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Continuity is the foundation, and differentiability is the strength that builds upon it. Continuity and differentiability class 12 solutions are important foundational steps for advanced calculus. Continuity of a function means that the function's graph can be drawn without any break or the graph can be drawn without lifting the pen. Differentiability means the function derivative exists at every point of the given interval, or we can also define differentiability if there is only a tangent to the given point in an interval. In the NCERT solutions for Class 12, Continuity and Differentiability, students will learn important concepts related to continuity, differentiability, and their interrelationships. The differentiation of inverse trigonometric functions, exponential and logarithmic functions is also discussed in this chapter.
A function without continuity is like a road with holes, and without differentiability, it is like a road full of sharp turns. These Class 12 Maths chapter 5 solutions are crucial for the final board examination and various competitive tests, such as JEE Mains, JEE Advanced, BITSAT, and others. At Careers360, experts with multiple years of experience have created these NCERT Solutions for Class 12 Maths, Continuity and Differentiability. It is advisable to work through all the NCERT problems, including examples and the miscellaneous exercises, to master this chapter. For syllabus, notes, and PDF, refer to this link: NCERT.
Students who wish to access the Class 12 Maths Chapter 5 NCERT Solutions can click on the link below to download the complete solution in PDF.
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.1 Page number: 116-118 Total questions: 34 |
Question 1: Prove that the function
Answer:
The given function is
Hence, the function is continuous at x = 0
Hence, the function is continuous at
Hence, the function is continuous at
Question 2: Examine the continuity of the function
Answer:
The given function is
at
Hence, the function is continuous at
Question 3: Examine the following functions for continuity.
Answer:
The given function is
Our function is defined for every real number, say k
and value at
And also,
Hence, the function
Question 3(b): Examine the following functions for continuity.
Answer:
The given function is
For every real number k,
We get,
Hence, function
Question 3(c): Examine the following functions for continuity.
Answer:
The given function is
For every real number k,
We get,
Hence, function
Question 3(d): Examine the following functions for continuity.
Answer:
The given function is
for
for
So, there are different cases.
case(i)
for every real number
Hence, function
case (ii)
for every real number
Hence, function
case(iii)
for
Hence, function
Hence, the function
Question 4: Prove that the function
Answer:
The given function is
The function
Hence, the function
Question 5: Is the function f defined by
continuous at x = 0? At x = 1? At x = 2?
Answer:
The given function is
The function is defined at
Hence, the given function is continuous at
The given function is defined for
Now, for
R.H.L
Therefore, the given function is not continuous at
Given function is defined for
Hence, the given function is continuous at
Question 6: Find all points of discontinuity of f, where f is defined by
Answer:
The given function is
The given function is defined for every real number k
There are different cases for the given function.
case(i)
Hence, the given function is continuous for each value of
case(ii)
Hence, the given function is continuous for each value of
case(iii)
Right 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:
The given function is
The given function is defined for every real number k
There are different cases.
case (i)
Hence, the given function is continuous for every value of k < -3
case(ii)
Hence, the given function is continuous for x = -3
case(iii)
Hence, for every value of k in -3 < k < 3 given function is continuous.
case(iv)
Hence,
case(v)
Hence, the given function is continuous for every value of k > 3
Question 8: Find all points of discontinuity of f, where f is defined by
Answer:
The given function is
if x > 0 ,
if x < 0 ,
The given function is defined for every real number k
Now,
case(i) k < 0
Hence, the given function is continuous for every value of k < 0
case(ii) k > 0
Hence, the 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:
The 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 every value of x
Hence, no point in discontinuity
Question 10: Find all points of discontinuity of f, where f is defined by
Answer:
The given function is
The given function is defined for every real number
There are different cases for the given function.
case(i)
Hence, the given function is continuous for each value of
case(ii)
Hence, the given function is continuous for each value of
case(iii)
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:
The given function is
The given function is defined for every real number
There are different cases for the given function.
case(i)
Hence, the given function is continuous for each value of
case(ii)
Hence, the given function is continuous for each value of
case(iii)
Hence, the given function is continuous at
There is no point of discontinuity
Question 12: Find all points of discontinuity of f, where f is defined by
Answer:
The given function is
The given function is defined for every real number k
There are different cases for the given function.
case(i)
Hence, the given function is continuous for each value of k > 1
case(ii)
Hence, the 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:
The given function is
The given function is defined for every real number
There are different cases for the given function.
