3Examine the following functions for continuity.
(a)f(x)=x5

Answer:

Given function is
f(x)=x5
Our function is defined for every real number, say k
and value at x=k , f(k)=k5
and also,
limxkf(x)=k5

limxkf(x)=f(k)
Hence, the function f(x)=x5 is continuous at every real number

Question 3 b) Examine the following functions for continuity.

f(x)=1x5,x5

Answer:

Given function is
f(x)=1x5
For every real number k, k5
We get,
f(k)=1k5

limxkf(x)=1k5

limxkf(x)=f(k)
Hence, function f(x)=1x5 continuous for every real value of x, x5

Question 3 c) Examine the following functions for continuity.

f(x)=x225x+5,x5

Answer:

Given function is
f(x)=x225x+5
For every real number k, k5
We get,
f(k)=k252k+5=(k+5)(k5)k+5=k5

limxkf(x)=k252k+5

=(k+5)(k5)k+5=k5

limxkf(x)=f(k)
Hence, function f(x)=x225x+5 continuous for every real value of x , x5

Question 3 d) Examine the following functions for continuity. f(x)=|x5|

Answer:

Given function is
f(x)=|x5|
for x>5,f(x)=x5
for x<5,f(x)=5x
So, there are different cases.
case(i) x>5
for every real number k>5 , f(x)=x5 is defined
f(k)=k5

limxkf(x)=k5

limxkf(x)=f(k)
Hence, function f(x)=x5 is continuous for x>5
case (ii) x<5
for every real number k<5 , f(x)=5x is defined
f(k)=5klimxkf(x)=5klimxkf(x)=f(k)
Hence, function f(x)=5x is continuous for x<5
case(iii) x=5
for x=5 , f(x)=x5 is defined
f(5)=55=0limx5f(x)=55=0limx5f(x)=f(5)
Hence, function f(x)=x5 is continous for x=5

Hence, the function f(x)=|x5| is continuous for every real number.

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

Answer:

Given function is
f(x)=xn
The function f(x)=xn is defined for all positive integer, n
f(n)=nn

limxnf(x)=nn

limxnf(x)=f(n)
Hence, the function f(x)=xn is continuous at x = n, where n is a positive integer

Question 5.Is the function f defined by
f(x)={x,ifx15ifx1
continuous at x = 0? At x = 1? At x = 2?

Answer:

Given function is
f(x)={x,ifx15ifx1
The function is defined at x=0 and its value is 0
f(0)=0

limx0f(x)=f(x)=0

limx0f(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)=1limx1f(x)=f(x)=1

limx1+f(x)=f(5)=5
R.H.L 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)=2limx2f(x)=f(5)=5limx2f(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)={2x+3ifx22x3ifx2

Answer:

Given function is
f(x)={2x+3ifx22x3ifx2
Given function is defined for every real number k
There are different cases for the given function.
case(i) k>2
f(k)=2k3limxkf(x)=2k3

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k>2

case(ii) k<2
f(k)=2k+3limxkf(x)=2k+3

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k<2

case(iii) x=2
limx2f(x)=2x+3=2×2+3=4+3=7

limx2+f(x)=2x3=2×23=43=1
Right hand limit at x= 2 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)={|x|+3ifx32xif3<x<36x+2ifx3

Answer:

Given function is
f(x)={|x|+3ifx32xif3<x<36x+2ifx3
Given function is defined for every real number k
There are different cases.
case (i) k<3
f(k)=k+3limxkf(x)=k+3

limxkf(x)=f(k)
Hence, the given function is continuous for every value of k < -3

case(ii) k=3
f(3)=(3)+3=6limx3f(x)=k+3=(3)+3=6

limx3+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)=2klimxkf(x)=2k

limxkf(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=6×3+2=18+2=20

limx3f(x)=2x=2(3)=6

limx3+f(x)=6x+2=6×3+2=20

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

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

limxkf(x)=6k+2

limxkf(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)={|x|xifx00ifx=0

Answer:

Given function is
f(x){|x|xifx00ifx=0
if x > 0 , f(x)=xx=1
if x < 0 , f(x)=(x)x=1
Given function is defined for every real number k
Now,
case(i) k < 0
f(k)=1limxkf(x)=1

limxkf(x)=f(k)
Hence, the given function is continuous for every value of k < 0
case(ii) k > 0
f(k)=1limxkf(x)=1

limxkf(x)=f(k)
Hence, the given function is continuous for every value of k > 0
case(iii) x = 0
f(0)=0limx0f(x)=1

limx0+f(x)=1f(0)R.H.L.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)={x|x|ifx<01ifx0

Answer:

Given function is
f(x)={x|x|ifx<01ifx0
if x < 0 , f(x)=x|x|=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)={x+1ifx1x2+1ifx<1

Answer:

Given function is
f(x)={x+1ifx1x2+1ifx<1
Given function is defined for every real number k
There are different cases for the given function.
case(i) k>1
f(k)=k+1limxkf(x)=k+1

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k>1

case(ii) k<1
f(k)=k2+1limxkf(x)=k2+1

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k<1

case(iii) x=1

limx1f(x)=x2+1=12+1=1+1=2

limx1+f(x)=x+1=1+1=2f(1)=12+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)={x33ifx2x2+1ifx>2

Answer:

Given function is
f(x)={x33ifx2x2+1ifx>2
Given function is defined for every real number k
There are different cases for the given function.
case(i) k>2
f(k)=k2+1limxkf(x)=k2+1

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k>2

case(ii) k<2
f(k)=k33limxkf(x)=k33

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k<2

case(iii) x=2
limx2f(x)=x33=233=83=5

limx2+f(x)=x2+1=22+1=4+1=5

f(2)=233=83=5f(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)={x101ifx1x2x>1

Answer:

Given function is
f(x)={x101ifx1x2x>1
Given function is defined for every real number k
There are different cases for the given function.
case(i) k>1
f(k)=k2limxkf(x)=k2

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k > 1

case(ii) k<1
f(k)=k101

limxkf(x)=k101

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k < 1

case(iii) x = 1
limx1f(x)=x101=1101=11=0

limx1+f(x)=x2=12=1

f(1)=x101=0 f(1)=L.H.L.R.H.L.
Hence, x = 1 is the point of discontinuity.

Question 13. Is the function defined by

f(x)={x+5ifx1x5ifx>1

A continuous function?

Answer:

Given function is
f(x)={x+5ifx1x5ifx>1
Given function is defined for every real number k
There are different cases for the given function.
case(i) k>1
f(k)=k5limxkf(x)=k5

limxkf(x)=f(k)
Hence, the given function is continuous for each value of k>1

case(ii) k<1
f(k)=k+5limxkf(x)=k+5limxkf(x)=f(k)
Hence, the given function is continuous for each value of k<1

case(iii) x=1
limx1f(x)=x+5=1+5=1+5=6limx1+f(x)=x5=15=4f(1)=x+5=1+5=6L.H.L.=f(1)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){3if0x14if1<x<35if3x10

Answer:

Given function is
f(x){3if0x14if1<x<35if3x10
Given function is defined for every real number k
Different cases are their
case (i) k<1
f(k)=3limxkf(x)=3limxkf(x)=f(k)
Hence, the given function is continuous for every value of k<1

case(ii) k=1
f(1)=3limx1f(x)=3limx1+f(x)=4R.H.L.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)=4limxkf(x)=4limxkf(x)=f(k)
Hence, for every value of k in 1<k<3 given function is continuous.

case(iv) k=3
f(3)=5limx3f(x)=4limx3+f(x)=5R.H.L.=f(3)L.H.L.
Hence, x=3 is the point of discontinuity

case(v) k>3
f(k)=5limxkf(x)=5limxkf(x)=f(k)
Hence, the given function is continuous for every value of k>3
case(vi) when k<3

f(k)=4limxkf(x)=4limxkf(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){2xifx<00if0x14xifx>1

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)={2ifx12xif1<x12ifx>1

Answer:

Given function is
f(x)={2ifx12xif1<x12ifx>1
Given function is defined for every real number k
Different cases are their
case (i) k<1
f(k)=2limxkf(x)=2limxkf(x)=f(k)
Hence, the given function is continuous for every value of k<1
case(ii) k = -1
f(1)=2limx1f(x)=2limx1+f(x)=2x=2(1)=2R.H.L.=L.H.L.=f(1)
Hence, the given function is continuous at x=1
case(iii) k>1
f(k)=2klimxkf(x)=2klimxkf(x)=f(k)
Hence, the given function is continuous for all values of x>1

case(vi) 1<k<1
f(k)=2klimxkf(x)=2klimxkf(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)=2limx1f(x)=2x=2(1)=2limx1+f(x)=2R.H.L.=f(1)=L.H.L.
Hence, at x =1 function is continuous

case(vi) k>1
f(k)=2limxkf(x)=2limxkf(x)=f(k)
Hence, the given function is continuous for every value of k > 1
case(vii) when k<1
f(k)=2klimxkf(x)=2klimxkf(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)={ax+1,ifx<3bx+3ifx>3
is continuous at x = 3.

Answer:

Given function is
f(x)={ax+1,ifx<3bx+3ifx>3
For the function to be continuous at x=3, R.H.L. must be equal to L.H.L.
limx3f(x)=ax+1=3a+1limx3+f(x)=bx+3=3b+3
For the function to be continuous
limx3f(x)=limx3+f(x)3a+1=3b+33(ab)=2ab=23a=b+23

Question:18.For what value of l is the function defined by
f(x)={λ(x22x)ifx04x+1ifx>0
continuous at x = 0? What about continuity at x = 1?

Answer:

Given function is
f(x)={λ(x22x)ifx04x+1ifx>0
For the function to be continuous at x=0, R.H.L. must be equal to L.H.L.
limx0f(x)=λ(x22x)=0limx0+f(x)=4x+1=1
For the function to be continuous
limx0f(x)=limx0+f(x)01
Hence, if no value of the function is continuous at x=0

For x=1
f(1)=4x+1=4(1)+1=5limx1f(x)=4+1=5 limx1f(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
limxkf(x)=k(k1)=kk+1=1limxk+f(x)=kk=0limxkf(x)limxk+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)=x2sinx+5 continuous at x = π ?

Answer:

Given function is
f(x)=x2sinx+5
Clearly, the Given function is defined at x = π
f(π)=π2sinπ+5=π20+5=π2+5limxπf(x)=π2sinπ+5=π20+5=π2+5limxπf(x)=f(π)
Hence, the function defined by f(x)=x2sinx+5 continuous at x = π

Question 21.Discuss the continuity of the following functions:
a) f(x)=sinx+cosx

Answer:

Given function is
f(x)=sinx+cosx
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)=sinx and h(x)=cosx
Let's suppose x=c+h
if xc, then h0
g(c)=sinclimxcg(x)

=limxcsinx=limh0sin(c+h)

We know that sin(a+b)=sinacosb+cosasinb

limh0sin(c+h)= limh0(sinccosh+coscsinh)

=limh0sinccosh+limh0coscsinh
=sinccos0+coscsin0=sinc
limxcg(x)=g(c)
Hence, function g(x)=sinx is a continuous function
Now,
h(x)=cosx
Let's suppose x=c+h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)

 We know that cos(a+b)=cosacosb+sinasinb

limh0cos(c+h)=limh0(cosccosh+sincsinh)

=limh0cosccosh+limh0sincsinh
=cosccos0+sincsin0=cosc
limxch(x)=h(c)
Hence, function h(x)=cosx is a continuous function
We proved independently that sinx and cosx are continuous functions.
So, we can say that
f(x)=g(x)+h(x)=sinx+cosx is also a continuous function

Question 21. b)Discuss the continuity of the following functions:
f(x)=sinxcosx

Answer:

Given function is
f(x)=sinxcosx
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)=sinx and h(x)=cosx
Let's suppose x=c+h
if xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)

We know thatsin(a+b)=sinacosb+cosasinb

limh0sin(c+h)=limh0(sinccosh+coscsinh)

=limh0sinccosh+limh0coscsinh
=sinccos0+coscsin0=sinc
limxcg(x)=g(c)
Hence, function g(x)=sinx is a continuous function
Now,
h(x)=cosx
Let's suppose x=c+h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)

We know thatcos(a+b)=cosacosb+sinasinb

limh0cos(c+h)=limh0(cosccosh+sincsinh)

=limh0cosccosh+limh0sincsinh
=cosccos0+sincsin0=cosc
limxch(x)=h(c)
Hence, function h(x)=cosx 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)=sinxcosx is also a continuous function.

