VMC VIQ Scholarship Test
Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.
NCERT solutions for Class 12 Maths chapter 5 continuity and differentiability-Solving the book of NCERT for Class 12 students is a must. NCERT Class 12 Mathematics book covers everything that will come in the board exams and helps in creating a base of the topics. Therefore, if one wants to earn the topic for exams and also for higher education, solving the NCERT questions is a must. We are here to provide you with all the NCERT Exemplar Class 12 Math solutions chapter 5. This is one of the most crucial chapters that need to be covered by the student to score well. That is why we are trying to make sure that the students have the answers with them for practice and reference.
Also, read - NCERT Class 12 Maths Solutions.
Question:1
Examine the continuity of the function
Answer:
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can present it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
is continuous if -
But we can see that,
Similarly, we proceed for RHL-
Question:2
Find which of the functions is continuous or discontinuous at the indicated points:
Answer:
Given,
We need to check its continuity at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 2 if -
Similarly, we proceed for RHL-
Now from equation 2, 3 and 4 we can conclude that
Question:3
Find which of the functions is continuous or discontinuous at the indicated points:
Answer:
Given,
We need to check its continuity at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can present it as
As this limit can be evaluated directly by putting value of h because it is taking indeterminate form (0/0)
As we know,
Again, using sandwich theorem, we get -
And,
f (0) = 5 …(4)
From equation 2, 3 and 4 we can conclude that
∴ f(x) is discontinuous at x = 0
Question:4
Find which of the functions is continuous or discontinuous at the indicated points:
at x = 2
Answer:
Given,
We need to check its continuity at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 2 if -
Then,
Similarly, we proceed for RHL-
From the above equation 2 , 3 and 4 we can say that
∴ f(x) is continuous at x = 2
Question:5
Find which of the functions is continuous or discontinuous at the indicated points:
at x = 4
Answer:
Given,
...(1)
We need to check its continuity at x = 4
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 4 if -
Clearly,
Similarly, we proceed for RHL-
∴ f(x) is discontinuous at x = 4
Question:6
Answer:
Given,
We need to check its continuity at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 4 if -
As cos (1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = 0 × (finite value) = 0 …(2)
Similarly, we proceed for RHL-
And,
f(0) = 0 {using eqn 1} …(4)
From the above equation 2 , 3 and 4 we can conclude that
∴ f(x) is continuous at x = 0
Question:7
Answer:
Given,
We need to check its continuity at x = a
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = a if -
Clearly,
As sin (-1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = 0 × (finite value) = 0 …(2)
Similarly we proceed for RHL-
.
h > 0 as defined above.
∴ |h| = h
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHL = 0 × (finite value) = 0 …(3)
And,
f(a) = 0 {using eqn 1} …(4)
From equation 2, 3 and 4 we can conclude that
Question:8
Find which of the functions is continuous or discontinuous at the indicated points:
at x = 0
Answer:
Given,
We need to check its continuity at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 4 if -
Similarly, we proceed for RHL-
And,
f(0) = 0 {using eqn 1} …(4)
from equation 2 , 3 and 4 we can conclude that
∴ f (x) is discontinuous at x = 0
Question:9
Find which of the functions is continuous or discontinuous at the indicated points:
Answer:
Given,
We need to check its continuity at x = 1
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 1 if -
Similarly, we proceed for RHL-
And,
Question:10
Find which of the functions is continuous or discontinuous at the indicated points:
at x = 1
Answer:
Given,
We need to check its continuity at x = 1
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 1 if -
Then,
Similarly, we proceed for RHL-
From equation 2,3 and 4 we can conclude that
F(x) is continuous at x=1
Question:11
Find the value of k so that the function f is continuous at the indicated point:
Answer:
Given,
We need to find the value of k such that f(x) is continuous at x = 5.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 5.
As we have to find k so pick out a combination so that we get k in our equation.
In this question we take LHL = f(5)
Question:12
Find the value of k so that the function f is continuous at the indicated point:
Answer:
Given,
We need to find the value of k such that f(x) is continuous at x = 2.
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 2.
Now to find k we have to pick out a combination so that we get k in our equation.
In this question we take LHL = f(5)
As the limit can't be evaluated directly as it is taking 0 / 0 form.
