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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 Ramraj Saini | Updated on Sep 14, 2023 07:52 PM IST | #CBSE Class 12th
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NCERT Continuity And Differentiability Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 5 are provided here. These NCERT solutions are created bu expert team at careers360 considering the latest syllabus of CBSE 2023-24. Questions based on the topics like continuity, differentiability, and relations between them are covered in the NCERT solutions for class 12 maths chapter 5. In NCERT Class 12 maths book, there are 48 solved examples to understand the concepts of continuity and differentiability class 12. If you are finding difficulties in solving them, you can take help from NCERT maths chapter 5 class 12 solutions.

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  1. NCERT Continuity And Differentiability Class 12 Questions And Answers
  2. NCERT Continuity And Differentiability Class 12 Questions And Answers PDF Free Download
  3. Continuity And Differentiability Class 12 NCERT Solutions - Important Formulae
  4. NCERT Continuity And Differentiability Class 12 Questions And Answers (Intext Questions and Exercise)
  5. Topics of NCERT class 12 maths chapter 5 Continuity and Differentiability
  6. NCERT class 12 maths ch 5 question answer - Topics
  7. NCERT solutions for class 12 maths - Chapter wise
  8. NCERT solutions for class 12 subject wise
  9. NCERT Solutions class wise
  10. NCERT Books and NCERT Syllabus
  11. Tips to Use NCERT Solutions for Class 12 Maths Chapter 5
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

In NCERT class 11 Maths solutions, you have already learned the differentiation of certain functions like polynomial functions and trigonometric functions. In this chapter, you will get NCERT solutions for class 12 maths chapter 5 continuity and differentiability. If you are interested in the chapter 5 class 12 maths NCERT solutions then you can check NCERT solutions for class 12 other subjects.

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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).

  • lim(x → a) f(x) exists.

  • lim(x → a) f(x) = f(a).

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

Algebra of Continuous Functions:

Sum, difference, product, and quotient of continuous functions are continuous.

Differentiation:

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

Chain Rule:

If f = v o u, where t = u(x), and if both dt/dx and dv/dx exist, then: df/dx = dv/dt * dt/dx.

Derivatives of Some Standard Functions:

  • d/dx(xn) = nxn-1

  • d/dx(sin x) = cos x

  • d/dx(cos x) = -sin x

  • d/dx(tan x) = sec2 x

  • d/dx(cot x) = -csc2 x

  • d/dx(sec x) = sec x * tan x

  • d/dx(csc x) = -csc x * cot x

  • d/dx(ax) = ax * ln(a)

  • d/dx(ex) = ex

  • d/dx(ln x) = 1/x

Mean Value Theorem:

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)) / (b - a).

Rolle's Theorem:

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

Lagrange's Mean Value Theorem:

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

f'(c) = (f(b) - f(a)) / (b - a).

Free download Continuity And Differentiability Class 12 NCERT Solutions for CBSE Exam.

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

NCERT Continuity And Differentiability Class 12 Solutions : Excercise: 5.1

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, function is continous at x = 0

f(3)=5(3)3=153=18limx3f(x)=5(3)3=153=18limx3f(x)=f(3)
Hence, function is continous at x = -3

f(5)=5(5)3=253=22limx5f(x)=5(5)3=253=22limx5f(x)=f(5)
Hence, 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=17limx3f(x)=2(3)21=2×91=181=17
limx3f(x)=f(3)
Hence, function is continous at x = 3

Question:3 Examine 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)=k5limxkf(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)=1k5limxkf(x)=1k5limxkf(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 gwt,
f(k)=k252k+5=(k+5)(k5)k+5=k5limxkf(x)=k252k+5=(k+5)(k5)k+5=k5limxkf(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) = 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)=k5limxkf(x)=k5limxkf(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)=5klimxkf(x)=5klimxkf(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)=55=0limx5f(x)=55=0limx5f(x)=f(5)
Hence, function f(x) = x - 5 is continous for x = 5

Hence, the function f(x)=|x5| is continuous for each and 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)=nnlimxnf(x)=nnlimxnf(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
function is defined at x = 0 and its value is 0
f(0)=0limx0f(x)=f(x)=0limx0f(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)=1limx1f(x)=f(x)=1limx1+f(x)=f(5)=5
R.H.L 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)=2limx2f(x)=f(5)=5limx2f(x)=f(2)
Hence, given function is continous 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)=2k3limxkf(x)=f(k)
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
f(k)=2k+3limxkf(x)=2k+3limxkf(x)=f(k)
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

limx2f(x)=2x+3=2×2+3=4+3=7limx2+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
Different cases are their
case (i) k < -3
f(k)=k+3limxkf(x)=k+3limxkf(x)=f(k)
Hence, given function is continuous for every value of k < -3