Question 21. c)Discuss the continuity of the following functions:
f(x)=sinxcosx

Answer:

Given function is
f(x)=sinxcosx
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)=sinx and h(x)=cosx
Let's suppose x=c+h
if xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)

We know thatsin(a+b)=sinacosb+cosasinb

limh0sin(c+h)=limh0(sinccosh+coscsinh)

=limh0sinccosh+limh0coscsinh
=sinccos0+coscsin0=sinc
limxcg(x)=g(c)
Hence, function g(x)=sinx is a continuous function
Now,
h(x)=cosx
Let's suppose x=c+h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)

We know thatcos(a+b)=cosacosb+sinasinb

limh0cos(c+h)=limh0(cosccosh+sincsinh)

=limh0cosccosh+limh0sincsinh
=cosccos0+sincsin0=cosc
limxch(x)=h(c)
Hence, function h(x)=cosx 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)=sinx.cosx 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.
g(x)h(x),h(x)0 is continuous

1h(x),h(x)0 is continuous1g(x),g(x)0 is continuous
Lets take g(x)=sinx and h(x)=cosx
Let's suppose x=c+h
if xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)

We know that sin(a+b)=sinacosb+cosasinb

limh0sin(c+h)=limh0(sinccosh+coscsinh)

=limh0sinccosh+limh0coscsinh
=sinccos0+coscsin0=sinc
limxcg(x)=g(c)
Hence, function g(x)=sinx is a continuous function
Now,
h(x)=cosx
Let's suppose x=c+h
if xc, then h0
h(c)=cosclimxch(x)

=limxccosx

=limh0cos(c+h)

We know that cos(a+b)=cosacosb+sinasinb

limh0cos(c+h)

=limh0(cosccosh+sincsinh)

=limh0cosccosh+limh0sincsinh
=cosccos0+sincsin0=cosc
limxch(x)=h(c)
Hence, the function h(x)=cosx is a continuous function
We proved independently that sinx and cosx a continuous functions.
So, we can say that
cosec x = 1sinx=1g(x) is also continuous except at x=nπ
sec x = 1cosx=1h(x) is also continuous except at x=(2n+1)π2
cot x = cosxsinx=h(x)g(x) is also continuous except at x=nπ

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

f(x)={sinxxifx<0x+1ifx>0

Answer:

Given function is
f(x)={sinxxifx<0x+1ifx>0
limx0f(x)=limx0sinxx=1limx0+f(x)=x+1=1limx0f(x)=limx0+f(x)
Hence, the function is continuous.
Therefore, no point of discontinuity

Question 24.Determine if f is defined by
f(x)={x2sin1/xifx00ifx=0
Is it a continuous function?

Answer:

Given function is
f(x)={x2sin1/xifx00ifx=0
Given function is defined for all real numbers k
when x = 0
f(0)=0limx0f(x)=limx0(x2sin1x)=limx0(x.sin1x1x)=0(1)=0      (limx0sinxx=1)
limx0f(x)=f(0)
Hence, the function is continuous at x = 0
when x0
f(k)=k2sin1klimxkf(x)=limxk(x2sin1x)=k2sin1klimxk=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)={sinxcosxifx01ifx=0

Answer:

Given function is
f(x)=sinxcosx
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)=sinx and h(x)=cosx
Let's suppose x=c+h
if xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)

We know thatsin(a+b)=sinacosb+cosasinb

limh0sin(c+h)=limh0(sinccosh+coscsinh)

=limh0sinccosh+limh0coscsinh
=sinccos0+coscsin0=sinc
limxcg(x)=g(c)
Hence, function g(x)=sinx is a continuous function
Now,
h(x)=cosx
Let's suppose x=c+h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx

=limh0cos(c+h)

We know thatcos(a+b)=cosacosb+sinasinb

limh0cos(c+h)=limh0(cosccosh+sincsinh)

=limh0cosccosh+limh0sincsinh
=cosccos0+sincsin0=cosc
limxch(x)=h(c)
Hence, function h(x)=cosx is a continuous function
We proved independently that sinx and cosx are continuous functions.
So, we can say that
f(x)=g(x)h(x)= sinx - cosx is also a continuous function

When x=0
f(0)=(1)limx0f(x)

=sin0cos0=1

limx0+f(x)=sin0cos0=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)={kcosxπ2xifxπ/23ifx=π/2atx=π/2

Answer:

Given function is
f(x)={kcosxπ2xifxπ/23ifx=π/2
When x=π2
f(π2)=3let x=π+hlimxπ2f(x)=limh0kcos(π2+h)π2(π2+h)=k.limh0sinh2h=k2
For the function to be continuous
limxπ2f(x)=f(π2)k2=3k=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)={kx2ifx23ifx>2atx=2

Answer:

Given function is
f(x)={kx2ifx23ifx>2
When x=2
For the function to be continuous
f(2)=R.H.L.=LH.L.
f(2)=4klimx2f(x)=4klimx2+f(x)=3f(2)=limx2f(x)=limx2+f(x)4k=3k=34
Hence, the values of k so that the function f is continuous at x= 2 is 34

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

f(x)={kx+1ifxπcosxifx>πatx=π

Answer:

Given function is
f(x)={kx+1ifxπcosxifx>π
When x = π
For the function to be continuous
, f( π ) = R.H.L. = LH.L.
f(π)=kπ+1limxπf(x)=kπ+1limxπ+f(x)=cosπ=1f(π)=limxπf(x)=limxπ+f(x)kπ+1=1k=2π
Hence, the values of k so that the function f is continuous at x= π is 2π

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

f(x)={kx+1ifx53x5ifx>5atx=5

Answer:

Given function is
f(x)={kx+1ifx53x5ifx>5
When x=5
For the function to be continuous
f(5)=R.H.L.=LH.L.
f(5)=5k+1limx5f(x)=5k+1limx5+f(x)=3(5)5=155=10f(5)=limx5f(x)=limx5+f(x)5k+1=10k=95
Hence, the values of k so that the function f is continuous at x=5 is 95

Question:30Find the values of a and b such that the function defined by
f(x)={5ifx2ax+bif2<x<1021,ifx>10
It is a continuous function.

Answer:

Given that continuous function is
f(x)={5ifx2ax+bif2<x<1021,ifx>10
The function is continuous so
limx2f(x)=limx2+f(x)andlimx10f(x)=limx10+f(x)
limx2f(x)=5limx2+f(x)=ax+b=2a+b

2a+b=5      (i)

limx10f(x)=ax+b=10a+blimx10+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)={5ifx2ax+bif2<x<1021,ifx>10 is a continuous function is 2 and 1 respectively

Question:31.Show that the function defined by f(x)=cos(x2) is a continuous function.

Answer:

Given function is
f(x)=cos(x2)
Given function is defined for all real values of x
Let x = k + h
if xk, then h0
f(k)=cosk2limxkf(x)=limxkcosx2=limh0cos(k+h)2

=cosk2limxkf(x)=f(k)
Hence, the function f(x)=cos(x2) is a continuous function

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

Answer:

Given function is
f(x)=|cosx|
Given function is defined for all values of x
f = g o h , g(x)=x and h(x)=cosx
Now,
g(x){x if x<00 if x=0x if x>0
g(x) is defined for all real numbers k
case(i) k<0
g(k)=klimxkg(x)=klimxkg(x)=g(k)
Hence, g(x) is continuous when k<0

case (ii) k>0
g(k)=klimxkg(x)=klimxkg(x)=g(k)
Hence, g(x) is continuous when k > 0

case (iii) k=0
g(0)=0limx0g(x)=x=0

limx0+g(x)=x=0

limx0g(x)=g(0)=limx0+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)=cosx
Let's suppose x=c+h
if xc, then h0
h(c)=cosclimxch(x)

=limxccosx

=limh0cos(c+h)

We know that cos(a+b)=cosacosb+sinasinb

limh0cos(c+h)=limh0(cosccosh+sincsinh)

=limh0cosccosh+limh0sincsinh
=cosccos0+sincsin0=cosc
limxch(x)=h(c)
Hence, function h(x)=cosx is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x)=goh is also continuous

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

Answer:

Given function is
f(x)=sinx
f(x) = h o g , h(x) =sinx and g(x) = |x|
Now,
g(x){x if x<00 if x=0x if x>0
g(x) is defined for all real numbers k
case(i) k<0
g(k)=klimxkg(x)=klimxkg(x)=g(k)
Hence, g(x) is continuous when k < 0

case (ii) k>0
g(k)=klimxkg(x)=klimxkg(x)=g(k)
Hence, g(x) is continuous when k > 0

case (iii) k=0
g(0)=0limx0g(x)=x=0

limx0+g(x)=x=0

limx0g(x)=g(0)=limx0+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) =sinx
Let's suppose x=c+h
if xc, then h0
h(c)=sinc

limxch(x)=limxcsinx

=limh0sin(c+h)

 We know that sin(a+b)=sinacosb+cosasinb

limh0sin(c+h)=limh0(sinccosh+coscsinh)

=limh0sinccosh+limh0coscsinh
=sinccos0+coscsin0=sinc
limxch(x)=h(c)
Hence, function h(x)=sinx is a continuous function
g(x) is continuous , h(x) is continuous
Therefore, f(x)=hog 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){x if x<00 if x=0x if x>0
g(x) is defined for all real numbers k
case(i) k < 0
g(k)=klimxkg(x)=klimxkg(x)=g(k)
Hence, g(x) is continuous when k < 0

case (ii) k > 0
g(k)=klimxkg(x)=klimxkg(x)=g(k)
Hence, g(x) is continuous when k > 0

case (iii) k = 0
g(0)=0limx0g(x)=x=0

limx0+g(x)=x=0

limx0g(x)=g(0)=limx0+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){(x+1) if x<10 if x=1(x+1) if x>1
g(x) is defined for all real numbers k
case(i) k < -1
h(k)=(k+1)limxkh(x)=(k+1)limxkh(x)=h(k)
Hence, h(x) is continuous when k < -1

case (ii) k > -1
h(k)=k+1limxkh(x)=k+1limxkh(x)=h(k)
Hence, h(x) is continuous when k > -1

case (iii) k = -1
h(1)=0limx1h(x)=(x1)

=0limx1+h(x)=x+1=0

limx1h(x)=h(0)=limx1+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(x2+5)

Answer:

Given function is
f(x)=sin(x2+5)
When we differentiate it w.r.t. x.
Let's take t=x2+5. then,
f(t)=sint
df(t)dx=df(t)dt.dtdx (By chain rule)
df(t)dt=d(sint)dt=cost=cos(x2+5)
dtdx=d(x2+5)dx=2x
Now,
df(t)dx=df(t)dt.dtdx=cos(x2+5).2x
Therefore, the answer is 2xcos(x2+5)

Question 2.Differentiate the functions with respect to x in

cos(sinx)

Answer:

Given function is
f(x)=cos(sinx)
Let’s take t=sinx then,
f(t)=cost
df(t)dx=df(t)dt.dtdx ( By chain rule)
df(t)dt=d(cost)dt=sint=sin(sinx)
dtdx=d(sinx)dt=cosx
Now,
df(t)dx=df(t)dt.dtdx=sin(sinx).cosx
Therefore, the answer is sin(sinx).cosx

