NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability: In CBSE class 11 you have already learnt the differentiation of certain functions like polynomial functions and trigonometric functions. In this article, you will get NCERT solutions for class 12 maths chapter 5 continuity and differentiability. Questions based on the topics like continuity, differentiability, and relations between them are covered in the NCERT solutions for class 12 maths chapter 5 continuity and differentiability. In the NCERT textbook, there are 48 solved examples to understand the concepts. First, try to solve examples and understand the concept, then it will be easy to solve NCERT exercise questions. If you are finding difficulties in solving them, you can take help from NCERT solutions for class 12 maths chapter 5 continuity and differentiability article. Here, you will also learn the differentiation of inverse trigonometric functions, exponential functions, and logarithmic functions. Check all NCERT solutions at a single place which will help you to learn CBSE science and maths. Here you will get NCERT solutions for class 12 also.
This chapter "continuity and differentiability" is a continuation of the differentiation of functions that you have already learnt in NCERT class XI. This chapter alone has 9% weightage in the 12th board final examination and the next chapters of calculus(44 % weightage in the final exam) also depend on the concepts of this chapter. This chapter is very important for the exam point of view as well as the application point of view. To get command on this chapter you should solve the miscellaneous exercise also. In this NCERT solutions for class 12 maths chapter 5 continuity and differentiability article, you will get solutions for miscellaneous exercise too. This chapter contains 8 exercises with 121 questions. You will find all these NCERT solutions for class 12 maths chapter 5 continuity and differentiability explained in a detailed manner.
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 . A function f is differentiable at a 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]. NCERT class 12 maths chapter 5 continuity and differentiability is one of the very important and time-consuming chapters. 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 exercise.
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
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Excercise: 5.1
Question:1 . Prove that the function is continuous at and at
Answer:
Given function is
Hence, function is continous at x = 0
Hence, function is continous at x = -3
Hence, function is continuous at x = 5
Question:2 . Examine the continuity of the function
Answer:
Given function is
at x = 3
Hence, function is continous at x = 3
Question:3 Examine the following functions for continuity.
Answer:
Given function is
Our function is defined for every real number say k
and value at x = k ,
and also,
Hence, the function is continuous at every real number
Question:3 b) Examine the following functions for continuity.
Answer:
Given function is
For every real number k ,
We get,
Hence, function continuous for every real value of x,
Question:3 c) Examine the following functions for continuity.
Answer:
Given function is
For every real number k ,
We gwt,
Hence, function continuous for every real value of x ,
Question:3 d) Examine the following functions for continuity.
Answer:
Given function is
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
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
Hence, function f(x) = 5 - x is continous for x < 5
case(iii) x = 5
for x = 5 , f(x) = x - 5 is defined
Hence, function f(x) = x - 5 is continous for x = 5
Hence, the function is continuous for each and every real number
Question:4 . Prove that the function is continuous at x = n, where n is a positive integer
Answer:
GIven function is
the function is defined for all positive integer, n
Hence, the function is continuous at x = n, where n is a positive integer
Question:5. Is the function f defined by
continuous at x = 0? At x = 1? At x = 2?