case(i)
Hence, the given function is continuous for each value of
case(ii)
Hence, the given function is continuous for each value of
case(iii)
Hence,
Question 14: Discuss the continuity of the function f, where f is defined by
Answer:
The given function is
The given function is defined for every real number k
Different cases are there
case (i)
Hence, the given function is continuous for every value of
case(ii)
Hence, the given function is discontinuous at
Therefore,
case(iii)
Hence, for every value of
case(iv)
Hence,
case(v)
Hence, the given function is continuous for every value of
case(vi) when
Hence, for every value of
Question 15: Discuss the continuity of the function f, where f is defined by
Answer:
Given function is satisfied for all real values of
case (i)
Hence, the function is continuous for all values of
case (ii)
L.H.L at
R.H.L. at
L.H.L. = R.H.L. =
Hence, the function is continuous at
case (iii)
L.H.L. = R.H.L. =
Hence, the function is continuous for all values of x > 0
case (iv) k < 1
Hence, the function is continuous for all values of x < 1
case (v) k > 1
Hence, the function is continuous for all values of x > 1
case (vi) x = 1
Hence, the function is not continuous at x = 1
Question 16: Discuss the continuity of the function f, where f is defined by
Answer:
The given function is
The given function is defined for every real number
Different cases are there
case (i)
Hence, the given function is continuous for every value of
case(ii) k = -1
Hence, the given function is continuous at
case(iii)
Hence, the given function is continuous for all values of
case(vi)
Hence, for every value of
case(v)
Hence, at x =1 function is continuous
case(vi)
Hence, the given function is continuous for every value of k > 1
case(vii) when
Hence, for every value of k in
Therefore, continuous at all points
Question 17: Find the relationship between a and b so that the function f is defined by
is continuous at x = 3.
Answer:
The given function is
For the function to be continuous at
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
For the function to be continuous
Hence, if no value of the function is continuous at
For
Hence, the given function is continuous at
Answer:
Given function is
Given is defined for all real numbers k
Hence, by this, we can say that the function defined by
Question 20: Is the function defined by
Answer:
Given function is
Clearly, the Given function is defined at x =
Hence, the function defined by
Question 21: Discuss the continuity of the following functions:
a)
Answer:
Given function is
The given function is defined for all real numbers.
We, know that if two function
Lets take
Let's suppose
if
Hence, function
Now,
Let's suppose
if
Hence, function
We proved independently that
So, we can say that
Question 21(b):Discuss the continuity of the following functions:
Answer:
Given function is
The given function is defined for all real numbers.
We know that if two functions g(x) and h(x) are continuous then
Lets take
Let's suppose
if
Hence, function
Now,
Let's suppose
if
Hence, function
Question 21(c): Discuss the continuity of the following functions:
Answer:
Given function is
The given function is defined for all real numbers.
We know that if two functions
Lets take
Let's suppose
if
Hence, function
Now,
Let's suppose
if
Hence, function
We proved independently that sin x and cos x are continuous functions.
So, we can say that
Question 22: Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Answer:
We know that if two functions g(x) and h(x) are continuous, then.
Lets take
Let's suppose
if
Hence, function
Now,
Let's suppose
if
Hence, the function
We proved independently that
So, we can say that
cosec x =
sec x =
cot x =
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 is defined by
Is it a continuous function?
Answer:
The given function is
The given function is defined for all real numbers k
when x = 0
Hence, the 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:
The given function is
The given function is defined for all real numbers.
We know that if two functions
Lets take
Let's suppose
if
Hence, function
Now,
Let's suppose
if
Hence, function
We proved independently that
So, we can say that
When
Hence, the function is also continuous at
Question 26: Find the values of k so that the function f is continuous at the indicated point in Exercises
Answer:
The given function is
When
For the function to be continuous
Therefore, the value of k so that the function f is continuous at 6
Question 27: Find the values of k so that the function f is continuous at the indicated point in Exercises
Answer:
The given function is
When
For the function to be continuous
Hence, the values of k so that the function f is continuous at x=2 are
Question 28: Find the values of k so that the function f is continuous at the indicated point in Exercises
Answer:
The given function is
When x =
For the function to be continuous
, f(
Hence, the values of k so that the function f is continuous at x=
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
For the function to be continuous
Hence, the values of k so that the function f is continuous at
Question 30: Find the values of a and b such that the function defined by
It is a continuous function.