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)=sint
df(t)dx=df(t)dt.dtdx (By chain rule)
df(t)dt=d(sint)dt=cost=cos(ax+b)
dtdx=d(ax+b)dx=a
Now,
df(t)dx=df(t)dt.dtdx=cos(ax+b).a
Therefore, the answer is acos(ax+b)

Question 4. Differentiate the functions with respect to x in

sec(tan(x))

Answer:

Given function is
f(x)=sec(tan(x))
When we differentiate it w.r.t. x.
Let's take t=x. then,
f(t)=sec(tant)
take tant=k . then,
f(k)=seck
df(k)dx=df(k)dk.dkdt.dtdx (By chain rule)
df(k)dk=d(seck)dk=secktank=sec(tanx)tan(tanx)
(k=tant and t=x)
df(t)dt=d(tant)dt=sec2t=sec2(x)      (t=x)
dtdx=d(x)dx=12x
Now,
df(k)dx=df(k)dk.dkdt.dtdx

=sec(tanx)tan(tanx).sec2(x).12x
Therefore, the answer is sec(tanx).tan(tanx).sec2(x)2x

Question 5.Differentiate the functions with respect to x in

sin(ax+b)cos(cx+d)

Answer:

Given function is
f(x)=sin(ax+b)cos(cx+d)=g(x)h(x)
We know that,
f(x)=g(x)h(x)g(x)h(x)h2(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)=sinu and cos(cx+d)=cosc
g(x)=d(g(x))dx=d(g(x))du.dudx (By chain rule)
d(g(x))du=d(sinu)du=cosu=cos(ax+b)         (u=ax+b)
dudx=d(ax+b)dx=a
g(x)=acos(ax+b) -(i)
Similarly,
h(x)=d(h(x))dx=d(h(x))dv.dvdx
d(h(x))dv=d(cosv)dv=sinv=sin(cx+d)       (v=(cx+d))
dvdx=d(cx+d)dv=c
h(x)=csin(cx+d) -(ii)
Now, put (i) and (ii) in
f(x)=g(x)h(x)g(x)h(x)h2(x)=acos(ax+b).cos(cx+d)sin(ax+b).(c.sin(cx+d))cos2(cx+d)
=acos(ax+b).cos(cx+d)cos2(cx+d)+sin(ax+b).c.sin(cx+d)cos2(cx+d)
=acos(ax+b).sec(cx+d)+csin(ax+b).tan(cx+d).sec(cx+d)
Therefore, the answer is acos(ax+b).sec(cx+d)+csin(ax+b).tan(cx+d).sec(cx+d)

Question 6.Differentiate the functions with respect to x in

cosx3.sin2(x5)

Answer:

Given function is
f(x)=cosx3.sin2(x5)
Differentiation w.r.t. x is
f(x)=g(x).h(x)+g(x).h(x)
g(x)=cosx3 and h(x)=sin2(x5)
Lets take u=x3 and v=x5
Our functions become,
cosx3=cosu and sin2(x5)=sin2v
Now,
g(x)=d(g(x))dx=d(g(u))du.dudx ( By chain rule)
d(g(u))du=d(cosu)du=sinu=sinx3    (u=x3)
dudx=d(x3)dx=3x2
g(x)=sinx3.3x2 -(i)
Similarly,
h(x)=d(h(x))dx=d(h(v))dv.dvdx
d(h(v))dv=d(sin2v)dv=2sinvcosv=2sinx5cosx5

(v=x5)
dvdx=d(x5)dx=5x4
h(x)=2sinx5cosx5.5x4=10x4sinx5cosx5 -(ii)
Put (i) and (ii) in
f(x)=g(x).h(x)+g(x).h(x)

=3x2sinx3.sin2x5+cosx3.10x4sinx5cosx5
Therefore, the answer is 10x4sinx5cosx5.cosx33x2sinx3.sin2x5

Question 7.Differentiate the functions with respect to x in

2cot(x2)

Answer:

Give function is
f(x)=2cot(x2)
Let's take t=x2
Now, take cott=k2
f(k)=2k
Differentiation w.r.t. x
d(f(k))dx=d(f(k))dk.dkdt.dtdx -(By chain rule)
d(f(k))dk=d(2k)dk=2
dkdt=d(cott)dt=12cott.(cosec2t)=cosec2x22cotx2(t=x2)
dtdx=d(x2)dx=2x
So,
d(f(k))dx=2.cosec2x22cotx2.2x=22xsin2x22sinx2cosx2sin2x2 ( Multiply and divide by 2 and multiply and divide cotx2 by sinx2
(becausecotx=cosxsinx andcscx=1sinx)
=22xsinx2sin2x2(2sinxcosx=sin2x)
There, the answer is 22xsinx2sin2x2

Question 8Differentiate the functions with respect to x in

cos(x)

Answer:

Let us assume : y = cos(x)

Differentiating y with respect to x, we get :

dydx = d(cos(x))dx

or = sinx.d(x)dx

or = sinx2x

Question 9. Prove that the function f given by f(x)=|x1|,xϵR is not differentiable at x = 1.

Answer:

Given function is
f(x)=|x1|,xϵR
We know that any function is differentiable when both.
limh0f(c+h)f(c)h and limh0+f(c+h)f(c)h are finite and equal
The required condition for the function to be differential at x = 1 is

limh0f(1+h)f(1)h=limh0+f(1+h)f(1)h
Now, the Left-hand limit of a function at x = 1 is
limh0f(1+h)f(1)h=limh0|1+h1||11|h=limh0|h|0h
=limh0hh=1    (h<0)
The right-hand limit of a function at x = 1 is
limh0+f(1+h)f(1)h=limh0+|1+h1||11|h=limh0+|h|0h
=limh0hh=1
Now, it is clear that.
R.H.L. at x= 1 L.H.L. at x= 1
Therefore, function f(x)=|x1| 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.
limh0f(c+h)f(c)h and limh0+f(c+h)f(c)h are finite and equal
The required condition for the function to be differential at x = 1 is

limh0f(1+h)f(1)h=limh0+f(1+h)f(1)h
Now, the Left-hand limit of the function at x = 1 is
limh0f(1+h)f(1)h=limh0[1+h][1]h=limh001h
=limh01h=     (h<01+h<1,[1+h]=0)
The right-hand limit of the function at x = 1 is
limh0+f(1+h)f(1)h=limh0+[1+h][1]h=limh0+11h
=limh00h=0     (h>01+h>1,[1+h]=1)
Now, it is clear that.
R.H.L. at x= 1 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

limh0f(2+h)f(2)h=limh0+f(2+h)f(2)h
Now, the Left-hand limit of the function at x = 2 is
limh0f(2+h)f(2)h=limh0[2+h][2]h=limh012h
=limh01h=    (h<02+h<2,[2+h]=1)
The right-hand limit of the function at x = 1 is
limh0+f(2+h)f(2)h=limh0+[2+h][2]h=limh0+22h
=limh00h=0     (h>02+h>2,[2+h]=2)
Now, it is clear that.
R.H.L. at x= 2 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 dydx in the following:

2x+3y=sinx

Answer:

Given function is
2x+3y=sinx
We can rewrite it as
3y=sinx2x
Now, differentiation w.r.t. x is
3dydx=d(sinx2x)dx=cosx2
dydx=cosx23
Therefore, the answer is cosx23

Question 2.Find dydx in the following: 2x+3y=siny

Answer:

Given function is
2x+3y=siny
We can rewrite it as
siny3y=2x
Now, differentiation w.r.t. x is
dydx(siny3y)=d(2x)dx

(cosydydx3dydx)=2
dydx=2cosy3
Therefore, the answer is 2cosy3

Question 3.Find dydx in the following: ax+by2=cosy

Answer:

Given function is
ax+by2=cosy
We can rewrite it as
by2cosy=ax
Now, differentiation w.r.t. x is
dydx(2by(siny))=d(ax)dx=a
dydx=a2by+siny
Therefore, the answer is a2by+siny

Question 4.Find dydx in the following:

xy+y2=tanx+y

Answer:

Given function is
xy+y2=tanx+y
We can rewrite it as
xy+y2y=tanx
Now, differentiation w.r.t. x is
y+dydx(x+2y1)=d(tanx)dx=sec2x
dydx=sec2xyx+2y1
Therefore, the answer is sec2xyx+2y1

Question 5.Find dydx in the following: x2+xy+y2=100

Answer:

Given function is
x2+xy+y2=100
We can rewrite it as
xy+y2=100x2
Now, differentiation w.r.t. x is
y+dydx(x+2y)=d(100x2)dx=2x
dydx=2xyx+2y
Therefore, the answer is 2xyx+2y

Question 6Find dydx in the following:

x3+x2y+xy2+y3=81

Answer:

Given function is
x3+x2y+xy2+y3=81
We can rewrite it as
x2y+xy2+y3=81x3
Now, differentiation w.r.t. x is
d(x2y+xy2+y3)dx=d(81x3)dx
2xy+y2+dydx(x2+2xy+3y2)=3x2dydx=(3x2+2xy+y2)(x2+2xy+3y2
Therefore, the answer is (3x2+2xy+y2)(x2+2xy+3y2

Question 7. Find dydx in the following: sin2y+cosxy=k

Answer:

Given function is
sin2y+cosxy=k
Now, differentiation w.r.t. x is
d(sin2y+cosxy)dx=d(k)dx
2sinycosydydx+(sinxy)(y+xdydx)=0

dydx(2sinycosyxsinxy)=ysinxydydx

=ysinxy2sinycosyxsinxy

=ysinxysin2yxsinxy      (2sinxcosy=sin2x)
Therefore, the answer is ysinxysin2yxsinxy

Question 8.Find dydx in the following:

sin2x+cos2y=1

Answer:

Given function is
sin2x+cos2y=1
We can rewrite it as
cos2y=1sin2x
Now, differentiation w.r.t. x is
d(cos2y)dx=d(1sin2x)dx
2cosy(siny)dydx=2sinxcosxdydx=2sinxcosx2sinycosy=sin2xsin2y      (2sinacosa=sin2a)
Therefore, the answer is sin2xsin2y

Question 9Find dydx in the following:

y=sin1(2x1+x2)

Answer:

Given function is
y=sin1(2x1+x2)
Lets consider x=tant
Then,
d(x)dx=d(tant)dt.dtdx         (by chain rule)
1=sec2t.dtdxdtdx=1sec2t=11+tan2t=11+x2       (sec2t=1+tan2t and x=tant)
Now,
2x1+x2=2tant1+tan2t=sin2t      (sin2x=2tanx1+tan2x)
Our equation reduces to
y=sin1(sin2t)
y=2t
Now, differentiation w.r.t. x is
d(y)dx=d(2t)dt.dtdx
dydx=2.11+x2=21+x2
Therefore, the answer is 21+x2

Question 10.Find dydx in the following:
y=tan1(3xx313x2),13<x<13

Answer:

Given function is
y=tan1(3xx313x2)
Lets consider x=tant
Then,
d(x)dx=d(tant)dt.dtdx         (by chain rule)
1=sec2t.dtdxdtdx=1sec2t=11+tan2t=11+x2       (sec2t=1+tan2t and x=tant)
Now,
3xx313x2=3tanttan3t13tan2t=tan3t      (tan3x=3tanxtan3x13tan2x)
Our equation reduces to
y=tan1(tan3t)
y=3t
Now, differentiation w.r.t. x is
d(y)dx=d(3t)dt.dtdx
dydx=3.11+x2=31+x2
Therefore, the answer is 31+x2

Question 11.Find dydx in the following:

y=cos1(1x21+x2),0<x<1

Answer:

Given function is
y=cos1(1x21+x2)
Let's consider x=tant
Then,
d(x)dx=d(tant)dt.dtdx         (by chain rule)
1=sec2t.dtdxdtdx=1sec2t=11+tan2t=11+x2       (sec2t=1+tan2t and x=tant)
Now,
1x21+x2=1tan2t1+tan2t=cos2t      (cos2x=1tan2x1+tan2x)
Our equation reduces to
y=cos1(cos2t)
y=2t
Now, differentiation w.r.t. x is
d(y)dx=d(2t)dt.dtdx
dydx=2.11+x2=21+x2
Therefore, the answer is 21+x2

Question 12. Find dydx in the following: y=sin1(1x21+x2),0<x<1

Answer:

Given function is
y=sin1(1x21+x2)
We can rewrite it as
siny= (1x21+x2)
Let's consider x=tant
Then,
d(x)dx=d(tant)dt.dtdx         (by chain rule)
1=sec2t.dtdxdtdx=1sec2t=11+tan2t=11+x2       (sec2t=1+tan2t and x=tant)
Now,
1x21+x2=1tan2t1+tan2t=cos2t      (cos2x=1tan2x1+tan2x)
Our equation reduces to
siny=cos2t
Now, differentiation w.r.t. x is
d(siny)dx=d(cos2t)dt.dtdx
cosydydx=2(sin2t).11+x2=2sin2t1+x2 =2.2tant1+tan2t1+x2=2.2x1+x21+x2=4x(1+x2)2
(sin2x=2tanx1+tan2x and x=tant)
siny= (1x21+x2)cosy=2x1+x2
2x1+x2dydx=4x(1+x2)2
dydx=2(1+x2)
Therefore, the answer is 21+x2

Question 13.Find dydx in the following:

y=cos1(2x1+x2),1<x<1

Answer:

Given function is
y=cos1(2x1+x2)
We can rewrite it as
cosy=(2x1+x2)
Let's consider x=tant
Then,
d(x)dx=d(tant)dt.dtdx         (by chain rule)
1=sec2t.dtdxdtdx=1sec2t=11+tan2t=11+x2       (sec2t=1+tan2t and x=tant)
Now,
2x1+x2=2tant1+tan2t=sin2t      (sin2x=2tanx1+tan2x)
Our equation reduces to
cosy=sin2t
Now, differentiation w.r.t. x is
d(cosy)dx=d(sin2t)dt.dtdx
(siny)dydx=2(cos2t).11+x2=2cos2t1+x2 =2.1tan2t1+tan2t1+x2=2.1x21+x21+x2=2(1x2)(1+x2)2
(cos2x=1tan2x1+tan2x and x=tant)
cosy= (2x1+x2)siny=1x21+x2
1x21+x2dydx=2(1x2)(1+x2)2
dydx=2(1+x2)
Therefore, the answer is 21+x2

Question 14. Find dydx in the following:

y=sin1(2x1x2),12<x12

Answer:

Given function is
y=sin1(2x1x2)
Let's take x=sint
Then,
d(x)dx=(sint)dt.dtdx     (by chain rule)
1=cost.dtdx
dtdx=1cost=11sin2t=11x2
(cosx=1sin2x and x=sint)
And
2x1x2=2sint1sin2t=2sintcos2t

=2sintcost=sin2t
(cosx=1sin2x and  2sinxcosx=sin2x)
Now, our equation reduces to
y=sin1(sin2t)
y=2t
Now, differentiation w.r.t. x
d(y)dx=d(2t)dt.dtdx
dydx=2.11x2=21x2
Therefore, the answer is 21x2

Question 15. Find dydx in the following:

y=sec1(12x21),0<x<1/2

Answer:

Given function is
y=sec1(12x21)
Let's take x=cost
Then,
d(x)dx=(cost)dt.dtdx     (by chain rule)
1=sint.dtdx
dtdx=1sint=11cos2t=11x2
(sinx=1cos2x and x=cost)
And
12x21=12cos2t1=1cos2t=sec2t
(cos2x=2cos2x1 and cosx=1secx)

Now, our equation reduces to
y=sec1(sec2t)
y=2t
Now, differentiation w.r.t. x
d(y)dx=d(2t)dt.dtdx
dydx=2.11x2=21x2
Therefore, the answer is 21x2

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:

exsinx

Answer:

Given function is
f(x)=exsinx
We differentiate with the help of the Quotient rule.
f(x)=d(ex)dx.sinxex.(sinx)dxsin2x
=ex.sinxex.cossin2x=ex(sinxcosx)sin2x
Therefore, the answer is ex(sinxcosx)sin2x

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

esin1x

Answer:

Given function is
f(x)=esin1x
Let g(x)=sin1x
Then,
f(x)=eg(x)
Now, differentiation w.r.t. x
f(x)=g(x).eg(x) -(i)
g(x)=sin1xg(x)=11x2
Put this value in our equation (i)
f(x)=11x2.esin1x=esin1x1x2

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

ex3

Answer:

Given function is
f(x)=ex3
Let g(x)=x3
Then,
f(x)=eg(x)
Now, differentiation w.r.t. x
f(x)=g(x).eg(x) -(i)
g(x)=x3g(x)=3x2
Put this value in our equation (i)
f(x)=3x2.ex3
Therefore, the answer is 3x2.ex3

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

sin(tan1ex)

Answer:

Given function is
f(x)=sin(tan1ex)
Let's take g(x)=tan1ex
Now, our function reduces to
f(x)=sin(g(x))
Now,
f(x)=g(x)cos(g(x)) -(i)
And
g(x)=tan1exg(x)=d(tan1ex)dx.d(ex)dx=11+(ex)2.ex=ex1+e2x
Put this value in our equation (i)
f(x)=ex1+e2xcos(tan1ex)
Therefore, the answer is ex1+e2xcos(tan1ex)

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

log(cosex)

Answer:

Given function is
f(x)=log(cosex)
Let's take g(x)=cosex
Now, our function reduces to
f(x)=log(g(x))
Now,
f(x)=g(x).1g(x) -(i)
And
g(x)=cosexg(x)=d(cosex)dx.d(ex)dx=(sinex).ex=ex.sinex
Put this value in our equation (i)
f(x)=ex.sinex.1cosex=ex.tanex     (sinxcosx=tanx)
Therefore, the answer is ex.tanex,   ex(2n+1)π2,  nN

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

ex+ex2+.....ex5

Answer:

Given function is
f(x)=ex+ex2+.....ex5
Now, differentiation w.r.t. x is
f(x)=d(ex)dx.d(x)dx+d(ex2)dx.d(x2)dx+d(ex3)dx.d(x3)dx+d(ex4)dx.d(x4)dx+d(ex5)dx.d(x5)dx
=ex.1+ex2.2x+ex3.3x2+ex4.4x3+ex5.5x4
=ex+2xex2+3x2ex3+4x3ex4+5x4ex5
Therefore, answer is ex+2xex2+3x2ex3+4x3ex4+5x4ex5

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

ex,x>0

Answer:

Given function is
f(x)=ex
Let's take g(x)=x
Now, our function reduces to
f(x)=eg(x)
Now,
f(x)=g(x).12eg(x).d(eg(x))dx=g(x).12eg(x).eg(x)=g(x).eg(x)2.eg(x)=g(x).ex2.ex -(i)
And
g(x)=xg(x)=(x)dx=12x
Put this value in our equation (i)
f(x)=ex2x.2.ex=ex4xex
Therefore, the answer is ex4xex.  x>0

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

Answer:

Given function is
f(x)=log(logx)
Let's take g(x)=logx
Now, our function reduces to
f(x)=log(g(x))
Now,
f(x)=g(x).1g(x) -(i)
And
g(x)=logxg(x)=1x
Put this value in our equation (i)
f(x)=1x.1logx=1xlogx
Therefore, the answer is 1xlogx,  x>1

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

cosxlogx,x>0

Answer:

Given function is
f(x)=cosxlogx
We differentiate with the help of the Quotient rule.
f(x)=d(cosx)dx.logxcosx.(logx)dx(logx)2
=(sinx).logxcosx.1x(logx)2=(xsinxlogx+cosx)x(logx)2
Therefore, the answer is (xsinxlogx+cosx)x(logx)2

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

cos(logx+ex),x>0

Answer:

Given function is
f(x)=cos(logx+ex)
Let's take g(x)=(logx+ex)
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)=(logx+ex)
g(x)=d(logx)dx+d(ex)dx=1x+ex
Put this value in our equation (i)
f(x)=(1x+ex)sin(logx+ex)
Therefore, the answer is (1x+ex)sin(logx+ex),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. cosx.cos2x.cos3x

Answer:

Given function is
y=cosx.cos2x.cos3x
Now, take a look at both sides.
logy=log(cosx.cos2x.cos3x)

logy=logcosx+logcos2x+logcos3x
Now, differentiation w.r.t. x
logy=log(cosx.cos2x.cos3x)

d(logy)dx=logcosxdx+logcos2xdx+logcos3xdx

1y.dydx=

(sinx)1cosx+(2sin2x)1cos2x+(3sin3x).1cos3x

1ydydx=(tanx+tan2x+tan3x)

(sinxcosx=tanx)

dydx=y(tanx+tan2x+tan3x)

dydx=cosxcos2xcos3x(tanx+tan2x+tan3x)
There, the answer is cosxcos2xcos3x(tanx+tan2x+tan3x)

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

(x1)(x2)(x3)(x4)(x5)

Answer:

Given function is
y=(x1)(x2)(x3)(x4)(x5)
Take logs on both sides.
logy=12log((x1)(x2)(x3)(x4)(x5))

logy=

12(log(x1)+log(x2)log(x3)log(x4)

log(x5))
Now, differentiation w.r.t. x is
d(logy)dx=12(d(log(x1))dx+d(log(x2))dxd(log(x3))dx

d(log(x4))dxd(log(x5))dx)
1ydydx=12(1x1+1x21x31x41x5)dydx

=y12(1x1+1x21x31x41x5)

dydx=12(x1)(x2)(x3)(x4)(x5)

(1x1+1x21x31x41x5)
Therefore, the answer is 12(x1)(x2)(x3)(x4)(x5)(1x1+1x21x31x41x5)

Question 3Differentiate the functions w.r.t. x. (logx)cosx

Answer:

Given function is
y=(logx)cosx
Take logs on both sides.
logy=cosxlog(logx)
Now, differentiation w.r.t x is

d(logy)dx=d(cosxlog(logx))dx

1ydydx=(sinx)(log(logx))+cosx(1logx1x)

dydx=y(cosx.1logx.1xsinxlog(logx))

dydx=(logx)cosx(cosxxlogxsinxlog(logx))
Therefore, the answer is (logx)cosx(cosxxlogxsinxlog(logx))

Question 4Differentiate the functions w.r.t. x. xx2sinx

Answer:

Given function is
y=xx2sinx
Let's take t=xx
Take logs on both sides.
logt=xlogx
Now, differentiation w.r.t x is
logt=xlogxd(logt)dt.dtdx=d(xlogx)dx       (by chain rule)

1t.dtdx=logx+1

dtdx=t(logx+1)

dtdx=xx(logx+1)               (t=xx)
Similarly, take k=2sinx
Now, take log on both sides and differentiate w.r.t. x
logk=sinxlog2d(logk)dk.dkdx=d(sinxlog2)dx       (by chain rule)

1k.dkdx=cosxlog2

dkdx=k(cosxlog2)

dkdx=2sinx(cosxlog2)

(k=2sinx)
Now,
dydx=dtdxdkdx

dydx=xx(logx+1)2sinx(cosxlog2)
Therefore, the answer is xx(logx+1)2sinx(cosxlog2)

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.
logy=log[(x+3)2.(x+4)3.(x+5)4]

logy=2log(x+3)+3log(x+4)+4log(x+5)
Now, differentiate w.r.t. x we get,
1y.dydx=2.1x+3+3.1x+4+4.1x+5

dydx=y(2x+3+3x+4+4x+5)

dydx=(x+3)2.(x+4)3.(x+5)4.(2x+3+3x+4+4x+5)

dydx=(x+3)2.(x+4)3.(x+5)4.