So, use the formula:
Divide the numerator and denominator by -h to match with the form in formula-
Question:13
Find the value of k so that the function f is continuous at the indicated point:
at x= 0
Answer:
Given,
We need to find the value of k such that f(x) is continuous at x = 0.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 0.
Now to find k pick out a combination using which we get k in our equation.
In this question we take LHL = f(0)
We can’t find the limit directly, because it is taking 0/0 form.
thus, we will rationalize it.
Question:14
Find the value of k so that the function f is continuous at the indicated point:
at x= 0
Answer:
Given,
We need to find the value of k such that f(x) is continuous at x = 0.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically, we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 0.
to find k we have to pick out a combination so that we get k in our equation.
In this question we take LHL = f(0)
As this limit can be evaluated directly by putting value of h because it is taking indeterminate form (0/0)
Thus, we use sandwich or squeeze theorem according to which -
Dividing and multiplying by to match the form in formula we have-
Question:15
Prove that the function f defined by
remains discontinuous at x=0, regardless the choice of k.
Answer:
Given,
We need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically, we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now, we need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k
If we show that,
Then there will not be involvement of k in the equation & we can easily prove it.
So let’s take LHL first -
Now Let's find RHL,
From the equation 2 and 3, conclude that
LHL ≠ RHL
Hence,
f(x) is discontinuous at x = 0 irrespective of the value of k.
Question:16
Find the values of a and b such that the function f defined by
is a continuous function at x = 4.
Answer:
Given,
…(1)
We need to find the value of a & b such that f(x) is continuous at x = 4.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 4.
to find a & b, we have to pick out a combination so that we get a or b in our equation.
In this question first we take LHL = f(4)
{using equation 1}
h > 0 as defined in theory above.
a - 1 = a + b
b = -1
Now, taking other combination,
RHL = f(4)
{using equation 1}
h > 0 as defined in theory above.
|h| = h
⇒ b + 1 = a + b
∴ a = 1
Hence,
a = 1 and b = -1
Question:17
Given the function . Find the point of discontinuity of the composite function y = f(f(x)).
Answer:
Given,
we have to find: Points discontinuity of composite function f(f(x))
As f(x) is not defined at x = -2 as denominator becomes 0, at x = -2.
x = -2 is a point of discontinuity
And f(f(x)) is not defined at x = -5/2 as denominator becomes 0, at x = -5/2.
∴ x = -5/2 is another point of discontinuity
Thus f (f(x)) has 2 points of discontinuity at x = -2 and x = -5/2
Question:18
Find all points of discontinuity of the function .
Answer:
We have to find: Points discontinuity of function f(t) where
As t is not defined at x = 1 as denominator becomes 0, at x = 1.
x = 1 is a point of discontinuity
f(t) =
The f(t) is not going to be defined whenever denominator is 0 and thus will give a point of discontinuity.
∴ Solution of the following equation gives other points of discontinuities.
Hence,
f(t) is discontinuous at x = 1, x = 2 and x = 1/2
Question:19
Show that the function f(x) = |sin x + cos x| is continuous at x = .
Answer:
Given,
We need to prove that f(x) is continuous at x = π
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = π if -
Now,
LHL =
⇒ LHL = {using eqn 1}
Similarly, we proceed for RHL-
{using eqn 1}
Now from equation 2, 3 and 4 we can conclude that
∴ f(x) is continuous at x = π is proved
Question:20
Examine the differentiability of f, where f is defined by
.
Answer:
Given,
…(1)
We need to check whether f(x) is continuous and differentiable at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative (LHD at x = c) = Right hand derivative (RHD at x = c) = f(c).
Mathematically we can represent it as-
Finally, we can state that for a function to be differentiable at x = c
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 2 if -
Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].
E.g. [1.29] = 1; [-4.65] = -5 ; [9] = 9
[2-h] is just less than 2 say 1.9999 so [1.999] = 1
⇒ LHL = (2-0) ×1
∴ LHL = 2 …(2)
Similarly,
RHL =
⇒ RHL =
And, f(2) = (2-1)(2) = 2 …(4) {using equation 1}
From equation 2,3 and 4 we observe that:
∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory-
f(x) is differentiable at x = 2 if -
LHD =
⇒ LHD =
Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].