case(ii) k = -3
f(3)=(3)+3=6limx3f(x)=k+3=(3)+3=6limx3+f(x)=2x=2(3)=6R.H.L.=L.H.L.=f(3)
Hence, given function is continous for x = -3

case(iii) -3 < k < 3
f(k)=2klimxkf(x)=2klimxkf(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=6×3+2=18+2=20limx3f(x)=2x=2(3)=6limx3+f(x)=6x+2=6×3+2=20R.H.L.=f(3)L.H.L.
Hence . x = 3 is the point of discontinuity

case(v) k > 3
f(k)=6k+2limxkf(x)=6k+2limxkf(x)=f(k)
Hence, given function is continuous for each and 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)=1limxkf(x)=f(k)
Hence, given function is continuous for every value of k < 0
case(ii) k > 0
f(k)=1limxkf(x)=1limxkf(x)=f(k)
Hence, given function is continuous for every value of k > 0
case(iii) x = 0
f(0)=0limx0f(x)=1limx0+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 each and every value of x
Hence, no point of 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+1limxkf(x)=f(k)
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
f(k)=k2++1limxkf(x)=k2+1limxkf(x)=f(k)
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

limx1f(x)=x2+1=12+1=1+1=2limx1+f(x)=x+1=1+1=2f(1)=12+1=2R.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+1limxkf(x)=f(k)
Hence, given function is continuous for each value of k > 2

case(ii) k < 2
f(k)=k33limxkf(x)=k33limxkf(x)=f(k)
Hence, given function is continuous for each value of k < 2

case(iii) x = 2

limx2f(x)=x33=233=83=5limx2+f(x)=x2+1=22+1=4+1=5f(2)=233=83=5f(2)=R.H.L.=L.H.L.
Hence, given function is continuous at x = 2
There, 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)=k2limxkf(x)=f(k)
Hence, given function is continuous for each value of k > 1

case(ii) k < 1
f(k)=k101limxkf(x)=k101limxkf(x)=f(k)
Hence, given function is continuous for each value of k < 1

case(iii) x = 1

limx1f(x)=x101=1101=11=0limx1+f(x)=x2=12=1f(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)=k5limxkf(x)=f(k)
Hence, 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, 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, given function is continous 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, given function is discontinous at x = 1
Therefore, x = 1 is he point od 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 continous

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, given function is continous for each and 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 continous

Question:15 Discuss the continuity of the function f, where f is defined by f(x){2xifx<00if0x14xifx>1

Answer:

Given function is
gif
Given function is satisfies for the all real values of x
case (i) k < 0
gif
gif
Hence, function is continuous for all values of x < 0

case (ii) x = 0
gif
L.H.L at x= 0
gif
R.H.L. at x = 0
gif
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii) k > 0
gif
gif
Hence , function is continuous for all values of x > 0

case (iv) k < 1
gif
gif
Hence , function is continuous for all values of x < 1

case (v) k > 1
gif
gif
Hence , function is continuous for all values of x > 1

case (vi) x = 1
gif
gif
gif
Hence, 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, 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, given function is continous at x = -1

case(iii) k > -1
f(k)=2klimxkf(x)=2klimxkf(x)=f(k)
Hence, given function is continous 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 continous

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 continous

case(vi) k > 1
f(k)=2limxkf(x)=2limxkf(x)=f(k)
Hence, given function is continous for each and 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 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, for no value of 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, 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, 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
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 xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)We know thatsin(a+b)=sinacosb+cosasinblimh0sin(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) = cos x
Let suppose x = c + h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)We know thatcos(a+b)=cosacosb+sinasinblimh0cos(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 is continous function
So, we can say that
f(x) = g(x) + h(x) = sin x + cos x is also a continuous function

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

Answer:

Given function is
f(x)=sinxcosx
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 xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)We know thatsin(a+b)=sinacosb+cosasinblimh0sin(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) = cos x
Let suppose x = c + h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)We know thatcos(a+b)=cosacosb+sinasinblimh0cos(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 is continous function
So, we can say that
f(x) = g(x) - h(x) = sin x - cos x is also a continuous function

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

Answer:

Given function is
f(x)=sinx.cosx
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 xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)We know thatsin(a+b)=sinacosb+cosasinblimh0sin(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) = cos x
Let suppose x = c + h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)We know thatcos(a+b)=cosacosb+sinasinblimh0cos(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 is continous function
So, we can say that
f(x) = g(x).h(x) = sin x .cos x is also a continuous function

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

Answer:

We, know that if two function g(x) and h(x) are continuous then
g(x)h(x),h(x)0 is continuous1h(x),h(x)0 is continuous1g(x),g(x)0 is continuous
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)We know thatsin(a+b)=sinacosb+cosasinblimh0sin(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) = cos x
Let suppose x = c + h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)We know thatcos(a+b)=cosacosb+sinasinblimh0cos(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 sin x and cos x is a continous function
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 defined by
f(x)={x2sin1/xifx00ifx=0
is 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, 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
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 xc, then h0
g(c)=sinclimxcg(x)=limxcsinx=limh0sin(c+h)We know thatsin(a+b)=sinacosb+cosasinblimh0sin(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) = cos x
Let suppose x = c + h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)We know thatcos(a+b)=cosacosb+sinasinblimh0cos(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 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)=1limx0f(x)=sin0cos0=1limx0+f(x)=sin0cos0=1R.H.L.=L.H.L.=f(0)
Hence, 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 values of k so that the function f is continuous is 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:29 Find 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:30 Find the values of a and b such that the function defined by
f(x)={5ifx2ax+bif2<x<1021,ifx>10
is a continuous function.

Answer:

Given 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+b2a+b=5           (i)andlimx10f(x)=ax+b=10a+blimx10+f(x)=2110a+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 a continuous function.

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) = cos 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=0limx0+g(x)=x=0limx0g(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) = cos x
Let suppose x = c + h
if xc, then h0
h(c)=cosclimxch(x)=limxccosx=limh0cos(c+h)We know thatcos(a+b)=cosacosb+sinasinblimh0cos(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) = g o h is also continuous

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

Answer:

Given function is
f(x) = sin |x|
f(x) = h o g , h(x) = sin x and g(x) = |x|
Now,

g(x){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=0limx0+g(x)=x=0limx0g(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) = sin x
Let suppose x = c + h
if xc, then h0
h(c)=sinclimxch(x)=limxcsinx=limh0sin(c+h)We know thatsin(a+b)=sinacosb+cosasinblimh0sin(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) = h o g is also continuous

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

Answer:

Given function is
f(x)=|x||x+1|
Let g(x) = |x| and h(x) = |x+1|
Now,
g(x){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=0limx0+g(x)=x=0limx0g(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=0limx1h(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: Excercise: 5.2

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.
Lets 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)
Lets 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.
Lets 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.
Lets 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)
Lets 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)
Differentitation 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 $\sqrt{\sin x^2$ )
(cotx=cosxsinx and cosecx=1sinx)
=22xsinx2sin2x2    (2sinxcosx=sin2x)
There, the answer is 22xsinx2sin2x2

Question:8 Differentiate 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
Required condition for function to be differential at x = 1 is

limh0f(1+h)f(1)h=limh0+f(1+h)f(1)h
Now, 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)
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
Required condition for function to be differential at x = 1 is

limh0f(1+h)f(1)h=limh0+f(1+h)f(1)h
Now, 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)
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
Similary, for x = 2
Required condition for function to be differential at x = 2 is

limh0f(2+h)f(2)h=limh0+f(2+h)f(2)h
Now, 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)
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

Question:1. Find dy/dx 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 dy/dx 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 dy/dx 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 dy/dx 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 dy/dx 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:6 Find dy/dx 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 dy/dx 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)=0dydx(2sinycosyxsinxy)=ysinxydydx=ysinxy2sinycosyxsinxy=ysinxysin2yxsinxy      (2sinxcosy=sin2x)
Therefore, the answer is ysinxysin2yxsinxy

Question:8. Find dy/dx 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:9 Find dy/dx 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 dy/dx 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 dy/dx 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 dy/dx 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 dy/dx 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 dy/dx in the following:

y=sin1(2x1x2),12<x12

Answer:

Given function is
y=sin1(2x1x2)
Lets 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 dy/dx 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

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

exsinx

Answer:

Given function is
f(x)=exsinx
We differentiate with the help of 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:

Double exponent: use braces to clarify

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

Answer:

Given function is
f(x)=log(logx)
Lets 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 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)
Lets 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

Question:1 Differentiate the functions w.r.t. x. cosx.cos2x.cos3x

Answer:

Given function is
y=cosx.cos2x.cos3x
Now, take log on 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+logcos3xdx1y.dydx=(sinx)1cosx+(2sin2x)1cos2x+(3sin3x).1cos3x1ydydx=(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 log on both the 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))dxd(log(x4))dx d(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:3 Differentiate the functions w.r.t. x. (logx)cosx

Answer:

Given function is
y=(logx)cosx
take log on both the sides
logy=cosxlog(logx)
Now, differentiation w.r.t x is
d(logy)dx=d(cosxlog(logx))dx1y.dydx=(sinx)(log(logx))+cosx.1logx.1xdydx=y(cosx.1logx.1xsinxlog(logx))dydx=(logx)cosx(cosxxlogxsinxlog(logx))
Therefore, the answer is (logx)cosx(cosxxlogxsinxlog(logx))