Answer:
Given function is
function is defined at x = 0 and its value is 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
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
Hence, given function is continous at x = 2
Question:6. Find all points of discontinuity of f, where f is defined by
Answer:
Given function is
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
Hence, given function is continuous for each value of k > 2
case(ii) k < 2
Hence, given function is continuous for each value of k < 2
case(iii) x = 2
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
Answer:
Given function is
GIven function is defined for every real number k
Different cases are their
case (i) k < -3
Hence, given function is continuous for every value of k < -3
case(ii) k = -3
Hence, given function is continous for x = -3
case(iii) -3 < k < 3
Hence, for every value of k in -3 < k < 3 given function is continous
case(iv) k = 3
Hence . x = 3 is the point of discontinuity
case(v) k > 3
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
Answer:
Given function is
if x > 0 ,
if x < 0 ,
given function is defined for every real number k
Now,
case(i) k < 0
Hence, given function is continuous for every value of k < 0
case(ii) k > 0
Hence, given function is continuous for every value of k > 0
case(iii) x = 0
Hence, 0 is the only point of discontinuity
Question:9. Find all points of discontinuity of f, where f is defined by
Answer:
Given function is
if x < 0 ,
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
Answer:
Given function is
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
Hence, given function is continuous for each value of k > 1
case(ii) k < 1
Hence, given function is continuous for each value of k < 1
case(iii) x = 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
Answer:
Given function is
given function is defined for every real number k
There are different cases for the given function
case(i) k > 2
Hence, given function is continuous for each value of k > 2
case(ii) k < 2
Hence, given function is continuous for each value of k < 2
case(iii) x = 2
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
Answer:
Given function is
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
Hence, given function is continuous for each value of k > 1
case(ii) k < 1
Hence, given function is continuous for each value of k < 1
case(iii) x = 1
Hence, x = 1 is the point of discontinuity
Question:13. Is the function defined by
Answer:
Given function is
given function is defined for every real number k
There are different cases for the given function
case(i) k > 1
Hence, given function is continuous for each value of k > 1
case(ii) k < 1
Hence, given function is continuous for each value of k < 1
case(iii) x = 1
Hence, x = 1 is the point of discontinuity
Question:14. Discuss the continuity of the function f, where f is defined by
Answer:
Given function is
GIven function is defined for every real number k
Different cases are their
case (i) k < 1
Hence, given function is continous for every value of k < 1
case(ii) k = 1
Hence, given function is discontinous at x = 1
Therefore, x = 1 is he point od discontinuity
case(iii) 1 < k < 3
Hence, for every value of k in 1 < k < 3 given function is continous
case(iv) k = 3
Hence. x = 3 is the point of discontinuity
case(v) k > 3
Hence, given function is continous for each and every value of k > 3
case(vi) when k < 3
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
Answer:
Given function is
Given function is satisfies for the all real values of x
case (i) k < 0
Hence, function is continuous for all values of x < 0
case (ii) x = 0
L.H.L at x= 0
R.H.L. at x = 0
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0
case (iii) k > 0
Hence , function is continuous for all values of x > 0
case (iv) k < 1
Hence , function is continuous for all values of x < 1
case (v) k > 1
Hence , function is continuous for all values of x > 1
case (vi) x = 1
Hence, function is not continuous at x = 1
Question:16. Discuss the continuity of the function f, where f is defined by
Answer:
Given function is
GIven function is defined for every real number k
Different cases are their
case (i) k < -1
Hence, given function is continuous for every value of k < -1
case(ii) k = -1
Hence, given function is continous at x = -1
case(iii) k > -1
Hence, given function is continous for all values of x > -1
case(vi) -1 < k < 1
Hence, for every value of k in -1 < k < 1 given function is continous
case(v) k = 1
Hence.at x =1 function is continous
case(vi) k > 1
Hence, given function is continous for each and every value of k > 1
case(vii) when k < 1
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
is continuous at x = 3.
Answer:
Given function is
For the function to be continuous at x = 3 , R.H.L. must be equal to L.H.L.
For the function to be continuous
Question:18. For what value of l is the function defined by
continuous at x = 0? What about continuity at x = 1?
Answer:
Given function is
For the function to be continuous at x = 0 , R.H.L. must be equal to L.H.L.
For the function to be continuous
Hence, for no value of function is continuous at x = 0
For x = 1
Hence, given function is continuous at x =1
Answer:
Given function is
Given is defined for all real numbers k
Hence, by this, we can say that the function defined by is discontinuous at all integral points
Question:20. Is the function defined by continuous at x = ?
Answer:
Given function is
Clearly, Given function is defined at x =
Hence, the function defined by continuous at x =
Question:21. Discuss the continuity of the following functions:
a)
Answer:
Given function is
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
Hence, function is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if
Hence, function 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:
Answer:
Given function is
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
Hence, function is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if
Hence, function 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:
Answer:
Given function is
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
Hence, function is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if
Hence, function 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
Lets take g(x) = sin x and h(x) = cos x
Let suppose x = c + h
if
Hence, function is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if
Hence, the function is a continuous function
We proved independently that sin x and cos x is a continous function
So, we can say that
cosec x = is also continuous except at
sec x = is also continuous except at
cot x = is also continuous except at
Question:23. Find all points of discontinuity of f, where
Answer:
Given function is
Hence, the function is continuous
Therefore, no point of discontinuity
Question:24. Determine if f defined by
is a continuous function?