(2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)(x+3)(x+4)(x+5))

dydx=(x+3)(x+4)2(x+5)3

(9x2+70x+133)
Therefore, the answer is (x+3)(x+4)2(x+5)3(9x2+70x+133)

Question 6Differentiate the functions w.r.t. x. (x+1x)x+x1+1x

Answer:

Given function is
y=(x+1x)x+x1+1x
Let's take t=(x+1x)x
Now, take a look at both sides.
logt=xlog(x+1x)
Now, differentiate w.r.t. x
we get,
1t.dtdx=log(x+1x)+x(11x2).1(x+1x)

=x21x2+1+log(x+1x)

dtdx=t((x21x2+1)+log(x+1x))

dtdx=(x+1x)x((x21x2+1)+log(x+1x))
Similarly, take k=x1+1x
Now, take a look at both sides.
logk=(1+1x)logx
Now, differentiate w.r.t. x
We get,
1k.dkdx=1x(1+1x)+(1x2).logx

=x2+1x2+1x2.logx

dkdx=t(x2+1x2+(1x2)logx)

dkdx=xx+1x(x2+1logxx2)
Now,
dydx=dtdx+dkdx
dydx=(x+1x)x((x21x2+1)+

log(x+1x))+xx+1x(x2+1logxx2)
Therefore, the answer is (x+1x)x((x21x2+1)+log(x+1x))+xx+1x(x2+1logxx2)

Question 7Differentiate the functions w.r.t. x. (logx)x+xlogx

Answer:

Given function is
y=(logx)x+xlogx
Let's take t=(logx)x
Now, take a look at both sides.
logt=xlog(logx)
Now, differentiate w.r.t. x
we get,
1tdtdx=log(logx)+x.1x.1logx

=log(logx)+1logx

dtdx=t.(log(logx)+1logx)

dtdx=(logx)x(log(logx))+(logx)x.1logx

=(logx)x(log(logx))+(logx)x1
Similarly, take k=xlogx
Now, take a look at both sides.
logk=logxlogx=(logx)2
Now, differentiate w.r.t. x
We get,
1kdkdx=2(logx).1x

dtdx=k.(2(logx).1x)

dtdx=xlogx.(2(logx).1x)=2xlogx1.logx
Now,
dydx=dtdx+dkdx
dydx=(logx)x(log(logx))+(logx)x1+2xlogx1.logx
Therefore, the answer is (logx)x(log(logx))+(logx)x1+2xlogx1.logx

Question 8: Differentiate the functions w.r.t. x. (sinx)x+sin1x

Answer:

Given function is
(sinx)x+sin1x
Let's take t=(sinx)x
Now, take a look at both sides.
logt=xlog(sinx)
Now, differentiate w.r.t. x
we get,
1tdtdx=log(sinx)+x.cosx.1sinx

=log(sinx)+x.cotx

   (cosxsinx=cotx)

dtdx=t.(log(sinx)+x.cotx)

dtdx=(sinx)x(log(sinx)+xcotx)
Similarly, take k=sin1x
Now, differentiate w.r.t. x
We get,
dkdt=11(x)2.12x

=12xx2

dkdt=12xx2
Now,
dydx=dtdx+dkdx
dydx=(sinx)x(log(sinx)+xcotx)+12xx2
Therefore, the answer is (sinx)x(log(sinx)+xcotx)+12xx2

Question 9Differentiate the functions w.r.t. x y=xsinx+(sinx)cosx

Answer:

Given function is
y=xsinx+(sinx)cosx

Now, take t=xsinx
Now, take a look at both sides.
logt=sinxlogx
Now, differentiate it w.r.t. x
we get,
1tdtdx=cosxlogx+1x.sinx

dtdx=t(cosxlogx+1x.sinx)

dtdx=xsinx(cosxlogx+1x.sinx)
Similarly, take k=(sinx)cosx
Now, take a look at both sides.
logk=cosxlog(sinx)
Now, differentiate it w.r.t. x
we get,
1kdkdt=(sinx)(log(sinx))+cosx.1sinx.cosx

=sinxlog(sinx)+cotx.cosxdkdt

=k(sinxlog(sinx)+cotx.cosx)

dkdt=(sinx)cosx(sinxlog(sinx)+cotx.cosx)
Now,
dydx=xsinx(cosxlogx+1x.sinx)+

(sinx)cosx(sinxlog(sinx)+cotx.cosx)
Therefore, the answer is xsinx(cosxlogx+1x.sinx)+

(sinx)cosx(sinxlog(sinx)+cotx.cosx)

Question 10: Differentiate the functions w.r.t. x. xxcosx+x2+1x21

Answer:

Given function is
y=xsinx+(sinx)cosx

Now, take t=xsinx
Now, take a look at both sides.
logt=sinxlogx
Now, differentiate it w.r.t. x
we get,
1tdtdx=cosxlogx+1x.sinx

dtdx=t(cosxlogx+1x.sinx)

dtdx=xsinx(cosxlogx+1x.sinx)
Similarly, take k=(sinx)cosx
Now, take a look at both sides.
logk=cosxlog(sinx)
Now, differentiate it w.r.t. x
we get,
1kdkdt=(sinx)(log(sinx))+cosx.1sinx.cosx

=sinxlog(sinx)+cotx.cosxdkdt

=k(sinxlog(sinx)+cotx.cosx)

dkdt=(sinx)cosx(sinxlog(sinx)+cotx.cosx)
Now,
dydx=xsinx(cosxlogx+1x.sinx)+

(sinx)cosx(sinxlog(sinx)+cotx.cosx)
Therefore, the answer is xsinx(cosxlogx+1x.sinx)+

(sinx)cosx(sinxlog(sinx)+cotx.cosx)

Question 11: Differentiate the functions w.r.t. x. (xcosx)x+(xsinx)1/x

Answer:

Given function is
f(x)=(xcosx)x+(xsinx)1/x
Let's take t=(xcosx)x
Now, take a look at both sides.
logt=xlog(xcosx)=x(logx+logcosx)
Now, differentiate w.r.t. x
we get,
1tdtdx=(logx+logcosx)+x(1x+1cosx.(sinx))

dtdx=t(logx+logcosx+1xtanx)

         (sinxcosx=tanx)

dtdx=(xcosx)x(logx+logcosx+1xtanx)

dtdx=(xcosx)x(+1xtanx+log(xcosx))
Similarly, take k=(xsinx)1x
Now, take a look at both sides.
logk=1x(logx+logsinx)
Now, differentiate w.r.t. x
we get,
1kdkdx=(1x2)(logx+logsinx)+

1x(1x+1sinx.(cosx))

dkdx=kx2(logxlogsinx+1x2+cotxx)

         (cosxsinx=cotx)

dkdx=(xsinx)1xx2(logxlogsinx+1x2+cotxx)

dkdx=(xsinx)1x(xcotx+1(logxsinx))x2
Now,
dydx=dtdx+dkdx
dydx=(xcosx)x(+1xtanx+log(xcosx))+

(xsinx)1x(xcotx+1(logxsinx))x2
Therefore, the answer is (xcosx)x(1xtanx+log(xcosx))+

(xsinx)1x(xcotx+1(logxsinx))x2

Question 12Find dydx of the functions given in Exercises 12 to 15

xy+yx=1 .

Answer:

Given function is
f(x)=xy+yx=1
Now, take t=xy
Take logs on both sides.
logt=ylogx
Now, differentiate w.r.t x
we get,
1tdtdx=dydx(logx)+y1x=dydx(logx)+yx

dtdx=t(dydx(logx)+yx)

dtdx=(xy)(dydx(logx)+yx)
Similarly, take k=yx
Now, take a look at both sides.
logk=xlogy
Now, differentiate w.r.t. x
we get,
1kdkdx=(logy)+x1ydydx=logy+xydydx

dkdx=k(logy+xydydx)

dkdx=(yx)(logy+xydydx)
Now,
f(x)=dtdx+dkdx=0

(xy)(dydx(logx)+yx)+(yx)(logy+xydydx)=0

dydx(xy(logx)+xyx1)=(yxy1+yx(logy))

dydx=(yxy1+yx(logy))(xy(logx)+xyx1)

Therefore, the answer is (yxy1+yx(logy))(xy(logx)+xyx1)

Question 13 Find dydx of the functions given in Exercises 12 to 15.

yx=xy

Answer:

Given function is
f(x)xy=yx
Now, take t=xy
Take logs on both sides.
logt=ylogx
Now, differentiate w.r.t x
we get,
1tdtdx=dydx(logx)+y1x=dydx(logx)+yx

dtdx=t(dydx(logx)+yx)

dtdx=(xy)(dydx(logx)+yx)
Similarly, take k=yx
Now, take a look at both sides.
logk=xlogy
Now, differentiate w.r.t. x
we get,
1kdkdx=(logy)+x1ydydx=logy+xydydx

dkdx=k(logy+xydydx)

dkdx=(yx)(logy+xydydx)
Now,
f(x)dtdx=dkdx

(xy)(dydx(logx)+yx)=(yx)(logy+xydydx)

dydx(xy(logx)xyx1)=(yx(logy)yxy1)

dydx=yx(logy)yxy1(xy(logx)xyx1)

=xy(yxlogyxylogx)

Therefore, the answer is xy(yxlogyxylogx)

Question 14Find dydx of the functions given in Exercises 12 to 15. (cosx)y=(cosy)x

Answer:

Given function is
f(x)(cosx)y=(cosy)x
Now, take log on both sides.
ylogcosx=xlogcosy
Now, differentiate w.r.t x
dydx(logcosx)ytanx=logcosyxtanydydx
By taking similar terms on the same side
We get,
(dydx(logcosx)ytanx)

=(logcosyxtanydydx)dydx(logcosx+(cosy)x.xtany))

=(logcosy+(cosx)y.ytanx)dydx

=(logcosy+ytanx)(logcosx+xtany))

=ytanx+logcosyxtany+logcosx

Therefore, the answer is ytanx+logcosyxtany+logcosx

Question 15Find dydx of the functions given in Exercises 12 to 15. xy=exy

Answer:

Given function is
f(x)xy=exy
Now, take a look at both sides.
logx+ logy=(xy)(1)            (loge=1)logx+ logy=(xy)
Now, differentiate w.r.t x
1x+1ydydx=1dydx
By taking similar terms on the same side
We get,
(1y+1)dydx=11xy+1y.dydx=x1xdydx=yx.x1y+1
Therefore, the answer is yx.x1y+1

Question 16Find the derivative of the function given by f(x)=(1+x)(1+x2)(1+x4)(1+x8) and hence find

f ' (1)

Answer:

Given function is
y=(1+x)(1+x2)(1+x4)(1+x8)
Take logs on both sides.
logy=log(1+x)+log(1+x2)+log(1+x4)+log(1+x8)
NOW, differentiate w.r.t. x
1y.dydx=11+x+2x1+x2+4x31+x4+8x71+x8

dydx=y.(11+x+2x1+x2+4x31+x4+8x71+x8)

dydx=(1+x)(1+x2)(1+x4)(1+x8).(11+x+2x1+x2+4x31+x4+8x71+x8)
Therefore, f(x)=

(1+x)(1+x2)(1+x4)(1+x8).(11+x+2x1+x2+4x31+x4+8x71+x8)
Now, the value of f(1) is
f(1)=(1+1)(1+12)(1+14)(1+18).(11+1+2(1)1+12+4(1)31+14+8(1)71+18)

f(1)=16.152=120

Question 17 (1)Differentiate (x25x+8)(x3+7x+9) in three ways mentioned below:
(i) By using product rule

Answer:

Given function is
f(x)=(x25x+8)(x3+7x+9)
Now, we need to differentiate using the product rule.
f(x)=d((x25x+8))dx.(x3+7x+9)+(x25x+8).d((x3+7x+9))dx
=(2x5).(x3+7x+9)+(x25x+8)(3x2+7)

=2x4+14x2+18x5x335x45+3x415x3+24x2+7x235x+56

=5x420x3+45x252x+11
Therefore, the answer is 5x420x3+45x252x+11

Question 17 (2)Differentiate (x25x+8)(x3+7x+9) in three ways mentioned below:
(ii) by expanding the product to obtain a single polynomial.