Answer:
Given that 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
Question 31 Show that the function defined by
Answer:
Given function is
Given function is defined for all real values of x
Let x = k + h
if
Hence, the function
Question 32: Show that the function defined by
Answer:
Given function is
Given function is defined for all values of x
f = g o h ,
Now,
g(x) is defined for all real numbers k
case(i)
Hence, g(x) is continuous when
case (ii)
Hence, g(x) is continuous when k > 0
case (iii)
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
Let's suppose
if
Hence, function
g(x) is continuous ,
Therefore,
Question 33: Examine that sin | x| is a continuous function.
Answer:
The given function is
f(x) = h o g , h(x) =
Now,
g(x) is defined for all real numbers k
case(i)
Hence, g(x) is continuous when k < 0
case (ii)
Hence, g(x) is continuous when k > 0
case (iii)
Hence, g(x) is continuous when k = 0
Therefore, g(x) = |x| is continuous for all real values of x
Now,
h(x) =
Let's suppose
if
Hence, function
g(x) is continuous , h(x) is continuous
Therefore,
Question 34: Find all the points of discontinuity of f defined by
Answer:
The 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,
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.2 |
Question 1: Differentiate the functions with respect to x in
Answer:
The given function is
When we differentiate it w.r.t. x.
Let's take
Now,
Therefore, the answer is
Question 2: Differentiate the functions with respect to x in
Answer:
The given function is
Let’s take
Now,
Therefore, the answer is
Question 3: Differentiate the functions with respect to x in
Answer:
The given function is
When we differentiate it w.r.t. x.
Let's take
Now,
Therefore, the answer is
Question 4: Differentiate the functions with respect to x in
Answer:
The given function is
When we differentiate it w.r.t. x.
Let's take
take
Now,
Therefore, the answer is
Question 5: Differentiate the functions with respect to x in
Answer:
The given function is
We know that,
Let's take
Then,
Similarly,
Now, put (i) and (ii) in
Therefore, the answer is
Question 6: Differentiate the functions with respect to x in
Answer:
The given function is
Differentiation w.r.t. x is
Lets take
Our functions become,
Now,
Similarly,
Put (i) and (ii) in
Therefore, the answer is
Question 7: Differentiate the functions with respect to x in
Answer:
The give function is
Let's take
Now, take
Differentiation w.r.t. x
So,
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
Answer:
The given function is
We know that any function is differentiable when both.
The required condition for the function to be differentiable at x = 1 is
Now, the Left-hand limit of a function at x = 1 is
The right-hand limit of a function at x = 1 is
Now, it is clear that.
R.H.L. at x= 1
Therefore, function
Question 10: Prove that the greatest integer function defined by
Answer:
The given function is
We know that any function is differentiable when both.
The required condition for the function to be differentiable at x = 1 is
Now, the Left-hand limit of the function at x = 1 is
The right-hand limit of the function at x = 1 is
Now, it is clear that.
R.H.L. at x= 1
Therefore, function
Similarly, for x = 2
The required condition for the function to be differentiable at x = 2 is
Now, the Left-hand limit of the function at x = 2 is
The right-hand limit of the function at x = 1 is
Now, it is clear that.
R.H.L. at x= 2
Therefore, function
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.3 |
Question 1: Find
Answer:
The given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 2: Find
Answer:
The given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 3: Find
Answer:
The given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 4: Find
Answer:
The given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 5: Find
Answer:
The given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 6: Find
Answer:
The given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 7: Find
Answer:
The given function is
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 8: Find
Answer:
The given function is
We can rewrite it as
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 9: Find
Answer:
The 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
Answer:
The 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
Answer:
The 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
Answer:
The 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
Answer:
The 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
Answer:
The given function is
Let's take
Then,
And
Now, our equation reduces to
Now, differentiation with respect to. x
Therefore, the answer is
Question 15: Find
Answer:
The given function is
Let's take
Then,
And
Now, our equation reduces to
Now, differentiation with respect to. x
Therefore, the answer is
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.4 |
Question 1: Differentiate the following w.r.t. x:
Answer:
The given function is
We differentiate with the help of the Quotient rule.