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

Answer:

Given function is
y=xx2sinx
Let's take t=xx
take log on both the sides
logt=xlogx
Now, differentiation w.r.t x is
logt=xlogxd(logt)dt.dtdx=d(xlogx)dx       (by chain rule)1t.dtdx=logx+1dtdx=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=cosxlog2dkdx=k(cosxlog2)dkdx=2sinx(cosxlog2)               (k=2sinx)
Now,
dydx=dtdxdkdxdydx=xx(logx+1)2sinx(cosxlog2)

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

Question:5 Differentiate 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 log 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+5dydx=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:6 Differentiate 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 log on 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 log on both sides
logk=(1+1x)logx
Now, differentiate w.r.t. x
We get,
1k.dkdx=1x(1+1x)+(1x2).logx=x2+1x2+1x2.logxdkdx=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:7 Differentiate 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 log on both the sides
logt=xlog(logx)
Now, differentiate w.r.t. x
we get,
1tdtdx=log(logx)+x.1x.1logx=log(logx)+1logxdtdx=t.(log(logx)+1logx)dtdx=(logx)x(log(logx))+(logx)x.1logx=(logx)x(log(logx))+(logx)x1
Similarly, take k=xlogx
Now, take log on both sides
logk=logxlogx=(logx)2
Now, differentiate w.r.t. x
We get,
1kdkdx=2(logx).1xdtdx=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
Lets take t=(sinx)x
Now, take log on both the 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=12xx2dkdt=12xx2
Now,
dydx=dtdx+dkdx
dydx=(sinx)x(log(sinx)+xcotx)+12xx2
Therefore, the answer is (sinx)x(log(sinx)+xcotx)+12xx2

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

Answer:

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

Now, take t=xsinx
Now, take log on both sides
logt=sinxlogx
Now, differentiate it w.r.t. x
we get,
1tdtdx=cosxlogx+1x.sinxdtdx=t(cosxlogx+1x.sinx)dtdx=xsinx(cosxlogx+1x.sinx)
Similarly, take k=(sinx)cosx
Now, take log on both the 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
xxcosx+x2+1x21
Take t=xxcosx
Take log on both the sides
logt=xcosxlogx
Now, differentiate w.r.t. x
we get,
1tdtdx=cosxlogxxsinxlogx+1x.x.cosxdtdx=t.(logx(cosxxsinx)+cosx)=xxcosx(logx(cosxxsinx)+cosx)
Similarly,
take k=x2+1x21
Now. differentiate it w.r.t. x
we get,
dkdx=2x(x21)2x(x2+1)(x21)2=2x32x2x32x(x21)2=4x(x21)2
Now,
dydx=dtdx+dkdx
dydx=xxcosx(logx(cosxxsinx)+cosx)4x(x21)2
Therefore, the answer is xxcosx(cosx(logx+1)xsinxlogx)4x(x21)2

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 log on 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 log on both the 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:12 Find dy/dx 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 log on both sides
logt=ylogx
Now, differentiate w.r.t x
we get,
1tdtdx=dydx(logx)+y1x=dydx(logx)+yxdtdx=t(dydx(logx)+yx)dtdx=(xy)(dydx(logx)+yx)
Similarly, take k=yx
Now, take log on both sides
logk=xlogy
Now, differentiate w.r.t. x
we get,
1kdkdx=(logy)+x1ydydx=logy+xydydxdkdx=k(logy+xydydx)dkdx=(yx)(logy+xydydx)
Now,
f(x)=dtdx+dkdx=0

(xy)(dydx(logx)+yx)+(yx)(logy+xydydx)=0dydx(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 dy/dx of the functions given in Exercises 12 to 15.

yx=xy

Answer:

Given function is
f(x)xy=yx
Now, take t=xy
take log on both sides
logt=ylogx
Now, differentiate w.r.t x
we get,
1tdtdx=dydx(logx)+y1x=dydx(logx)+yxdtdx=t(dydx(logx)+yx)dtdx=(xy)(dydx(logx)+yx)
Similarly, take k=yx
Now, take log on both sides
logk=xlogy
Now, differentiate w.r.t. x
we get,
1kdkdx=(logy)+x1ydydx=logy+xydydxdkdx=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:14 Find dy/dx 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 the 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:15 Find dy/dx of the functions given in Exercises 12 to 15. xy=exy

Answer:

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

Question:16 Find 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 log 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+x8dydx=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 vale 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 log on both the 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 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 product rule we proved it

Now, by taking the log
Again take y=u.v.w
Now, take log on 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