Answer:
Given function is
Given function is defined for all real numbers k
when x = 0
Hence, function is continuous at x = 0
when
Hence, the given function is continuous for all points
Question:25 . Examine the continuity of f, where f is defined by
Answer:
Given function is
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
Hence, function is a continuous function
Now,
h(x) = cos x
Let suppose x = c + h
if
Hence, function 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
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
Answer:
Given function is
When
For the function to be continuous
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
Answer:
Given function is
When x = 2
For the function to be continuous
f(2) = R.H.L. = LH.L.
Hence, the values of k so that the function f is continuous at x= 2 is
Question:28 . Find the values of k so that the function f is continuous at the indicated point in Exercises
Answer:
Given function is
When x =
For the function to be continuous
f( ) = R.H.L. = LH.L.
Hence, the values of k so that the function f is continuous at x= is
Question:29 Find the values of k so that the function f is continuous at the indicated point in Exercises
Answer:
Given function is
When x = 5
For the function to be continuous
f(5) = R.H.L. = LH.L.
Hence, the values of k so that the function f is continuous at x= 5 is
Question:30 Find the values of a and b such that the function defined by
is a continuous function.
Answer:
Given continuous function is
The function is continuous so
By solving equation (i) and (ii)
a = 2 and b = 1
Hence, values of a and b such that the function defined by is a continuous function is 2 and 1 respectively
Question:31. Show that the function defined by is a continuous function.
Answer:
Given function is
given function is defined for all real values of x
Let x = k + h
if
Hence, the function is a continuous function
Question:32. Show that the function defined by is a continuous function.
Answer:
Given function is
given function is defined for all values of x
f = g o h , g(x) = |x| and h(x) = cos x
Now,
g(x) is defined for all real numbers k
case(i) k < 0
Hence, g(x) is continuous when k < 0
case (ii) k > 0
Hence, g(x) is continuous when k > 0
case (iii) k = 0
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
Hence, function 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) is defined for all real numbers k
case(i) k < 0
Hence, g(x) is continuous when k < 0
case (ii) k > 0
Hence, g(x) is continuous when k > 0
case (iii) k = 0
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
Hence, function 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
Answer:
Given function is
Let g(x) = |x| and h(x) = |x+1|
Now,
g(x) is defined for all real numbers k
case(i) k < 0
Hence, g(x) is continuous when k < 0
case (ii) k > 0
Hence, g(x) is continuous when k > 0
case (iii) k = 0
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
g(x) is defined for all real numbers k
case(i) k < -1
Hence, h(x) is continuous when k < -1
case (ii) k > -1
Hence, h(x) is continuous when k > -1
case (iii) k = -1
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 solutions for class 12 maths chapter 5 Continuity and Differentiability Excercise: 5.2
Question:1. Differentiate the functions with respect to x in
Answer:
Given function is
when we differentiate it w.r.t. x.
Lets take . then,
(By chain rule)
Now,
Therefore, the answer is
Question:2. Differentiate the functions with respect to x in
Answer:
Given function is
Lets take then,
( By chain rule)
Now,
Therefore, the answer is
Question:3. Differentiate the functions with respect to x in
Answer:
Given function is
when we differentiate it w.r.t. x.
Lets take . then,
(By chain rule)
Now,
Therefore, the answer is
Question:4 . Differentiate the functions with respect to x in
Answer:
Given function is
when we differentiate it w.r.t. x.
Lets take . then,
take . then,
(By chain rule)
Now,
Therefore, the answer is
Question:5. Differentiate the functions with respect to x in
Answer:
Given function is
We know that,
and
Lets take
Then,
(By chain rule)
-(i)
Similarly,
-(ii)
Now, put (i) and (ii) in
Therefore, the answer is
Question:6. Differentiate the functions with respect to x in
Answer:
Given function is
Differentitation w.r.t. x is
Lets take
Our functions become,
and
Now,
( By chain rule)
-(i)
Similarly,
-(ii)
Put (i) and (ii) in
Therefore, the answer is
Question:7. Differentiate the functions with respect to x in
Answer:
Give function is
Let's take
Now, take
Differentiation w.r.t. x
-(By chain rule)
So,
( Multiply and divide by and multiply and divide by )
There, the answer is
Question:8 Differentiate the functions with respect to x in
Answer:
Let us assume :
Differentiating y with respect to x, we get :
or
or
Question:9 . Prove that the function f given by is not differentiable at x = 1.