Answer:

Given function is
f(x)=(x25x+8)(x3+7x+9)
Multiply both to obtain a single higher-degree polynomial.
f(x)=x2(x3+7x+9)5x(x3+7x+9)+8(x3+7x+9)
=x5+7x3+9x25x435x245x+8x3+56x+72
=x55x4+15x326x2+11x+72
Now, differentiate w.r.t. x
we get,
f(x)=5x420x3+45x252x+11
Therefore, the answer is 5x420x3+45x252x+11

Question 17 (3)Differentiate (x25x+8)(x3+7x+9) in three ways mentioned below:
(iii) by logarithmic differentiation.
Do they all give the same answer?

Answer:

Given function is
y=(x25x+8)(x3+7x+9)
Now, take a look at both sides.
logy=log(x25x+8)+log(x3+7x+9)
Now, differentiate w.r.t. x
we get,
1y.dydx=1x25x+8.(2x5)+1x3+7x+9.(3x2+7)

dydx=y.((2x5)(x3+7x+9)+(3x2+7)(x25x+8)(x25x+8)(x3+7x+9))

dydx=(x25x+8)(x3+7x+9).((2x5)(x3+7x+9)+(3x2+7)(x25x+8)(x25x+8)(x3+7x+9))

dydx=(2x5)(x3+7x+9)+(3x2+7)(x25x+8)dydx=5x420x3+45x256x+11
Therefore, the answer is 5x420x3+45x256x+11
And yes, they all give the same answer.

Question 18 If u, v and w are functions of x, then show that ddx(u,v,w)=dudxv.w+u.dvdxv.w+u.dvdx.w+u.vdwdx 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,
dydx=dudx.(v.w)+d(v.w)dx.u -(i)
Now, again, by the product rule
d(v.w)dx=dvdx.w+dwdx.v
Put this in equation (i)
we get,
dydx=dudx.(v.w)+dvdx.(u.w)+dwdx.(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.
logy=logu+logv+logw
Now, differentiate w.r.t. x
we get,
1y.dydx=1u.dudx+1vdvdx+1w.dwdxdydx=y.(v.w.dudx+u.w.dvdx+u.v.dwdxu.v.w)dydx

=(u.v.w)(v.w.dudx+u.w.dvdx+u.v.dwdxu.v.w)
dydx=dudx.(v.w)+dvdx.(u.w)+dwdx.(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 dydx.

x=2at2,y=at4

Answer:

Given equations are
x=2at2,y=at4
Now, differentiate both w.r.t t
We get,
dxdt=d(2at2)dt=4at
Similarly,
dydt=d(at4)dt=4at3
Now, dydx=dydtdxdt=4at34at=t2
Therefore, the answer is dydx=t2

Question 2: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find dydx.

Answer:

Given equations are
x=acosθ,y=bcosθ
Now, differentiate both w.r.t θ
We get,
dxdθ=d(acosθ)dθ=asinθ
Similarly,
dydθ=d(bcosθ)dθ=bsinθ
Now, dydx=dydθdxdθ=bsinθasinθ=ba
Therefore, answer is dydx=ba

Question 3: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find dydx. x=sint,y=cos2t

Answer:

Given equations are
x=sint,y=cos2t
Now, differentiate both w.r.t t
We get,
dxdt=d(sint)dt=cost
Similarly,
dydt=d(cos2t)dt=2sin2t=4sintcost

     (sin2x=sinxcosx)
Now, dydx=dydtdxdt=4sintcostcost=4sint
Therefore, the answer is dydx=4sint

Question 4If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find  dydx

x=4t,y=4/t

Answer:

Given equations are
x=4t,y=4/t
Now, differentiate both w.r.t t
We get,
dxdt=d(4t)dt=4
Similarly,
dydt=d(4t)dt=4t2
Now, dydx=dydtdxdt=4t24=1t2
Therefore, the answer is dydx=1t2

Question 5: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find dydx x=cosθcos2θ,y=sinθsin2θ

Answer:

Given equations are
x=cosθcos2θ,y=sinθsin2θ
Now, differentiate both w.r.t θ
We get,
dxdθ=d(cosθcos2θ)dθ=sinθ(2sin2θ)=2sin2θsinθ
Similarly,
dydθ=d(sinθsin2θ)dθ=cosθ2cos2θ
Now, dydx=dydθdxdθ=cosθ2cos2θ2sin2θsinθ
Therefore, answer is dydx=cosθ2cos2θ2sin2θsinθ

Question 6If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dydx x=a(θsinθ),y=a(1+cosθ)

Answer:

Given equations are
x=a(θsinθ),y=a(1+cosθ)
Now, differentiate both w.r.t θ
We get,
dxdθ=d(a(θsinθ))dθ=a(1cosθ)
Similarly,
dydθ=d(a(1+cosθ))dθ=asinθ
Now, dydx=dydθdxdθ=asinθa(1cosθ)=sin1cosθ=cotθ2       (cotx2=sinx1cosx)
Therefore, the answer is dydx=cotθ2

Question 7If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dydx x=sin3tcos2t,y=cos3tcos2t

Answer:

Given equations are
x=sin3tcos2t,y=cos3tcos2t
Now, differentiate both w.r.t
We get,
dxdt=d(sin3tcos2t)dt

=cos2t.d(sin3t)dtsin3t.d(cos2t)dt(cos2t)2

=3sin2tcost.cos2tsin3t.12cos2t.(2sin2t)cos2t
=3sin2tcost.cos2t+sin3tsin2tcos2tcos2t
=sin3tsin2t(3cottcot2t+1)cos2tcos2t     (cossinx=cotx)
Similarly,
dydt=d(cos3tcos2t)dt

=cos2t.d(cos3t)dtcos3t.d(cos2t)dt(cos2t)2

=3cos2t(sint).cos2tcos3t.12cos2t.(2sin2t)(cos2t)2
=3cos2tsintcos2t+cos3tsin2tcos2tcos2t
=sin2tcos3t(13tantcot2t)cos2tcos2t
Now, dydx=dydtdxdt

=sin2tcos3t(13tantcot2t)cos2tcos2tsin3tsin2t(3cottcot2t+1)cos2tcos2t

=cot3t(13tantcot2t)(3cottcot2t+1)
=cos3t(13.sintcost.cos2tsin2t)sin3t(3.costsint.cos2tsin2t+1)

=cos2t(costsin2t3sintcos2t)sin2t(3costcos2t+sintsin2t)
=cos2t(cost.2sintcost3sint(2cos2t1))sin2t(3cost(12sin22t)+sint.2sintcost)
(sin2x=2sinxcosx and cos2x=2cos2x1 and cos2x=12sin2x)
=cos2t(2sintcos2t6sintcos2t+3sint)sin2t(3cost6costsin2t+2sin2cost)=sintcost(4cos3t+3cost)sintcost(3sint4sin3t)

dydx=4cos3t+3cost3sint4sin3t=cos3tsin3t=cot3t (sin3t=3sint4sin3t and cos3t=4cos3t3cost)

Therefore, the answer is dydx=cot3t

Question 8: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find dydx x=a(cost+logtant/2),y=asint

Answer:

Given equations are
x=a(cost+logtant2),y=asint
Now, differentiate both w.r.t t
We get,
dxdt=d(a(cost+logtant2))dt=a(sint+1tant2.sec2t2.12)
=a(sint+12.cost2sint2.1cos2t2)=a(sint+12sint2cost2)
=a(sint+1sin2.t2)=a(sin2t+1sint)=a(cos2tsint)
Similarly,
dydt=d(asint)dt=acost
Now, dydx=dydtdxdt=acosta(cos2tsint)=sintcost=tant
Therefore, the answer is dydx=tant

Question 9: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find dydx x=asecθ,y=b tanθ

Answer:

Given equations are
x=asecθ,y=b tanθ
Now, differentiate both w.r.t θ
We get,
dxdθ=d(asecθ)dθ=asecθtanθ
Similarly,
dydθ=d(btanθ)dθ=bsec2θ
Now, dydx=dydθdxdθ=bsec2θasecθtanθ=bsecθatanθ=b1cosθasinθcosθ=basinθ=bcosecθa
Therefore, the answer is dydx=bcosecθa

Question 10If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dydx x=a(cosθ+θsinθ),y=a(sinθθcosθ)

Answer:

Given equations are
x=a(cosθ+θsinθ),y=a(sinθθcosθ)
Now, differentiate both w.r.t θ
We get,
dxdθ=d(a(cosθ+θsinθ))dθ=a(sinθ+sinθ+θcosθ)=aθcosθ
Similarly,
dydθ=d(a(sinθθcosθ))dθ=a(cosθcosθ+θsinθ)=aθsinθ
Now, dydx=dydθdxdθ=aθsinθaθcosθ=tanθ
Therefore, the answer is dydx=tanθ

Question 11If x=asin1t,y=acos1t , show that dydx = - y /x$

Answer:

Given equations are
x=asin1t,y=acos1t

xy=asin1t+cos1tsincesin1x+cos1x=π2xy=aπ2=constant=c

Differentiating with respect to x

xdydx+y=0dydx=yx

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.

x2+3x+2

Answer:

Given function is
y=x2+3x+2
Now, differentiation w.r.t. x
dydx=2x+3
Now, second-order derivative
d2ydx2=2
Therefore, the second order derivative is d2ydx2=2

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

x20

Answer:

Given function is
y=x20
Now, differentiation w.r.t. x
dydx=20x19
Now, the second-order derivative is
d2ydx2=20.19x18=380x18
Therefore, second-order derivative is d2ydx2=380x18

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

xcosx

Answer:

Given function is
y=xcosx
Now, differentiation w.r.t. x
dydx=cosx+x(sinx)=cosxxsinx
Now, the second-order derivative is
d2ydx2=sinx(sinx+xcosx)=2sinxxsinx
Therefore, the second-order derivative is d2ydx2=2sinxxsinx

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

logx

Answer:

Given function is
y=logx
Now, differentiation w.r.t. x
dydx=1x
Now, the second order derivative is
d2ydx2=1x2
Therefore, second order derivative is d2ydx2=1x2

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

x3logx

Answer:

Given function is
y=x3logx
Now, differentiation w.r.t. x
dydx=3x2.logx+x3.1x=3x2.logx+x2
Now, the second-order derivative is
d2ydx2=6x.logx+3x2.1x+2x=6x.logx+3x+2x=x(6.logx+5)
Therefore, the second-order derivative is d2ydx2=x(6.logx+5)

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

exsin5x

Answer:

Given function is
y=exsin5x
Now, differentiation w.r.t. x
dydx=ex.sin5x+ex.5cos5x=ex(sin5x+5cos5x)
Now, the second order derivative is
d2ydx2=ex(sin5x+5cos5x)+ex(5cos5x+5.(5sin5x))
=ex(sin5x+5cos5x)+ex(5cos5x25sin5x)=ex(10cos5x24sin5x)
=2ex(5cos5x12sin5x)
Therefore, second order derivative is dydx=2ex(5cos5x12sin5x)