Question:1 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

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 dy/dx .

x=acosθ,y=bcosθ

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 dy/dx . 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:4 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx

x=4t,y=4/t

Answer:

Given equations are
x=4t,y=4/t
Now, differentiate both w.r.t t
We get,
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 dy/dx 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:6 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx 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:7 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x=sin3tcos2t,y=cos3tcos2t

Answer:

Given equations are
x=sin3tcos2t,y=cos3tcos2t
Now, differentiate both w.r.t 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     (cosxsinx=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 dy/dx 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 dy/dx 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:10 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx 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

Answer:

Given equations are
x=asin1t,y=acos1t

xy=asin1t+cos1tsince sin1x+cos1x=π2xy=aπ2=constant=c

differentiating with respect to x

xdydx+y=0dydx=yx


Class 12 Maths Chapter 5 NCERT solutions: Exercise: 5.7

Question:1 Find 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:2 Find 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:3 Find 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:4 Find 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, second order derivative is
d2ydx2=1x2
Therefore, second order derivative is d2ydx2=1x2

Question:5 Find 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:6 Find 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, 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:7 Find 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, 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:8 Find 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, 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:9 Find 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, 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:10 Find 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, 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:11 If 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:12 If 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, 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, 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)+3 cos(logx) +4sin(logx)
=0
Hence proved

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

Answer:

Given function is
y=Aemx+Benx
Now, differentiation w.r.t. x
dydx=d(Aemx+Benx)dx=mAemx+nBenx -(i)
Now, 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:15 If 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, 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:16 If 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, 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:17 If 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

Class 12 Maths Chapter 5 NCERT solutions: Excercise: 5.8

Question:1 Verify Rolle’s theorem for the function f(x)=x2+2x8,xϵ[4,2].

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c ϵ (x,y) such that f(c)=0
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
f(x)=x2+2x8
Now, being a polynomial function, f(x)=x2+2x8 is both continuous in [-4,2] and differentiable in (-4,2)
Now,
f(4)=(4)2+2(4)8=1688=1616=0
Similalrly,
f(2)=(2)2+2(2)8=4+48=88=0
Therefore, value of f(4)=f(2)=0 and value of f(x) at -4 and 2 are equal
Now,
According to roll's theorem their is point c , c ϵ(4,2) such that f(c)=0
Now,
f(x)=2x+2f(c)=2c+2f(c)=02c+2=0c=1
And c=1 ϵ (4,2)
Hence, Rolle's theorem is verified for the given function f(x)=x2+2x8

Question:2 (1) Examine if Rolle’s theorem is applicable to any of the following functions. Can
you say some thing about the converse of Rolle’s theorem from these example?
f(x)=[x]forxϵ[5,9]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c ϵ (x,y) such that f(c)=0
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
f(x)=[x]
It is clear that Given function f(x)=[x] is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n ϵ [5,9]
limh0f(n+h)f(n)h=limh0[n+h][n]h=limh0n1nh=limh01h =
([n+h]=n1h<0 therefore (n+h)<n)
Now,
R.H.L. at x = n , n ϵ [5,9]
limh0+f(n+h)f(n)h=limh0+[n+h][n]h=limh0+nnh=limh00h=0
([n+h]=nh>0 therefore (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, the function is not differential in (5,9)
Hence, Rolle's theorem is not applicable for given function f(x)=[x] , x ϵ [5,9]

Question:2 (2) Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?

f(x)=[x]forxϵ[2,2]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c ϵ (x,y) such that f(c)=0
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
f(x)=[x]
It is clear that Given function f(x)=[x] is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n ϵ [2,2]
limh0f(n+h)f(n)h=limh0[n+h][n]h=limh0n1nh=limh01h =
([n+h]=n1h<0 therefore (n+h)<n)
Now,
R.H.L. at x = n , n ϵ [2,2]
limh0+f(n+h)f(n)h=limh0+[n+h][n]h=limh0+nnh=limh00h=0
([n+h]=nh>0 therefore (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Rolle's theorem is not applicable for given function f(x)=[x] , x ϵ [2,2]

Question:2 (3) Examine if Rolle’s theorem is applicable to any of the following functions. Can
you say some thing about the converse of Rolle’s theorem from these example?
f(x)=x21forxϵ[1,2]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then there exist a c ϵ (x,y) such that f(c)=0
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
f(x)=x21
Now, being a polynomial , function f(x)=x21 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=121=11=0
And
f(2)=221=41=3
Therefore, f(1)f(2)
Therefore, All conditions are not satisfied
Hence, Rolle's theorem is not applicable for given function f(x)=[x] , x ϵ [2,2]

Question:3 If f;[5,5]R is a differentiable function and if f(x) does not vanish
anywhere, then prove that f(5)f(5)