Therefore, the answer is
Question 2: Differentiate the following w.r.t. x:
Answer:
The given function is
Let
Then,
Now, differentiation with respect to. x
Put this value in our equation (i)
Question 3: Differentiate the following w.r.t. x:
Answer:
The given function is
Let
Then,
Now, differentiation with respect to. x
Put this value in our equation (i)
Therefore, the answer is
Question 4: Differentiate the following w.r.t. x:
Answer:
The given function is
Let's take
Now, our function reduces to
Now,
And
Put this value in our equation (i)
Therefore, the answer is
Question 5: Differentiate the following w.r.t. x:
Answer:
The given function is
Let's take
Now, our function reduces to
Now,
And
Put this value in our equation (i)
Therefore, the answer is
Question 6: Differentiate the following w.r.t. x:
Answer:
The given function is
Now, differentiation w.r.t. x is
Therefore, answer is
Question 7: Differentiate the following w.r.t. x:
Answer:
The given function is
Let's take
Now, our function reduces to
Now,
And
Put this value in our equation (i)
Therefore, the answer is
Question 8: Differentiate the following w.r.t. x:
Answer:
The given function is
Let's take
Now, our function reduces to
Now,
And
Put this value in our equation (i)
Therefore, the answer is
Question 9: Differentiate the following w.r.t. x:
Answer:
The given function is
We differentiate with the help of the Quotient rule.
Therefore, the answer is
Question 10: Differentiate the following w.r.t. x:
Answer:
The given function is
Let's take
Then, our function reduces to
Now, differentiation w.r.t. x is
And
Put this value in our equation (i)
Therefore, the answer is
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.5 |
Question 1: Differentiate the functions w.r.t. x.
Answer:
The given function is
Now, take a look at both sides.
Now, differentiation with respect to. x
There, the answer is
Question 2: Differentiate the functions with respect to. x.
Answer:
The given function is
Take logs on both sides.
Now, differentiation w.r.t. x is
Therefore, the answer is
Question 3: Differentiate the functions w.r.t. x.
Answer:
The given function is
Take logs on both sides.
Now, differentiation w.r.t x is
Therefore, the answer is
Question 4: Differentiate the functions w.r.t. x.
Answer:
The given function is
Let's take
Take logs on both sides.
Now, differentiation w.r.t x is
Similarly, take
Now, take the log on both sides and differentiate with respect to. x
Now,
Therefore, the answer is
Question 5: Differentiate the functions w.r.t. x.
Answer:
The given function is
Take logs 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:
The given function is
Let's take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Similarly, take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Now,
Therefore, the answer is
Question 7: Differentiate the functions with respect to. x.
Answer:
The given function is
Let's take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Similarly, take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Now,
Therefore, the answer is
Question 8: Differentiate the functions with respect to. x.
Answer:
The given function is
Let's take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Similarly, take
Now, differentiate with respect. x
We get,
Now,
Therefore, the answer is
Question 9: Differentiate the functions w.r.t. x
Answer:
The given function is
Now, take
Now, take a look at both sides.
Now, differentiate it w.r.t. x
We get,
Similarly, take
Now, take a look at both sides.
Now, differentiate it w.r.t. x
We get,
Now,
Therefore, the answer is
Question 10: Differentiate the functions with respect to. x.
Answer:
The given function is
Now, take
Now, take a look at both sides.
Now, differentiate it w.r.t. x
We get,
Similarly, take
Now, take a look at both sides.
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 a look at both sides.
Now, differentiate w.r.t. x
we get,
Similarly, take
Now, take a look at both sides.
Now, differentiate w.r.t. x
we get,
Now,
Therefore, the answer is
Question 12: Find
Answer:
The given function is
Now, take
Take logs on both sides.
Now, differentiate w.r.t x
We get,
Similarly, take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Now,
Therefore, the answer is
Question 13: Find
Answer:
The given function is
Now, take
Take logs on both sides.
Now, differentiate w.r.t x
We get,
Similarly, take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Now,
Therefore, the answer is
Question 14: Find
Answer:
The given function is
Now, take the log on both sides.
Now, differentiate w.r.t x
By taking similar terms on the same side
We get,
Therefore, the answer is
Question 15: Find
Answer:
The given function is
Now, take a look at both sides.
Now, differentiate w.r.t x
By taking similar terms on the same side
We get,
Therefore, the answer is
Question 16: Find the derivative of the function given by
Answer:
The given function is
Take logs on both sides.
NOW, differentiate with respect. x
Therefore,
Now, the value of
Question 17(1): Differentiate
(i) By using the product rule
Answer:
The given function is
Now, we need to differentiate using the product rule.