Answer:
Given function is
We know that any function is differentiable when both
and are finite and equal
Required condition for function to be differential at x = 1 is
Now, Left-hand limit of a function at x = 1 is
Right-hand limit of a function at x = 1 is
Now, it is clear that
R.H.L. at x= 1 L.H.L. at x= 1
Therefore, function is not differentiable at x = 1
Question:10. Prove that the greatest integer function defined by is not differentiable at
Answer:
Given function is
We know that any function is differentiable when both
and are finite and equal
Required condition for function to be differential at x = 1 is
Now, Left-hand limit of the function at x = 1 is
Right-hand limit of the function at x = 1 is
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 is not differentiable at x = 1
Similary, for x = 2
Required condition for function to be differential at x = 2 is
Now, Left-hand limit of the function at x = 2 is
Right-hand limit of the function at x = 1 is
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 is not differentiable at x = 2
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Exercise: 5.3
Question:1. Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:2. Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:3. Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:4. Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:5. Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:6 Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:7 . Find dy/dx in the following:
Answer:
Given function is
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:8. Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:9 Find dy/dx in the following:
Answer:
Given function is
Lets consider
Then,
Now,
Our equation reduces to
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:10. Find dy/dx in the following:
Answer:
Given function is
Lets consider
Then,
Now,
Our equation reduces to
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:11. Find dy/dx in the following:
Answer:
Given function is
Let's consider
Then,
Now,
Our equation reduces to
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:12 . Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Let's consider
Then,
Now,
Our equation reduces to
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:13. Find dy/dx in the following:
Answer:
Given function is
We can rewrite it as
Let's consider
Then,
Now,
Our equation reduces to
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:14 . Find dy/dx in the following:
Answer:
Given function is
Lets take
Then,
And
Now, our equation reduces to
Now, differentiation w.r.t. x
Therefore, the answer is
Question:15 . Find dy/dx in the following:
Answer:
Given function is
Let's take
Then,
And
Now, our equation reduces to
Now, differentiation w.r.t. x
Therefore, the answer is
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Exercise 5.4
Question:1. Differentiate the following w.r.t. x:
Answer:
Given function is
We differentiate with the help of Quotient rule
Therefore, the answer is
Question:2 . Differentiate the following w.r.t. x:
Answer:
Given function is
Let
Then,
Now, differentiation w.r.t. x
-(i)
Put this value in our equation (i)
Question:3 . Differentiate the following w.r.t. x:
Answer:
Given function is
Let
Then,
Now, differentiation w.r.t. x
-(i)
Put this value in our equation (i)
Therefore, the answer is
Question:4. Differentiate the following w.r.t. x:
Answer:
Given function is
Let's take
Now, our function reduces to
Now,
-(i)
And
Put this value in our equation (i)
Therefore, the answer is
Question:5 . Differentiate the following w.r.t. x:
Answer:
Given function is
Let's take
Now, our function reduces to
Now,
-(i)
And
Put this value in our equation (i)
Therefore, the answer is
Question:6 . Differentiate the following w.r.t. x:
Answer:
Given function is
Now, differentiation w.r.t. x is
Therefore, answer is
Question:7 . Differentiate the following w.r.t. x:
Answer:
Given function is
Lets take
Now, our function reduces to
Now,
-(i)
And
Put this value in our equation (i)
Therefore, the answer is
Question:8 Differentiate the following w.r.t. x:
Answer:
Given function is
Lets take
Now, our function reduces to
Now,
-(i)
And
Put this value in our equation (i)
Therefore, the answer is
Question:9. Differentiate the following w.r.t. x:
Answer:
Given function is
We differentiate with the help of Quotient rule
Therefore, the answer is
Question:10. Differentiate the following w.r.t. x:
Answer:
Given function is
Lets take
Then , our function reduces to
Now, differentiation w.r.t. x is
-(i)
And
Put this value in our equation (i)
Therefore, the answer is
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Exercise: 5.5
Question:1 Differentiate the functions w.r.t. x.