Answer:

It is given that
f;[5,5]R is a differentiable function
Now, f is a differential function. So, f is also a continuous function
We obtain the following results
a ) f is continuous in [-5,5]
b ) f is differentiable in (-5,5)
Then, by Mean value theorem we can say that there exist a c in (-5,5) such that
f(c)=f(b)f(a)ba
f(c)=f(5)f(5)5(5)f(c)=f(5)f(5)1010f(c)=f(5)f(5)
Now, it is given that f(x) does not vanish anywhere
Therefore,
10f(c)0f(5)f(5)0f(5)f(5)
Hence proved

Question:4 Verify Mean Value Theorem, if f(x)=x24x3 in the interval [a, b], where
a = 1 and b = 4.

Answer:

Condition for M.V.T.
If f;[a,b]R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, there exist a c in (a,b) such that
f(c)=f(b)f(a)ba
It is given that
f(x)=x24x3 and interval is [1,4]
Now, f is a polynomial function , f(x)=x24x3 is continuous in[1,4] and differentiable in (1,4)
And
f(1)=124(1)3=17=6
and
f(4)=424(4)3=16163=1619=3
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
f(c)=f(b)f(a)ba
f(c)=f(4)f(1)41f(c)=3(6)3f(c)=33f(c)=1
Now,
f(x)=2x4f(c)2c41=2c42c=5c=52
And c=52 ϵ (1,4)
Hence, mean value theorem is verified for the function f(x)=x24x3

Question:5 Verify Mean Value Theorem, if f(x)=x35x23x in the interval [a, b], where
a = 1 and b = 3. Find all cϵ(1,3) for which f '(c) = 0.

Answer:

Condition for M.V.T.
If f;[a,b]R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, their exist a c in (a,b) such that
f(c)=f(b)f(a)ba
It is given that
f(x)=x35x23x and interval is [1,3]
Now, f being a polynomial function , f(x)=x35x23x is continuous in[1,3] and differentiable in (1,3)
And
f(1)=135(1)23(1)=153=18=7
and
f(3)=335(3)23(3)=275.99=1845=27
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
f(c)=f(b)f(a)ba
f(c)=f(3)f(1)31f(c)=27(7)2f(c)=202f(c)=10
Now,
f(x)=3x210x3f(c)=3c210c310=3c210c33c210c+7=03c23c7c+7=0(c1)(3c7)=0c=1   and   c=73
And c=1,73 and 73 ϵ (1,3)
Hence, mean value theorem is varified for following function f(x)=x35x23x and c=73 is the only point where f '(c) = 0

Question:6 Examine the applicability of Mean Value Theorem for all three functions given in
the above exercise 2.

Answer:

According to Mean value theorem function
f:[a,b]R must be
a ) continuous in given closed interval say [a,b]
b ) differentiable in given open interval say (a,b)
Then their exist a c ϵ (x,y) such that
f(c)=f(b)f(a)ba
If all these conditions are satisfies then we can verify mean value theorem
Given function is
f(x)=[x]
It is clear that Given function f(x)=[x] is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n ϵ [5,9]
limh0f(n+h)f(n)h=limh0[n+h][n]h=limh0n1nh=limh01h =
([n+h]=n1h<0 therefore (n+h)<n)
Now,
R.H.L. at x = n , n ϵ [5,9]
limh0+f(n+h)f(n)h=limh0+[n+h][n]h=limh0+nnh=limh00h=0
([n+h]=nh>0 therefore (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (5,9)
Hence, Mean value theorem is not applicable for given function f(x)=[x] , x ϵ [5,9]

Similaly,
Given function is
f(x)=[x]
It is clear that Given function f(x)=[x] is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n ϵ [2,2]
limh0f(n+h)f(n)h=limh0[n+h][n]h=limh0n1nh=limh01h =
([n+h]=n1h<0 therefore (n+h)<n)
Now,
R.H.L. at x = n , n ϵ [2,2]
limh0+f(n+h)f(n)h=limh0+[n+h][n]h=limh0+nnh=limh00h=0
([n+h]=nh>0 therefore (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Mean value theorem is not applicable for given function f(x)=[x] , x ϵ [2,2]

Similarly,
Given function is
f(x)=x21
Now, being a polynomial , function f(x)=x21 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=121=11=0
And
f(2)=221=41=3
Now,
f(c)=f(b)f(a)ba
f(c)=f(2)f(1)21f(c)=301f(c)=3
Now,
f(x)=2xf(c)=2c3=2cc=32
And c=32 ϵ (1,2)
Therefore, mean value theorem is applicable for the function f(x)=x21


NCERT class 12 continuity and differentiability ncert solutions Miscellaneous Excercise

Question:1 Differentiate 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:2 Differentiate 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)

Question:3 Differentiate w.r.t. x the function in Exercises 1 to 11.