E.g. [1.29] = 1 ; [-4.65] = -5 ; [9] = 9
[2-h] is just less than 2 say 1.9999 so [1.999] = 1
⇒ LHD =
⇒ LHD =
Now,
RHD =
⇒ RHD =
⇒ RHD =
∴ RHD =
Clearly from equation 5 and 6, we can conclude that-
(LHD at x=2) ≠ (RHD at x = 2)
∴ f(x) is not differentiable at x = 2
Question:21
Examine the differentiability of f, where f is defined by
Answer:
Given,
We need to check whether f(x) is continuous and differentiable at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-
Finally, we can state that for a function to be differentiable at x = c
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 0 if -
As sin (-1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = × (finite value) = 0
∴ LHL = 0 …(2)
Similarly,
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHL = (finite value) = 0 …(3)
And, f(0) = 0 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:
∴ f(x) is continuous at x = 0. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory-
f(x) is differentiable at x = 0 if -
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHD = 0×(some finite value) = 0
∴ LHD = 0 …(5)
Now,
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHD = 0×(some finite value) = 0
∴ RHD = 0 …(6)
Clearly from equation 5 and 6, we can conclude that-
(LHD at x=0) = (RHD at x = 0)
∴ f(x) is differentiable at x = 0
Question:22
Examine the differentiability of f, where f is defined by
Answer:
Given,
We need to check whether f(x) is continuous and differentiable at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
.
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-
Finally, we can state that for a function to be differentiable at x = c
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 2 if -
And, f(2) = 1 + 2 = 3 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:
∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory-
f(x) is differentiable at x = 2 if -
Clearly from equation 5 and 6,we can conclude that-
(LHD at x=2) ≠ (RHD at x = 2)
∴ f(x) is not differentiable at x = 2
Question:23
Show that f(x) = |x - 5| is continuous but not differentiable at x = 5.
Answer:
Given,
f(x) = |x - 5| …(1)
We need to prove that f(x) is continuous but not differentiable at x = 5
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-
Finally, we can state that for a function to be differentiable at x = c
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 5 if -
And, f(5) = |5-5| = 0 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:
∴ f(x) is continuous at x = 5. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory
f(x) is differentiable at x = 2 if -
As h > 0 as defined in theory above.
∴ |-h| = h
Now,
As h>0 as defined in theory above.
Clearly from equation 5 and 6,we can conclude that-
(LHD at x=5) ≠ (RHD at x = 5)
∴ f(x) is not differentiable at x = 5 but continuous at x = 5.
Hence proved.
Question:24
Answer:
Given f(x) is differentiable at x = 0 and f(x) ≠ 0
And f(x + y) = f(x)f(y) also f’(0) = 2
To prove: f’(x) = 2f(x)
As we know that,
As
f(x + y) = f(x)f(y)
put x = y = 0
Question:25
Differentiate each of the following w.r.t. x
Answer:
Given:
Let Assume
Now, Taking Log on both sides we get,
Now, substitute the value of y
Question:26
Differentiate each of the following w.r.t. x
Answer:
We have been given
Let us Assume
Now, taking log on both sides, we get
since we know,
since we know,
Now, Differentiate w.r.t x
Hence,
Question:27
Differentiate each of the following w.r.t. x
Answer:
Given:
Now, Differentiate w.r.t t
And,
Now, differentiate w.r.t x
Now, using chain rule, we get
Substitute the value of t
Question:28
Differentiate each of the following w.r.t. x
Answer:
Differentiate the given function w.r.t x
Let Assume
Let
Let Assume
Let Assume
Differentiate both side w.r.t x
Now, Bu using chain rule we get, Differentiation of
Hence, This the differentiation of given function.
Question:31
Differentiate each of the following w.r.t. x
Answer:
We have given
Let us Assume
And
Now, differentiate w.r.t v
And,
Now, again differentiate w.r.t. w
And, we know,
So, differentiate w w.r.t. x we get
Now, using chain rule we get,
Substitute the value of v and w
Hence, dy/dy is the differentiation of function.
Question:32
Differentiate each of the following w.r.t. x
Answer:
Let us Assume
Now, differentiate y w.r.t x
Question:33
Differentiate each of the following w.r.t. x
Answer:
Given
Let us assume
Substitute the value of t
Hence, this is the differentiation of .