Answer:
Given function is
Now, take log on both sides
Now, differentiation w.r.t. x
There, the answer is
Question:2. Differentiate the functions w.r.t. x.
Answer:
Given function is
Take log on both the sides
Now, differentiation w.r.t. x is
Therefore, the answer is
Question:3 Differentiate the functions w.r.t. x.
Answer:
Given function is
take log on both the sides
Now, differentiation w.r.t x is
Therefore, the answer is
Question:4 Differentiate the functions w.r.t. x.
Answer:
Given function is
Let's take
take log on both the sides
Now, differentiation w.r.t x is
Similarly, take
Now, take log on both sides and differentiate w.r.t. x
Now,
Therefore, the answer is
Question:5 Differentiate the functions w.r.t. x.
Answer:
Given function is
Take log on both sides
Now, differentiate w.r.t. x we get,
Therefore, the answer is
Question:6 Differentiate the functions w.r.t. x.
Answer:
Given function is
Let's take
Now, take log on both sides
Now, differentiate w.r.t. x
we get,
Similarly, take
Now, take log on both sides
Now, differentiate w.r.t. x
We get,
Now,
Therefore, the answer is
Question:7 Differentiate the functions w.r.t. x.
Answer:
Given function is
Let's take
Now, take log on both the sides
Now, differentiate w.r.t. x
we get,
Similarly, take
Now, take log on both sides
Now, differentiate w.r.t. x
We get,
Now,
Therefore, the answer is
Question:8 Differentiate the functions w.r.t. x.
Answer:
Given function is
Lets take
Now, take log on both the sides
Now, differentiate w.r.t. x
we get,
Similarly, take
Now, differentiate w.r.t. x
We get,
Now,
Therefore, the answer is
Question:9 Differentiate the functions w.r.t. x
Answer:
Given function is
Now, take
Now, take log on both sides
Now, differentiate it w.r.t. x
we get,
Similarly, take
Now, take log on both the sides
Now, differentiate it w.r.t. x
we get,
Now,
Therefore, the answer is
Question:10 Differentiate the functions w.r.t. x.
Answer:
Given function is
Take
Take log on both the sides
Now, differentiate w.r.t. x
we get,
Similarly,
take
Now. differentiate it w.r.t. x
we get,
Now,
Therefore, the answer is
Question:11 Differentiate the functions w.r.t. x.
Answer:
Given function is
Let's take
Now, take log on both sides
Now, differentiate w.r.t. x
we get,
Similarly, take
Now, take log on both the sides
Now, differentiate w.r.t. x
we get,
Now,
Therefore, the answer is
Question:12 Find dy/dx of the functions given in Exercises 12 to 15
Answer:
Given function is
Now, take
take log on both sides
Now, differentiate w.r.t x
we get,
Similarly, take
Now, take log on both sides
Now, differentiate w.r.t. x
we get,
Now,
Therefore, the answer is
Question:13 Find dy/dx of the functions given in Exercises 12 to 15.
Answer:
Given function is
Now, take
take log on both sides
Now, differentiate w.r.t x
we get,
Similarly, take
Now, take log on both sides
Now, differentiate w.r.t. x
we get,
Now,
Therefore, the answer is
Question:14 Find dy/dx of the functions given in Exercises 12 to 15.
Answer:
Given function is
Now, take log on both the sides
Now, differentiate w.r.t x
By taking similar terms on the same side
We get,
Therefore, the answer is
Question:15 Find dy/dx of the functions given in Exercises 12 to 15.
Answer:
Given function is
Now, take take log on both the sides
Now, differentiate w.r.t x
By taking similar terms on same side
We get,
Therefore, the answer is
Question:16 Find the derivative of the function given by and hence find
Answer:
Given function is
Take log on both sides
NOW, differentiate w.r.t. x
Therefore,
Now, the vale of is
Question:17 (1) Differentiate in three ways mentioned below:
(i) by using product rule
Answer:
Given function is
Now, we need to differentiate using the product rule
Therefore, the answer is
Question:17 (2) Differentiate in three ways mentioned below:
(ii) by expanding the product to obtain a single polynomial.
Answer:
Given function is
Multiply both to obtain a single higher degree polynomial
Now, differentiate w.r.t. x
we get,
Therefore, the answer is
Question:17 (3) Differentiate in three ways mentioned below:
(iii) by logarithmic differentiation.