(5x)3cos2x

Answer:

Given function is
y=(5x)3cos2x
Take, log on both the sides
logy=3cos2xlog5x
Now, differentiation w.r.t. x is
By using 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:4 Differentiate 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:5 Differentiate 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)2f(x)=11(x2)2.12.2x+7cos1x2.12.2x+7.22x+7f(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:6 Differentiate 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:7 Differentiate 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 log 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 log on both the 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:10 xx+xa+ax+aa , for some fixed a > 0 and x > 0

Answer:

Given function is
f(x)=xx+xa+ax+aa
Lets take
u=xx
Now, take log on 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 log on both the 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 log on both the 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 log on both the 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: 11 xx23+(x3)x2,forx>3

Answer:

Given function is
f(x)=xx23+(x3)x2,forx>3
take u=xx23
Now, take log on both the sides
logu=(x23)logx
Now, differentiate w.r.t x
1u.dudx=d(x23)dx.logx+(x23).d(logx)dx1u.dudx=2x.logx+(x23).1x1u.dudx=2x2logx+x23xdudx=u.(2x2logx+x23x)dudx=x(x23).(2x2logx+x23x) -(i)
Similarly,
take Double exponent: use braces to clarify
Now, take log on both the sides
logv=x2log(x3)
Now, differentiate w.r.t x
1v.dvdx=d(x2)dx.log(x3)+x2.d(log(x3))dx1v.dvdx=2x.log(x3)+x2.1(x3)1v.dvdx=2xlog(x3)+x2x3dvdx=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:12 Find dy/dx 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:13 Find dy/dx if y=sin1x+sin11x2,0<x<1

Answer:

Given function is
y=sin1x+sin11x2,0<x<1
Now, differentiatiate w.r.t. x
dydx=d(sin1x+sin11x2)dx=11x2+11(1x2)2.d(1x2)dxdydx=11x2+111+x2.121x2.(2x)dydx=11x211x2dydx=0
Therefore, differentiatiate w.r.t. x is 0

Question:14 If 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+y2x2y2=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:15 If (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:16 If 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))dxsinydydx=1.cos(a+y)+x.(sin(a+y)).dydxdydx.(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:17 If 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:18 If 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 the cases
f(x)=x3f(x)=3x2f(x)=6x
And
f(x)=x3f(x)=3x2f(x)=6x
In both, the cases f ''(x) exist
Hence, we can say that f ''(x) exists for all real x
and values are
f(x){6xx<06xx>0

Question:19 Using mathematical induction prove that ddx(xn)=nxn1 for all positive integers n.

Answer:

Given equation is
ddx(xn)=nxn1
We need to show that ddx(xn)=nxn1 for all positive integers n
Now,
For ( n = 1) d(x1)dx=1.x11=1.x0=1
Hence, true for n = 1
For (n = k) d(xk)dx=k.xk1
Hence, true for n = k
For ( n = k+1) d(xk+1)dx=d(x.xk)dx
=d(x)dx.xk+x.d(xk)dx
=1.xk+x.(k.xk1)=xk+k.xk=(k+1)xk
Hence, (n = k+1) is true whenever (n = k) is true
Therefore, by the principle of mathematical induction we can say that ddx(xn)=nxn1 is true for all positive integers n

Question:20 Using 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:21 Does 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 sum of two continuous function 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 is 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:22 If y=|f(x)g(x)h(x)lmnabc| , prove that dy/dx=|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:23 If y = e ^{a \cos ^{-1}x} , -1 \leqx \leq 1 , show that (1x2)d2ydx2xdydxa2y=0

Answer:

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

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.

Topics of NCERT class 12 maths chapter 5 Continuity and Differentiability

5.1 Introduction

5.2 Continuity

5.2.1 Algebra of continuous functions

5.3. Differentiability

5.3.1 Derivatives of composite functions

5.3.2 Derivatives of implicit functions

5.3.3 Derivatives of inverse trigonometric functions

5.4 Exponential and Logarithmic Functions

5.5. Logarithmic Differentiation

5.6 Derivatives of Functions in Parametric Form

The mathematical definition of Continuity and Differentiability -

Let f be a real function and c be a point in the domain of f. Then f is continuous at c if limxcf(x)=f(c) . A function f is differentiable at point c in its domain if it is continuous at point c. A function is said to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b].

'Continuity and differentiability' is one of the very important and time-consuming chapters of the NCERT Class 12 maths syllabus. It contains 8 exercises with 121 questions and also 23 questions in the miscellaneous exercise. In this article, you will find all NCERT solutions for class 12 maths chapter 5 continuity and differentiability including miscellaneous exercises.