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

e6xcos3x

Answer:

Given function is
y=e6xcos3x
Now, differentiation w.r.t. x
dydx=6e6x.cos3x+e6x.(3sin3x)=e6x(6cos3x3sin3x)
Now, the second order derivative is
d2ydx2=6e6x(6cos3x3sin3x)+e6x(6.(3sin3x)3.3cos3x)
=6e6x(6cos3x3sin3x)e6x(18sin3x+9cos3x)
e6x(27cos3x36sin3x)=9e6x(3cos3x4sin3x)
Therefore, second order derivative is dydx=9e6x(3cos3x4sin3x)

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

tan1x

Answer:

Given function is
y=tan1x
Now, differentiation w.r.t. x
dydx=d(tan1x)dx=11+x2
Now, the second order derivative is
d2ydx2=1(1+x2)2.2x=2x(1+x2)2
Therefore, second order derivative is d2ydx2=2x(1+x2)2

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

log(logx)

Answer:

Given function is
y=log(logx)
Now, differentiation w.r.t. x
dydx=d(log(logx))dx=1logx.1x=1xlogx
Now, the second order derivative is
d2ydx2=1(xlogx)2.(1.logx+x.1x)=(logx+1)(xlogx)2
Therefore, second order derivative is d2ydx2=(logx+1)(xlogx)2

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

sin(logx)

Answer:

Given function is
y=sin(logx)
Now, differentiation w.r.t. x
dydx=d(sin(logx))dx=cos(logx).1x=cos(logx)x
Now, the second order derivative is
Using Quotient rule
d2ydx2=sin(logx)1x.xcos(logx).1x2=(sin(logx)+cos(logx))x2
Therefore, second order derivative is d2ydx2=(sin(logx)+cos(logx))x2

Question 11If y=5cosx3sinx prove that d2ydx2+y=0

Answer:

Given function is
y=5cosx3sinx
Now, differentiation w.r.t. x
dydx=d(5cosx3sinx)dx

=5sinx3cosx
Now, the second-order derivative is
d2ydx2=d2(5sinx3cosx)dx2

=5cosx+3sinx
Now,
d2ydx2+y=5cosx+3sinx+5cosx3sinx=0
Hence proved

Question 12If y=cos1x Find d2ydx2 in terms of y alone.

Answer:

Given function is
y=cos1x
Now, differentiation w.r.t. x
dydx=d(cos1x)dx=11x2
Now, the second order derivative is
d2ydx2=d2(11x2)dx2=(1)(1x2)2.(2x)=2x1x2 -(i)
Now, we want d2ydx2 in terms of y
y=cos1x
x=cosy
Now, put the value of x in (i)
d2ydx2=2cosy1cos2y=2cosysin2y=2cotycosecy
(1cos2x=sin2x and cosxsinx=cotx and 1sinx=cosecx)
Therefore, answer is d2ydx2=2cotycosecy

Question 13 If y=3cos(logx)+4sin(logx) , show that x2y2+xy1+y=0

Answer:

Given function is
y=3cos(logx)+4sin(logx)
Now, differentiation w.r.t. x
y1=dydx=d(3cos(logx)+4sin(logx))dx=3sin(logx).1x+4cos(logx).1x
=4cos(logx)3sin(logx)x -(i)
Now, the second order derivative is
By using the Quotient rule
y2=d2ydx2=d2(4cos(logx)3sin(logx)x)dx2=(4sin(logx).1x3cos(logx).1x).x1.(4cos(logx)3sin(logx))x2
=sin(logx)+7cos(logx)x2 -(ii)
Now, from equation (i) and (ii), we will get y1 and y2
Now, we need to show.
x2y2+xy1+y=0
Put the value of y1 and y2 from equation (i) and (ii)
x2(sin(logx)+7cos(logx)x2)+x(4cos(logx)3sin(logx)x)+3cos(logx) +4sin(logx)
sin(logx)7cos(logx)+4cos(logx)3sin(logx)+3cos(logx) +4sin(logx)
=0
Hence proved

Question 14If y=Aemx+Benx , show that d2ydx2(m+n)dydx+many=0

Answer:

Given function is
y=Aemx+Benx
Now, differentiation w.r.t. x
dydx=d(Aemx+Benx)dx=mAemx+nBenx -(i)
Now, the second order derivative is
d2ydx2=d2(mAemx+nBenx)dx2=m2Aemx+n2Benx -(ii)
Now, we need to show.
d2ydx2(m+n)dydx+mny=0
Put the value of d2ydx2 and dydx from equation (i) and (ii)
m2Aemx+n2Benx(m+n)(mAemx+nBxnx)+mn(Aemx+Benx)
m2Aemx+n2Benxm2AemxmnBxnxmnAemxn2Benx+mnAemx +mnBenx
=0
Hence proved

Question 15If y=500e7x+600e7x , show that d2ydx2=49y
Answer:

Given function is
y=500e7x+600e7x
Now, differentiation w.r.t. x
dydx=d(500e7x+600e7x)dx=7.500e7x7.600e7x=3500e7x4200e7x -(i)
Now, the second order derivative is
d2ydx2=d2(3500e7x4200e7x)dx2
=7.3500e7x(7).4200e7x=24500e7x+29400e7x -(ii)
Now, we need to show.
d2ydx2=49y
Put the value of d2ydx2 from equation (ii)
24500e7x+29400e7x=49(500e7x+600e7x)
=24500e7x+29400e7x
Hence, L.H.S. = R.H.S.
Hence proved

Question 16If ey(x+1)=1 show that d2ydx2=(dydx)2

Answer:

Given function is
ey(x+1)=1
We can rewrite it as
ey=1x+1
Now, differentiation w.r.t. x
d(ey)dx=d(1x+1)dxey.dydx=1(x+1)21x+1.dydx=1(x+1)2         (ey=1x+1)dydx=1x+1 -(i)
Now, the second order derivative is
d2ydx2=d2(1x+1)dx2=(1)(x+1)2=1(x+1)2 -(ii)
Now, we need to show.
d2ydx2=(dydx)2
Put value of d2ydx2 and dydx from equation (i) and (ii)
1(x+1)2=(1x+1)2
=1(x+1)2
Hence, L.H.S. = R.H.S.
Hence proved

Question 17If y=(tan1x)2 show that (x2+1)2y2+2x(x2+1)y1=2

Answer:

Given function is
y=(tan1x)2
Now, differentiation w.r.t. x
y1=dydx=d((tan1x)2)dx=2.tan1x.11+x2=2tan1x1+x2 -(i)
Now, the second-order derivative is
By using the quotient rule
y2=d2ydx2=d2(2tan1x1+x2)dx2=2.11+x2.(1+x2)2tan1x(2x)(1+x2)2=24xtan1x(1+x2)2 -(ii)
Now, we need to show.
(x2+1)2y2+2x(x2+1)y1=2
Put the value from equation (i) and (ii)
(x2+1)2.24xtan1x(1+x2)2+2x(x2+1).2tan1xx2+124xtan1x+4xtan1x=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.

(3x29x+5)9

Answer:

Given function is
f(x)=(3x29x+5)9
Now, differentiation w.r.t. x is
f(x)=d(f(x))dx=d((3x29x+5)9)dx=9(3x29x+5)8.(6x9)
=27(2x3)(3x29x+5)8
Therefore, differentiation w.r.t. x is 27(3x29x+5)8(2x3)

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

sin3x+cos6x

Answer:

Given function is
f(x)=sin3x+cos6x
Now, differentiation w.r.t. x is
f(x)=d(f(x))dx=d(sin3x+cos6x)dx=3sin2x.d(sinx)dx+6cos5x.d(cosx)dx
=3sin2x.cosx+6cos5x.(sinx)
=3sin2xcosx6cos5xsinx=3sinxcosx(sinx2cos4x)

Therefore, differentiation w.r.t. x is 3sinxcosx(sinx2cos4x)

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

(5x)3cos2x

Answer:

Given function is
y=(5x)3cos2x
Take a log on both sides.
logy=3cos2xlog5x
Now, differentiation w.r.t. x is
By using the product rule
1y.dydx=3.(2sin2x)log5x+3cos2x.15x.5

=6sin2xlog5x+3cos2xxdydx

=y.(6sin2xlog5x+3cos2xx)

dydx=(5x)3cos2x.(6sin2xlog5x+3cos2xx)

Therefore, differentiation w.r.t. x is (5x)3cos2x.(3cos2xx6sin2xlog5x)

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

sin1(xx),0x1

Answer:

Given function is
f(x)=sin1(xx),0x1
Now, differentiation w.r.t. x is
f(x)=d(f(x))dx=d(sin1xx)dx=11(xx)2.d(xx)dx
=11x3.(1.x+x12x)
=11x3.(3x2)
=32.x1x3

Therefore, differentiation w.r.t. x is 32.x1x3

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

cos1x/22x+7,2<x<2

Answer:

Given function is
f(x)=cos1x/22x+7,2<x<2
Now, differentiation w.r.t. x is
By using the Quotient rule
f(x)=d(f(x))dx=d(cos1x22x+7)dx

=d(cos1x2)dx.2x+7cos1x2.d(2x+7)dx(2x+7)2

f(x)=11(x2)2.12.2x+7cos1x2.12.2x+7.22x+7

f(x)=[1(4x2)(2x+7)+cos1x2(2x+7)32]
Therefore, differentiation w.r.t. x is [1(4x2)(2x+7)+cos1x2(2x+7)32]

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

cot1[1+sinx+1sinx1+sinx1sinx],0<x<π/2

Answer:

Given function is
f(x)=cot1[1+sinx+1sinx1+sinx1sinx],0<x<π/2
Now, rationalize the part.
[1+sinx+1sinx1+sinx1sinx]

=[1+sinx+1sinx1+sinx1sinx.1+sinx+1sinx1+sinx+1sinx]
=(1+sinx+1sinx)2(1+sinx)2(1sinx)2

      (Using (ab)(a+b)=a2b2)
=((1+sinx)2+(1sinx)2+2(1+sinx)(1sinx))1+sinx1+sinx
(Using (a+b)2=a2+b2+2ab)
=1+sinx+1sinx+21sin2x2sinx
=2(1+cosx)2sinx=1+cosxsinx
=2cos2x22sinx2cosx2     (2cos2=1+cos2x and sin2x=2sinxcosx)
=2cosx22sinx2=cotx2
Given function reduces to
f(x)=cot1(cotx2)f(x)=x2
Now, differentiation w.r.t. x is
f(x)=d(f(x))dx

=d(x2)dx=12
Therefore, differentiation w.r.t. x is 12

Question 7Differentiate w.r.t. x the function in Exercises 1 to 11. (logx)logx,x>1

Answer:

Given function is
y=(logx)logx,x>1
Take logs on both sides.
logy=logxlog(logx)
Now, differentiate w.r.t.
1y.dydx=1x.log(logx)+logx.1logx.1x=logx+1x
dydx=y.(logx+1x)
dydx=(logx)logx.(logx+1x)
Therefore, differentiation w.r.t x is (logx)logx.(logx+1x)

Question 8 cos(acosx+bsinx), for some constant a and b.