Question:34
Differentiate each of the following w.r.t. x
Answer:
Given:
To Find: Differentiate w.r.t x
We have
Let
Now, Taking Log on both sides, we get
Log y = cos x.log(sin x)
Now, Differentiate both side w.r.t. x
Substitute the value of y, we get
Question:35
Differentiate each of the following w.r.t. x
Answer:
It is given
Taking log both side, we get
Differentiate w.r.t x
Question:36
Differentiate each of the following w.r.t. x
Answer:
We have given,
Let us Assume,
Taking log both side
Differentiate w.r.t x
Question:39
Differentiate each of the following w.r.t. x
Answer:
We have given
Let us Assume (sec x +tan x) =t
So,
Now, differentiate w.r.t t
Question:40
Differentiate each of the following w.r.t. x
Answer:
We have given
Now, divide by cos x in both numerator and denominator
Question:44
Find of each of the functions expressed in parametric form in
Answer:
We have given, two parametric equation,
Now, differentiate both equation w.r.t x
We know,
So,
and,
Now,
Question:45
Find of each of the functions expressed in parametric form
Answer:
We have two equations
Now, differentiate w.r.t θ
So,
Also,
Question:46
Find dy/dx of each of the functions expressed in parametric form in
Answer:
We have given,
Now, differentiate both the equation w.r.t. x then we get
Question:47
Find dy/dx of each of the functions expressed in parametric form in
Answer:
We have given two parametric equation:
Let us Assume t= tan θ
Question:48
Find dy/dx of each of the functions expressed in parametric form
Answer:
We have given,
Now, differentiate w.r.t t
Also,
Question:49
Answer:
Taking log on both sides to get,
Taking log on both sides we get,
log y = sin 2t
Differentiating w.r.t x,
Question:50
Answer:
Differentiate w.r.t t
Also,
y = bcos2t
Now, for dy/dx
Hence Proved.
Question:53
Answer:
We have
Let us Assume,
Hence Differentiation of w.r.t. is .
Question:54
Find when x and y are connected by the relation given
Answer:
We have,
Use chain rule and quotient rule to get:
Chain Rule
f(x) = g(h(x))
f’(x) = g’(h(x))h’(x)
By Quotient Rule
On differentiating both the sides with respect to x, we get
Question:55
Find when x and y are connected by the relation given
Answer:
We have,
sec(x + y) = xy
By the rules given below:
Chain Rule
f(x) = g(h(x))
f’(x) = g’(h(x))h’(x)
Product rule:
Question:58
Answer:
Given:
Differentiating the above with respect to x, we get
Now, we again differentiate eq (i) with respect to y, we get,
Now, multiplying Eq. (ii) and (iii), we get
Hence Proved
Question:61
Answer:
Given:
⇒ y = (cos x)^{y}
Taking log both the sides, we get
log y = y log(cos x)
On differentiating both the sides, we get
Hence Proved
Question:63
Answer:
Given:
Now, we know that
Now, we use some trigonometry formulas,
Hence Proved
Question:65
Verify the Rolle’s theorem for each of the functions
in [0, 1].
Answer:
Given:
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
On expanding
Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all
Hence, condition 1 is satisfied.
Condition 2:
Since, f(x) is a polynomial and every polynomial function is differentiable for all
Hence, condition 2 is satisfied.
Condition 3:
Hence, f(0) = f(1)
Hence, condition 3 is also satisfied.
Now, let us show that such that f’(c) = 0
On differentiating above with respect to x, we get
Put x = c in above equation, we get
, all the three conditions of Rolle’s theorem are satisfied
f’(c) = 0
On factorising, we get
Thus, Rolle’s theorem is verified.
Question:66
Verify the Rolle’s theorem for each of the functions
Answer:
Given:
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
Since, f(x) is a trigonometric function and trigonometric function is continuous everywhere
Hence, condition 1 is satisfied.
Condition 2:
On differentiating above with respect to x, we get
f(x) is differentiable at
Hence, condition 2 is satisfied.
Condition 3:
Hence, condition 3 is also satisfied.