Also read,

NCERT exemplar solutions class 12 maths chapter 5

NCERT class 12 maths ch 5 question answer - Topics

The main topics covered in chapter 5 maths class 12 are:

  • Continuity

A function is continuous at a given point if the left-hand limit, right-hand limit and value of function exist and are equal. In this class 12 NCERT topics elaborate concepts related to continuity, point of discontinuity, algebra of continuous function. Continuity and Differentiability class 12 solutions include a comprehensive module of quality questions.

  • Differentiability

This ch 5 maths class 12 discuss differentiability concepts of different functions including derivatives of composite functions, derivatives of implicit functions, derivatives of inverse trigonometric functions. To get command on these concepts you can refer to NCERT solutions for class 12 maths chapter 5.

  • Exponential and Logarithmic Functions

This ch 5 maths class 12 also includes concepts of exponential and logarithmic functions including natural log and their graphical representation. maths class 12 chapter 5 also contains fundamental properties of the logarithmic function. You can refer to class 12 NCERT solutions for questions about these concepts.

  • Logarithmic Differentiation

this class 12 ncert chapter discusses a special technique of differentiation known as logarithmic differentiation. to get command of these concepts you can go through the NCERT solution for class 12 maths chapter 5.

  • Derivatives of Functions in Parametric Forms

concepts to differentiate a function which is not implicit and explicit but given in the parametric form are explained in this chapter. Continuity and Differentiability class 12 solutions include problems to understand the concepts.

ch 5 maths class 12 also discuss in detail the concepts of second-order derivative, mean value theorem, Rolle's theorem. for questions on these concepts, you can browse NCERT solutions for class 12 chapter 5.

Topics enumerated in class 12 NCERT are very important and students are suggested to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the NCERT solutions for class 12 maths chapter 5

Also read,

NCERT solutions for class 12 maths - Chapter wise

NCERT solutions for class 12 subject wise

NCERT Solutions class wise

NCERT Books and NCERT Syllabus

Tips to Use NCERT 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 on this chapter.

  • You should make sure that concepts related to 'limit' are clear to you as it forms the base for continuity.
  • First, go for the theorem and solved 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 easy 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 practicing questions of NCERT.
  • Once you are good in continuity, then go for the differentiability. Practice more and more questions to get command on it.
  • Differentiation is mostly formula-based, so practice NCERT questions, it won't take much effort to remember the formulas.

Also check,

Happy learning!!!

Frequently Asked Questions (FAQs)

1. What are the important topics in chapter Continuity and Differentiability?

Basic concepts of continuity and differentiability, derivatives of composite functions, derivatives of implicit functions, derivatives of inverse trigonometric functions, exponential and logarithmic functions, logarithmic differentiation, derivatives of functions in parametric form are the important topics in this chapter. Practice these class 12 maths ch 5 question answer to command the concepts. 

2. What are the reasons to opt for ncert continuity and differentiability solution?

The maths chapter 5 class 12 NCERT solutions created by the experts at Careers360 offer numerous advantages to students preparing for their board exams. These solutions provide comprehensive explanations of each topic, which help students achieve high scores. Additionally, the solutions are based on the latest CBSE syllabus for the 2022-23 academic year. Furthermore, these solutions also assist students in preparing for other competitive exams such as JEE Main and JEE Advanced. For ease, Students can study continuity and differentiability pdf both online and offline

3. Which is the best book for CBSE class 12 maths ?

NCERT is the best book for CBSE class 12 maths. Most of the questions in CBSE board exam are directly asked from NCERT textbook. All you need to do is rigorous practice of all the problems given in the NCERT textbook.

4. How many exercises are included in ncert solutions class 12 maths chapter 5?

According to the given information, there are 8 exercises in NCERT Solutions for maths chapter 5 class 12 . The following is the number of questions in each exercise:

  1. Exercise 5.1: 34 questions

  2. Exercise 5.2: 10 questions

  3. Exercise 5.3: 15 questions

  4. Exercise 5.4: 10 questions

  5. Exercise 5.5: 18 questions

  6. Exercise 5.6: 11 questions

  7. Exercise 5.7: 17 questions

  8. Exercise 5.8: 6 questions

Additionally, there is a Miscellaneous Exercise with 23 questions.

5. What is the weightage of the chapter Continuity and Differentiability for CBSE board exam ?

Generally, Continuity and differntiability has 9% weightage in the 12th board final examination. if you want to obtain meritious marks or full marks then you should have good command on concepts that can be developed by practice therefore you should practice NCERT solutions and NCERT exercise solutions.

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Hello there! Thanks for reaching out to us at Careers360.

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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I hope this information helps you.







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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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