Therefore, the answer is
Question 17(2): Differentiate
(ii) by expanding the product to obtain a single polynomial.
Answer:
The given function is
Multiply both to obtain a single higher-degree polynomial.
Now, differentiate with respect. x
We get,
Therefore, the answer is
Question 17(3): Differentiate
(iii) by logarithmic differentiation.
Do they all give the same answer?
Answer:
The given function is
Now, take a look at both sides.
Now, differentiate with respect. 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
Answer:
It is given that u, v and w are the functions of x
Let
Now, we differentiate using the product rule with respect to x
First, take
Now,
Now, again, by the product rule
Put this in equation (i)
We get,
Hence, by the product rule, we proved it.
Now, by taking the log
Again take
Now, take a look at both sides.
Now, differentiate with respect. x
We get,
Hence, we proved it by taking the log.
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.6 |
Answer:
The given equations are
Now, differentiate both with respect to t
We get,
Similarly,
Now,
Therefore, the answer is
Question 2: If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, find
Answer:
The given equations are
Now, differentiate both w.r.t
We get,
Similarly,
Now,
Therefore, answer is
Answer:
The given equations are
Now, differentiate both w.r.t t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
The given equations are
Now, differentiate both w.r.t t
We get,
Similarly,
Now,
Therefore, the answer is
Answer:
The 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
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:
The given equations are
Now, differentiate both with respect to
We get,
Similarly,
Now,
Therefore, the answer is
Question 11: If
Answer:
The given equations are
Differentiating with respect to x
NCERT Continuity and Differentiability Class 12 Solutions: Exercise: 5.7 |
Question 1: Find the second-order derivatives of the functions given in Exercises 1 to 10.
Answer:
The given function is
Now, differentiation with respect to. x
Now, the 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:
The given function is
Now, differentiation with respect to. 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:
The given function is
Now, differentiation with respect to. 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:
The given function is
Now, differentiation with respect to. x
Now, the 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:
The given function is
Now, differentiation with respect. 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, the 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, the 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, the 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, the 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, the second order derivative is
Using Quotient rule
Therefore, second order derivative is
Question 11: If
Answer:
Given function is
Now, differentiation w.r.t. x
Now, the second-order derivative is
Now,
Hence proved
Question 12: If
Answer:
Given function is
Now, differentiation w.r.t. x
Now, the second order derivative is
Now, we want
Now, put the value of x in (i)
Therefore, answer is
Question 13: If
Answer:
The given function is
Now, differentiation with respect to. x
Now, the second-order derivative is
By using the Quotient rule
Now, from equation (i) and (ii), we will get
Now, we need to show.
Put the value of
Hence proved
Question 14: If
Answer:
The given function is
Now, differentiation with respect to. x
Now, the second-order derivative is
Now, we need to show.
Put the value of
Hence proved
Question 15: If
Answer:
The given function is
Now, differentiation w.r.t. x
Now, the second-order derivative is
Now, we need to show.
Put the value of
Hence, L.H.S. = R.H.S.
Hence proved
Question 16: If
Answer:
The given function is
We can rewrite it as
Now, differentiation with respect to. x
Now, the second-order derivative is
Now, we need to show.
Put value of
Hence, L.H.S. = R.H.S.
Hence proved
Question 17: If
Answer:
The given function is
Now, differentiation with respect to. x
Now, the second-order derivative is
By using the quotient rule
Now, we need to show.
Put the value from equations (i) and (ii)
Hence, L.H.S. = R.H.S.
Hence proved
NCERT Continuity and Differentiability Class 12 Solutions: Miscellaneous Exercise |
Question 1: Differentiate with respect to. x the function in Exercises 1 to 11.
Answer:
The given function is
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question 2: Differentiation with respect to. x the function in Exercises 1 to 11.
Answer:
The given function is
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question 3: Differentiate with respect. x the function in Exercises 1 to 11.
Answer:
The given function is
Take a log on both sides.
Now, differentiation w.r.t. x is
By using the product rule
Therefore, differentiation w.r.t. x is
Question 4: Differentiate with respect to. x the function in Exercises 1 to 11.
Answer:
The given function is
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question 5: Differentiate with respect to. x the function in Exercises 1 to 11.
Answer:
The 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 with respect to. x the function in Exercises 1 to 11.
Answer:
The given function is
Now, rationalise the part.
Given function reduces to
Now, differentiation w.r.t. x is
Therefore, differentiation w.r.t. x is
Question 7: Differentiate with respect to. x the function in Exercises 1 to 11.