Now, let us show that c ∈ such that f’(c) = 0
Put x = c in above equation, we get
all the three conditions of Rolle’s theorem are satisfied
f’(c) = 0
or Now, cos 2c = 0
Thus, Rolle’s theorem is verified.
Question:67
Verify the Rolle’s theorem for each of the functions
in [- 1, 1].
Answer:
Given:
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
Since, f(x) is a logarithmic function and logarithmic function is continuous for all values of x.
Hence, condition 1 is satisfied.
Condition 2:
Hence, condition 2 is satisfied.
Condition 3:
Hence, condition 3 is also satisfied.
Now, let us show that such that f(c)=0
Put x=c in above equation, we get
Question:68
Verify the Rolle’s theorem for each of the functions
in [-3, 0].
Answer:
Given:
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
Since, f(x) is multiplication of algebra and exponential function and is defined everywhere in its domain.
is continuous at x ∈ [-3,0]
Hence, condition 1 is satisfied.
Condition 2:
On differentiating f(x) with respect to x, we get,
⇒ f(x) is differentiable at [-3,0]
Hence, condition 2 is satisfied.
Condition 3:
all the three conditions of Rolle’s theorem are satisfied
f’(c) = 0
Thus, Rolle’s theorem is verified.
Question:69
Verify the Rolle’s theorem for each of the functions
Answer:
Given:
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
Firstly, we have to show that f(x) is continuous.
Here, f(x) is continuous because f(x) has a unique value for each x ∈ [-2,2]
Condition 2:
Now, we have to show that f(x) is differentiable
So, f(x) is differentiable on (-2,2)
Hence, Condition 2 is satisfied.
Condition 3:
Now, we have to show that f(a) = f(b)
so, f(a) = f(-2)
and
Hence, condition 3 is satisfied
Now, let us show that
On differentiating above with respect to x, we get
Put x=c in above equation, we get,
Thus, all the three conditions of Rolle's theorem is satisfied. Now we have to see that there exist such that
Hence, Rolle’s theorem is verified.
Question:70
Discuss the applicability of Rolle’s theorem on the function given by
Answer:
Given:
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
At x = 1
and
Hence, condition 1 is satisfied.
Condition 2:
Now, we have to check f(x) is differentiable
On differentiating with respect to x, we get
or
Now, let us consider the differentiability of f(x) at x = 1
LHD ⇒ f(x) = 2x = 2(1) = 2
RHD ⇒ f(x) = -1 = -1
LHD ≠ RHD
∴ f(x) is not differentiable at x = 1
Thus, Rolle’s theorem is not applicable to the given function.
Question:71
Find the points on the curve y = (cosx - 1) in [0, ], where the tangent is parallel to x-axis.
Answer:
Given: Equation of curve, y = cos x - 1
Firstly, we differentiate the above equation with respect to x, we get
Given tangent to the curve is parallel to the x - axis
This means, Slope of tangent = Slope of x - axis
Hence, the tangent to the curve is parallel to the x -axis at
(π, -2)
Question:72
Answer:
Given: y = x(x - 4)
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
On expanding we get
since, is a polynomial and we know that, every polynomial function is continuous for all
Hence, condition 1 is satisfied.
Condition 2
is differentiable at [0,4]
Hence, condition 2 is satisfied.
Condition 3:
Hence, condition 3 is also satisfied.
Now, there is atleast one value of c ∈ (0,4)
Given tangent to the curve is parallel to the x - axis
This means, Slope of tangent = Slope of x - axis
Hence, the tangent to the curve is parallel to the x -axis at (2, -4).
Question:73
Verify mean value theorem for each of the functions given
Answer:
Given:
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
Here,
On differentiating above with respect to x, we get
Hence, f(x) is differentiable in (1,4)
We know that,
Differentiability Continuity
Hence, f(x) is continuous in (1,4)
Thus, Mean Value Theorem is applicable to the given function
Now,
Now, let us show that there exist c ∈ (0,1) such that
but
So, value of
Thus, Mean Value Theorem is verified.
Question:74
Verify mean value theorem for each of the functions given
Answer:
Given:
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
Condition 1:
Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x ∈ R
is continuous at x ∈ [0,1]
Hence, condition 1 is satisfied.