Do they all give the same answer?
Answer:
Given function is
Now, take log on both the sides
Now, differentiate w.r.t. x
we get,
Therefore, the answer is
And yes they all give the same answer
Question:18 If u, v and w are functions of x, then show that 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
Now, we differentiate using product rule w.r.t x
First, take
Now,
-(i)
Now, again by the product rule
Put this in equation (i)
we get,
Hence, by product rule we proved it
Now, by taking the log
Again take
Now, take log on both sides
Now, differentiate w.r.t. x
we get,
Hence, we proved it by taking the log
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Exercise:5.6
Answer:
Given equations are
Now, differentiate both w.r.t t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
Given equations are
Now, differentiate both w.r.t
We get,
Similarly,
Now,
Therefore, answer is
Answer:
Given equations are
Now, differentiate both w.r.t t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
Given equations are
Now, differentiate both w.r.t t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
Given equations are
Now, differentiate both w.r.t
We get,
Similarly,
Now,
Therefore, answer is
Answer:
Given equations are
Now, differentiate both w.r.t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
Given equations are
Now, differentiate both w.r.t t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
Given equations are
Now, differentiate both w.r.t t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
Given equations are
Now, differentiate both w.r.t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
Given equations are
Now, differentiate both w.r.t
We get,
Similarly,
Now,
Therefore, the answer is
Question:11 If , show that
Answer:
Given equations are
differentiating with respect to x
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Exercise: 5.7
Question:1 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative
Therefore, the second order derivative is
Question:2 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, the second-order derivative is
Therefore, second-order derivative is
Question:3 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, the second-order derivative is
Therefore, the second-order derivative is
Question:4 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative is
Therefore, second order derivative is
Question:5 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, the second-order derivative is
Therefore, the second-order derivative is
Question:6 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative is
Therefore, second order derivative is
Question:7 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative is
Therefore, second order derivative is
Question:8 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative is
Therefore, second order derivative is
Question:9 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative is
Therefore, second order derivative is
Question:10 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative is
Using Quotient rule
Therefore, second order derivative is
Question:11 If prove that
Answer:
Given function is
Now, differentiation w.r.t. x
Now, the second-order derivative is
Now,
Hence proved
Question:12 If Find in terms of y alone.
Answer:
Given function is
Now, differentiation w.r.t. x
Now, second order derivative is
-(i)
Now, we want in terms of y
Now, put the value of x in (i)
Therefore, answer is
Question:13 If , show that
Answer:
Given function is
Now, differentiation w.r.t. x
-(i)
Now, second order derivative is
By using the Quotient rule
-(ii)
Now, from equation (i) and (ii) we will get
Now, we need to show
Put the value of from equation (i) and (ii)
Hence proved
Question:14 If , show that
Answer:
Given function is
Now, differentiation w.r.t. x
-(i)
Now, second order derivative is
-(ii)
Now, we need to show
Put the value of from equation (i) and (ii)
Hence proved
Question:15 If , show that
Answer:
Given function is
Now, differentiation w.r.t. x
-(i)
Now, second order derivative is
-(ii)
Now, we need to show
Put the value of from equation (ii)
Hence, L.H.S. = R.H.S.
Hence proved
Question:16 If show that
Answer:
Given function is
We can rewrite it as
Now, differentiation w.r.t. x
-(i)
Now, second order derivative is
-(ii)
Now, we need to show
Put value of from equation (i) and (ii)
Hence, L.H.S. = R.H.S.
Hence proved
Question:17 If show that
Answer:
Given function is
Now, differentiation w.r.t. x
-(i)
Now, the second-order derivative is
By using the quotient rule
-(ii)
Now, we need to show
Put the value from equation (i) and (ii)
Hence, L.H.S. = R.H.S.