Answer:
The given function is
Take logs on both sides.
Now, differentiate with respect.
Therefore, differentiation w.r.t x is
Question 8:
The given function is
Now, differentiation w.r.t x
Therefore, differentiation w.r.t x
Answer:
The given function is
Take logs on both sides.
Now, differentiate with respect. x
Therefore, differentiation w.r.t x is
Question 10:
Answer:
The given function is
Let's take
Now, take a look at both sides.
Now, differentiate w.r.t x
Similarly, take
Take logs on both sides.
Now, differentiate w.r.t x
Similarly, take
Take logs on both sides.
Now, differentiate w.r.t x
Similarly, take
Take logs on both sides.
Now, differentiate w.r.t x
Now,
Put values from equations (i), (ii),(iii) and (iv)
Therefore, differentiation w.r.t. x is
Answer:
The given function is
take
Now, take a look at both sides.
Now, differentiate w.r.t x
Similarly,
take
Now, take a look at both sides.
Now, differentiate w.r.t x
Now
Put the value from equations (i) and (ii)
Therefore, differentiation w.r.t x is
Question 12: Find
Answer:
The 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
Answer:
The given function is
Now, differentiate with respect. x
Therefore, differentiate w.r.t. x is 0
Question 14: If
Answer:
The given function is
Now, squaring both sides.
Now, differentiate w.r.t. x is
Hence proved
Question 15: If
Answer:
The given function is
Now, differentiate with respect. x
Now, the second derivative
Now, put values from equations (i) and (ii)
Now,
Which is independent of a and b
Hence proved
Question 16: If
Answer:
The given function is
Now, differentiate w.r.t x
Hence proved
Question 17: If
Answer:
Given functions are
Now, differentiate both the functions w.r.t. t independently.
We get
Similarly,
Now,
Now, the second derivative
Therefore,
Question 18: If
Answer:
The given function is
Now, differentiate in both cases.
And
In both cases, f ''(x) exists.
Hence, we can say that f ''(x) exists for all real x
And values are
Question 19: Using the fact that
Obtain the sum formula for cosines.
Answer:
The given function is
Now, differentiate with respect. 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 the sum of two continuous functions is also a continuous function.
Therefore, our function f(x) is continuous.
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
R.H.L. at x = 0
R.H.L. is not equal to L.H.L.
Hence. At x = 0, the function is not differentiable.
Now, Similarly
R.H.L. at x = -1
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 21: If
Answer:
Given that
We can rewrite it as
Now, differentiate w.r.t x
We will get
Hence proved
Question 22: If
Answer:
The given function is
Now, differentiate w.r.t x, we will get.
Now, again differentiate with respect to x
Now, we need to show that.
Put the values from equations (i) and (ii)
Hence proved
Also, read,
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Question: If
Solution:
Hence, the answer is
Here are the topics that are discussed in the NCERT Solutions for class 12 chapter 5, Continuity and Differentiability.
A function
The sum, difference, product, and quotient of continuous functions are continuous.
Differentiation:
The derivative of
Chain Rule:
If
The mean Value Theorem states that if
Rolle's Theorem states that if
Lagrange's Mean Value Theorem states that if
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Differentiation of Inverse Trigonometric Function (cos/sine/tan) |
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Differentiation of a Function wrt Another Function and Higher Order derivative of a Function |
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Non - Removable, Infinite and Oscillatory Type Discontinuity |
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Students can access all the Maths solutions from the NCERT book from the links below.
Also, read,
For subject-wise solutions, you can refer here
For the solution of other classes, you can refer here
Here, you can refer to the latest syllabus and NCERT Books
Yes, every differentiable function is always continuous, but the converse is not true; a function can be continuous without being differentiable. For example, the absolute value function, f(x)=|x|, is continuous everywhere but not differentiable at x=0.
In Chapter 5, applications of differentiation focus on understanding and using derivatives to analyse functions, including finding rates of change, determining increasing/decreasing intervals, locating extrema, and sketching curves.
Continuity refers to a function having no breaks or gaps in its graph, while differentiability means the function has a defined slope (derivative) at every point in its domain.
Formulas include the power rule, product rule, quotient rule, and chain rule for differentiation.
You can download the book and solutions from the Careers360 site for free.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
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!
It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.
Possible steps:
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I hope this information helps you.
Hi,
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hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
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