Answer:

Given function is
f(x)=cos(acosx+bsinx)
Now, differentiation w.r.t x
f(x)=d(f(x))dx=d(cos(acosx+bsinx))dx
=sin(acosx+bsinx).d(acosx+bsinx)dx
=sin(acosx+bsinx).(asinx+bcosx)
=(asinxbcosx)sin(acosx+bsinx).
Therefore, differentiation w.r.t x (asinxbcosx)sin(acosx+bsinx)

Question 9 (sinxcosx)(sinxcosx),,π4<x<3π4

Answer:

Given function is
y=(sinxcosx)(sinxcosx),,π4<x<3π4
Take logs on both sides.
logy=(sinxcosx)log(sinxcosx)
Now, differentiate w.r.t. x
1y.dydx=d(sinxcosx)dx.log(sinxcosx)+(sinxcosx).d(log(sinxcosx))dx
1y.dydx=(cosx(sinx)).log(sinxcosx)+(sinxcosx).(cosx(sinx))(sinxcosx)
dydx=y.(cosx+sinx)(log(sinxcosx)+1)
dydx=(sinxcosx)(sinxcosx).(cosx+sinx)(log(sinxcosx)+1)
Therefore, differentiation w.r.t x is (sinxcosx)(sinxcosx).(cosx+sinx)(log(sinxcosx)+1),sinx>cosx

Question 10xx+xa+ax+aa , for some fixed a > 0 and x > 0

Answer:

Given function is
f(x)=xx+xa+ax+aa
Let's take
u=xx
Now, take a look at both sides.
logu=xlogx
Now, differentiate w.r.t x
1u.dudx=dxdx.logx+x.d(logx)dx1u.dudx=1.logx+x.1xdudx=y.(logx+1)dudx=xx.(logx+1) -(i)
Similarly, take v=xa
Take logs on both sides.
logv=alogx
Now, differentiate w.r.t x
1v.dvdx=a.d(logx)dx=a.1x=axdvdx=v.axdvdx=xa.ax -(ii)

Similarly, take z=ax
Take logs on both sides.
logz=xloga
Now, differentiate w.r.t x
1z.dzdx=loga.d(x)dx=loga.1=logadzdx=z.logadzdx=ax.loga -(iii)

Similarly, take w=aa
Take logs on both sides.
logw=aloga= constant
Now, differentiate w.r.t x
1w.dwdx=a.d(aloga)dx=0dwdx=0 -(iv)
Now,
f(x)=u+v+z+w
f(x)=dudx+dvdx+dzdx+dwdx
Put values from equation (i), (ii),(iii) and (iv)
f(x)=xx(logx+1)+axa1+axloga
Therefore, differentiation w.r.t. x is xx(logx+1)+axa1+axloga

Question 11xx23+(x3)x2,forx>3

Answer:

Given function is
f(x)=xx23+(x3)x2,forx>3
take u=xx23
Now, take a look at both sides.
logu=(x23)logx
Now, differentiate w.r.t x
1u.dudx=d(x23)dx.logx+(x23).d(logx)dx

1u.dudx=2x.logx+(x23).1x

1u.dudx=2x2logx+x23x

dudx=u.(2x2logx+x23x)

dudx=x(x23).(2x2logx+x23x) -(i)
Similarly,
take v=(x3)x2
Now, take a look at both sides.
logv=x2log(x3)
Now, differentiate w.r.t x
1v.dvdx=d(x2)dx.log(x3)+x2.d(log(x3))dx

1v.dvdx=2x.log(x3)+x2.1(x3)

1v.dvdx=2xlog(x3)+x2x3

dvdx=v.(2xlog(x3)+x2x3)

dvdx=(x3)x2.(2xlog(x3)+x2x3) -(ii)
Now
f(x)=u+v
f(x)=dudx+dvdx
Put the value from equation (i) and (ii)
f(x)=x(x23).(2x2logx+x23x)+(x3)x2.(2xlog(x3)+x2x3)
Therefore, differentiation w.r.t x is x(x23).(2x2logx+x23x)+(x3)x2.(2xlog(x3)+x2x3)

Question 12Find dydx if y=12(1cost),x=10(tsint), π2<t<π2

Answer:

Given equations are
y=12(1cost),x=10(tsint),
Now, differentiate both y and x w.r.t t independently.
dydt=d(12(1cost))dt=12(sint)=12sint
And
dxdt=d(10(tsint))dt=1010cost
Now
dydx=dydtdxdt=12sint10(1cost)=65.2sint2cost22sin2t2=65.cost2sint2
(sin2x=2sinxcosx and 1cos2x=2sin2x)
dydx=65.cott2
Therefore, differentiation w.r.t x is 65.cott2

Question 13Find dydx if y=sin1x+sin11x2,0<x<1

Answer:

Given function is
y=sin1x+sin11x2,0<x<1
Now, differentiate w.r.t. x
dydx=d(sin1x+sin11x2)dx=11x2+11(1x2)2.d(1x2)dx

dydx=11x2+111+x2.121x2.(2x)

dydx=11x211x2

dydx=0
Therefore, differentiate w.r.t. x is 0

Question 14If x1+y+y1+x=0for,1<x<1provethatdydx=1(1+x)2

Answer:

Given function is
x1+y+y1+x=0
x1+y=y1+x
Now, squaring both sides.
(x1+y)2=(y1+x)2x2(1+y)=y2(1+x)

x2+x2y=y2x+y2

x2y2=y2xx2y(xy)(x+y)=xy(xy)

x+y=xyy=x1+x
Now, differentiate w.r.t. x is
dydx=d(x1+x)dx=1.(1+x)(x).(1)(1+x)2=1(1+x)2
Hence proved

Question 15If (xa)2+(yb)2=c2 , for some c > 0, prove that [1+(dydx)2]3/2d2ydx2 is a constant independent of a and b.

Answer:

Given function is
(xa)2+(yb)2=c2
(yb)2=c2(xa)2 - (i)
Now, differentiate w.r.t. x
d((xa)2)dx+((yb)2)dx=d(c2)dx2(xa)+2(yb).dydx=0dydx=axyb -(ii)
Now, the second derivative
d2ydx2=d(ax)dx.(yb)(ax).d(yb)dx(yb)2d2ydx2=(1).(yb)(ax).dydx(yb)2
Now, put values from equation (i) and (ii)
d2ydx2=(yb)(ax).axyb(yb)2d2ydx2=((yb)2+(ax)2)(yb)32=c2(yb)32 ((xa)2+(yb)2=c2)
Now,
[1+(dydx)2]3/2d2ydx2=(1+(xayb)2)32c2(yb)32=((yb)2+(xa)2)32(yb)32c2(yb)32=(c2)32c2=c3c2=c ((xa)2+(yb)2=c2)
Which is independent of a and b
Hence proved

Question 16If cosy=xcos(a+y) , with cosa±1 , prove that dydx=cos2(a+y)sina

Answer:

Given function is
cosy=xcos(a+y)
Now, differentiate w.r.t x
d(cosy)dx=dxdx.cos(a+y)+x.d(cos(a+y))dx

sinydydx=1.cos(a+y)+x.(sin(a+y)).dydx

dydx.(xsin(a+y)siny)=cos(a+y)

dydx.(cosycos(a+b).sin(a+y)siny)=cos(a+b)

     (x=cosycos(a+b))
dydx.(cosysin(a+y)sinycos(a+y))=cos2(a+b)

dydx.(sin(a+yy))=cos2(a+b)

       (cosAsinBsinAcosB=sin(AB))

dydx=cos2(a+y)sina
Hence proved

Question 17If x=a(cost+tsint) and y=a(sinttcost), find d2ydx2

Answer:

Given functions are
x=a(cost+tsint) and y=a(sinttcost)
Now, differentiate both the functions w.r.t. t independently.
We get
dxdt=d(a(cost+tsint))dt=a(sint)+a(sint+tcost)
=asint+asint+atcost=atcost
Similarly,
dydt=d(a(sinttcost))dt=acosta(cost+t(sint))
=acostacost+atsint=atsint
Now,
dydx=dydtdxdt=atsintatcost=tant
Now, the second derivative
d2ydx2=ddxdydx=sec2t.dtdx=sec2t.sectat=sec3tat
(dxdt=atcostdtdx=1atcost=sectat)
Therefore, d2ydx2=sec3tat

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){x3x<0x3x>0
Now, differentiate in both cases.
f(x)=x3f(x)=3x2f(x)=6x
And
f(x)=x3f(x)=3x2f(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){6xx<06xx>0

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

Answer:

Given function is
sin(A+B)=sinAcosB+cosAsinB
Now, differentiate w.r.t. x
d(sin(A+B))dx=dsinAdx.cosB+sinA.dcosBdx+dcosAdx.sinB+cosA.dsinBdx
cos(A+b)d(A+B)dx =dAdx(cosAcosBsinAcosB)+dBdx(cosAsinBsinAsinB)
=(cosAsinBsinAsinB).d(A+B)dx
cos(A+B)=cosAsinBsinAcosB
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
limh0f(x+h)f(x)h=limh0f(h)f(0)h=limh0|h|+|h+1||1|h
=limh0h(h+1)1h=0 (|h|=h because h<0)
R.H.L. at x = 0
limh0+f(x+h)f(x)h=limh0+f(h)f(0)h=limh0+|h|+|h+1||1|h
=limh0+h+h+11h=limh0+2hh=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
limh0+f(x+h)f(x)h=limh0+f(1+h)f(1)h=limh0+|1+h|+|h||1|h
=limh0+1h+h1h=limh0+0h=0 (|h|=h because h>0)
L.H.L. at x = -1
limh0f(x+h)f(x)h=limh0f(1+h)f(1)h=limh0|1+h|+|h||1|h
=limh1+1hh1h=limh0+2hh=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=|f(x)g(x)h(x)lmnabc| , prove that dydx = |f(x)g(x)h(x)lmnabc|$

Answer:

Given that
y=|f(x)g(x)h(x)lmnabc|
We can rewrite it as
y=f(x)(mcbn)g(x)(lcan)+h(x)(lbam)
Now, differentiate w.r.t x
We will get
dydx=f(x)(mcbn)g(x)(lcan)+h(x)(lbam)[f(x)g(x)h(x)lmnabc]
Hence proved

Question 22If y=eacos1x,11, show that (1x2)d2ydx2xdydxa2y=0

Answer:

Given function is
y=eacos1x,11

Now, differentiate w.r.t x we will get.
dydx=d(eacos1x)dx.d(acos1x)dx=eacos1x.a1x2 -(i)
Now, again differentiate w.r.t x
d2ydx2=ddxdydx=aeacos1x.a1x2.1x2+aeacos1x.1.(2x)21x2(1x2)2
= a2eacos1xaxeacos1x1x21x2 -(ii)
Now, we need to show that.
(1x2)d2ydx2xdydxa2y=0
Put the values from equation (i) and (ii)
(1x2).( a2eacos1xaxeacos1x1x21x2)x.(aeacos1x1x2)a2eacos1x
a2eacos1xaxeacos1x1x2+(axeacos1x1x2)a2eacos1x=0
Hence proved

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

Also read,

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

NCERT solutions for class 12 maths - Chapter-wise

NCERT solutions for class 12 subject-wise

For subject-wise solutions, you can refer here

NCERT Solutions class-wise

For the solution of other classes, you can refer here

NCERT Books and NCERT Syllabus

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

Importance of solving NCERT Questions and Solutions for 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.

Frequently Asked Questions (FAQs)

1. Is every differentiable function always continuous?

Yes, every differentiable function is always continuous, but the converse is not true; a function can be continuous without being differentiable. For example, the absolute value function, f(x)=|x|, is continuous everywhere but not differentiable at x=0.

2. What are the applications of differentiation covered in Chapter 5?

In Chapter 5, applications of differentiation focus on understanding and using derivatives to analyse functions, including finding rates of change, determining increasing/decreasing intervals, locating extrema, and sketching curves.

3. What is the difference between continuity and differentiability?

Continuity refers to a function having no breaks or gaps in its graph, while differentiability means the function has a defined slope (derivative) at every point in its domain.

4. What are the basic formulas used in Class 12 Maths Chapter 5?

Formulas include the power rule, product rule, quotient rule, and chain rule for differentiation.

5. Where can I find NCERT Solutions for Class 12 Maths Chapter 5 PDF?

You can download the book and solutions from the Careers360 site for free.

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Questions related to CBSE Class 12th

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Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

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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.

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

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