Hence proved
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Excercise: 5.8
Question:1 Verify Rolle’s theorem for the function
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 such that
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
Now, being a polynomial function, is both continuous in [-4,2] and differentiable in (-4,2)
Now,
Similalrly,
Therefore, value of and value of f(x) at -4 and 2 are equal
Now,
According to roll's theorem their is point c , such that
Now,
And
Hence, Rolle's theorem is verified for the given function
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 such that
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
It is clear that Given function is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n ,
Now,
R.H.L. at x = 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 ,
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 such that
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
It is clear that Given function is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n ,
Now,
R.H.L. at x = 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 ,
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 such that
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
Now, being a polynomial , function is continuous in [1,2] and differentiable in(1,2)
Now,
And
Therefore,
Therefore, All conditions are not satisfied
Hence, Rolle's theorem is not applicable for given function ,
Question:3 If is a differentiable function and if does not vanish
anywhere, then prove that
Answer:
It is given that
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
Now, it is given that does not vanish anywhere
Therefore,
Hence proved
Question:4 Verify Mean Value Theorem, if in the interval [a, b], where
a = 1 and b = 4.
Answer:
Condition for M.V.T.
If
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, there exist a c in (a,b) such that
It is given that
and interval is [1,4]
Now, f is a polynomial function , is continuous in[1,4] and differentiable in (1,4)
And
and
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
Now,
And
Hence, mean value theorem is verified for the function
Answer:
Condition for M.V.T.
If
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, their exist a c in (a,b) such that
It is given that
and interval is [1,3]
Now, f being a polynomial function , is continuous in[1,3] and differentiable in (1,3)
And
and
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
Now,
And
Hence, mean value theorem is varified for following function and is the only point where f '(c) = 0
Answer:
According to Mean value theorem function
must be
a ) continuous in given closed interval say [a,b]
b ) differentiable in given open interval say (a,b)
Then their exist a such that
If all these conditions are satisfies then we can verify mean value theorem
Given function is
It is clear that Given function is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n ,
Now,
R.H.L. at x = 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 ,
Similaly,
Given function is
It is clear that Given function is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n ,
Now,
R.H.L. at x = 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 ,
Similarly,
Given function is
Now, being a polynomial , function is continuous in [1,2] and differentiable in(1,2)
Now,
And
Now,
Now,
And
Therefore, mean value theorem is applicable for the function
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability Miscellaneous Excercise
Question:1 Differentiate w.r.t. x the function in Exercises 1 to 11.
Answer:
Given function is
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question:2 Differentiate w.r.t. x the function in Exercises 1 to 11.
Answer:
Given function is
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question:3 Differentiate w.r.t. x the function in Exercises 1 to 11.
Answer:
Given function is
Take, log on both the sides
Now, differentiation w.r.t. x is
By using product rule
Therefore, differentiation w.r.t. x is
Question:4 Differentiate w.r.t. x the function in Exercises 1 to 11.
Answer:
Given function is
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question:5 Differentiate w.r.t. x the function in Exercises 1 to 11.
Answer:
Given function is
Now, differentiation w.r.t. x is
By using the Quotient rule
Therefore, differentiation w.r.t. x is
Question:6 Differentiate w.r.t. x the function in Exercises 1 to 11.
Answer:
Given function is
Now, rationalize the [] part
Given function reduces to
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question:7 Differentiate w.r.t. x the function in Exercises 1 to 11.
Answer:
Given function is
Take log on both sides
Now, differentiate w.r.t.
Therefore, differentiation w.r.t x is
Question:8 , for some constant a and b.
Answer:
Given function is
Now, differentiation w.r.t x
Therefore, differentiation w.r.t x
Question: 9
Answer:
Given function is
Take log on both the sides
Now, differentiate w.r.t. x
Therefore, differentiation w.r.t x is
Question:10 , for some fixed a > 0 and x > 0
Answer:
Given function is
Lets take
Now, take log on both sides
Now, differentiate w.r.t x
-(i)
Similarly, take
take log on both the sides
Now, differentiate w.r.t x
-(ii)
Similarly, take
take log on both the sides
Now, differentiate w.r.t x
-(iii)
Similarly, take
take log on both the sides
Now, differentiate w.r.t x
-(iv)
Now,
Put values from equation (i) , (ii) ,(iii) and (iv)
Therefore, differentiation w.r.t. x is
Question: 11
Answer:
Given function is
take
Now, take log on both the sides
Now, differentiate w.r.t x
-(i)
Similarly,
take
Now, take log on both the sides
Now, differentiate w.r.t x
-(ii)
Now
Put the value from equation (i) and (ii)
Therefore, differentiation w.r.t x is
Question:12 Find dy/dx if
Answer:
Given equations are
Now, differentiate both y and x w.r.t t independently
And
Now
Therefore, differentiation w.r.t x is
Question:13 Find dy/dx if
Answer:
Given function is
Now, differentiatiate w.r.t. x
Therefore, differentiatiate w.r.t. x is 0
Question:14 If
Answer:
Given function is
Now, squaring both sides
Now, differentiate w.r.t. x is
Hence proved
Question:15 If , for some c > 0, prove that is a constant independent of a and b.
Answer:
Given function is
- (i)
Now, differentiate w.r.t. x
-(ii)
Now, the second derivative
Now, put values from equation (i) and (ii)
Now,
Which is independent of a and b
Hence proved
Question:16 If , with , prove that
Answer:
Given function is
Now, Differentiate w.r.t x
Hence proved
Question:17 If and find
Answer:
Given functions are
and
Now, differentiate both the functions w.r.t. t independently
We get
Similarly,
Now,
Now, the second derivative
Therefore,
Question:18 If , show that f ''(x) exists for all real x and find it.
Answer:
Given function is
Now, differentiate in both the cases
And
In both, the cases f ''(x) exist
Hence, we can say that f ''(x) exists for all real x
and values are
Question:19 Using mathematical induction prove that for all positive integers n.
Answer:
Given equation is
We need to show that for all positive integers n
Now,
For ( n = 1)
Hence, true for n = 1
For (n = k)
Hence, true for n = k
For ( n = k+1)
Hence, (n = k+1) is true whenever (n = k) is true
Therefore, by the principle of mathematical induction we can say that is true for all positive integers n
Question:20 Using the fact that and the differentiation,
obtain the sum formula for cosines.
Answer:
Given function is
Now, differentiate w.r.t. x
Hence, we get the formula by differentiation of sin(A + B)
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
R.H.L. at x = 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
L.H.L. at x = -1
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 , prove that
Answer:
Given that
We can rewrite it as
Now, differentiate w.r.t x
we will get
Hence proved
Question:23 If , show that
Answer:
Given function is
Now, differentiate w.r.t x
we will get
-(i)
Now, again differentiate w.r.t x
-(ii)
Now, we need to show that
Put the values from equation (i) and (ii)
Hence proved
NCERT solutions for class 12 maths - Chapter wise
chapter 1 | NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions |
chapter 2 | NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions |
chapter 3 | |
chapter 4 | |
chapter 5 | NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability |
chapter 6 | NCERT solutions for class 12 maths chapter 6 Application of Derivatives |
chapter 7 | |
chapter 8 | NCERT solutions for class 12 maths chapter 8 Application of Integrals |
chapter 9 | NCERT solutions for class 12 maths chapter 9 Differential Equations |
chapter 10 | NCERT solutions for class 12 maths chapter 10 Vector Algebra |
chapter 11 | NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry |
chapter 12 | NCERT solutions for class 12 maths chapter 12 Linear Programming |
chapter 13 |
NCERT solutions for class 12 subject wise
NCERT Solutions class wise
- NCERT solutions for class 12
- NCERT solutions for class 11
- NCERT solutions for class 10
- NCERT solutions for class 9
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 of the limit are clear to you as it forms the base for the 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, 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
Happy learning!!!
Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
Question: What are the important topics in chapter Continuity and Differentiability ?
Answer:
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.
Question: What is the weightage of the chapter Continuity and Differentiability for CBSE board exam ?
Answer:
Generally, Continuity and differntiability has 9% weightage in the 12th board final examination.
Question: How the NCERT solutions are helpful in the board exam ?
Answer:
As CBSE board exam paper is designed entirely based on NCERT textbooks so one must know the NCERT very well in order to get good marks in the exam. Students can take help from these NCERT solutions if they stuck while solving the problems. Also, they will get to know how to answer in the board exam in order to get good marks in the board exam.
Question: Does CBSE provides the solutions of NCERT class 12 maths ?
Answer:
No, CBSE doesn’t provided NCERT solutions for any class or subject.
Question: Where can I find the complete solutions of NCERT for class 12 maths ?
Answer:
A Here you will get the detailed NCERT solutions for class 12 maths by clicking on the link.
Question: Which is the best book for CBSE class 12 maths ?
Answer:
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.
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