Condition 2:
Since, f(x) is a polynomial and every polynomial function is differentiable for all x ∈ R
⇒ f(x) is differentiable at [0,1]
Hence, condition 2 is satisfied.
Thus, Mean Value Theorem is applicable to the given function
Now,
On differentiating above with respect to x, we get
Put x=c in above equation, we get
By Mean Value Theorem,
Thus, Mean Value Theorem is verified.
Question:75
Verify mean value theorem for each of the functions given
in
Answer:
Given:
f(x) = sinx - sin2x in [0,π]
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
Condition 1:
f(x) = sinx - sin 2x
Since, f(x) is a trigonometric function and we know that, sine function are defined for all real values and are continuous for all x ∈ R
⇒ f(x) = sinx - sin 2x is continuous at x ∈ [0,π]
Hence, condition 1 is satisfied.
Condition 2:
f(x) = sinx - sin 2x
f’(x) = cosx - 2 cos2x
⇒ f(x) is differentiable at [0,π]
Hence, condition 2 is satisfied.
Thus, Mean Value Theorem is applicable to the given function
Now,
Now, let us show that there exist c ∈ (0,1) such that
On differentiating above with respect to x, we get
Put x=c in above equation, we get
f’(c) = cos(c) - 2cos2c …(i)
By Mean Value Theorem,
Now, to find the factors of the above equation, we use
Thus, Mean Value Theorem is verified.
Question:76
Verify mean value theorem for each of the functions given
in [1,5]
Answer:
Given:
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
Condition 1:
Firstly, we have to show that f(x) is continuous.
Here, f(x) is continuous because f(x) has a unique value for each x ∈ [1,5]
Condition 2:
Now, we have to show that f(x) is differentiable
∴ f’(x) exists for all x ∈ (1,5)
So, f(x) is differentiable on (1,5)
Hence, Condition 2 is satisfied.
Thus, mean value theorem is applicable to given function.
Now,
Now, we will find f(a) and f(b)
and
Now, let us show that such that
On differentiating above with respect to x, we get
Put x=c in above equation, we get
By Mean Value theorem,
Hence, Mean Value Theorem is verified.
Question:77
Answer:
Given: Equation of curve,
Firstly, we differentiate the above equation with respect to x, we get
Given tangent to the curve is parallel to the chord joining the points (3, Q) and (4,1)
i.e
Put in , we have
Hence, the tangent to the curve is parallel to chord joining the points (3,0) and (4,1) at
Question:78
Answer:
Given: in [1,2]
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
Condition 1:
since, f(x) is a polynomial and we know that, every polynomial function is continuous for all
is continuous at
Hence, condition 1 is satisfied.
Condition 2
since, f(x) is a polynomial and every polynomial function is differentiable for all
is differentiable at [1, 2]
Hence, condition 2 is satisfied.
Thus, Mean Value Theorem is applicable to the given function.
Now,
Then, there exist such that
Put x=c in equation, we get
By Mean Value Theorem,
So, value of
Thus, Mean Value Theorem is verified.
Put in given equation we have
Question:79
Find the values of p and q so that
Is differentiable at x = 1.
Answer:
Given that,
Is differentiable at x = 1.
We know that, f(x) is differentiable at x =1 ⇔ Lf’(1) = Rf’(1).
Since, Lf’(1) = Rf’(1)
∴ 5 = q (i)
Now, we know that if a function is differentiable at a point,it is necessarily continuous at that point.
⇒ f(x) is continuous at x = 1.
⇒ f(1-) = f(1+) = f(1)
⇒ 1+3+p = q+2 = 1+3+p
⇒ p-q = 2-4 = -2
⇒ q-p = 2
Now substituting the value of ‘q’ from (i), we get
⇒ 5-p = 2
⇒ p = 3
∴ p = 3 and q = 5
Question:80
A.If prove that
B. If prove that
Answer:
A.
We have,
Taking log on both sides, we get
Taking log on both sides, we get
Differentiating both sides w.r.t x, we get
Hence proved.
B.
We have,
Hence Proved
Question:82
Answer:
We have,
Putting
Taking log on both sides, we get
Differentiating w.r.t x, we get
Now,
Differentiating w.r.t x, we get
Now, y=u+v
Question:83
If f(x) = 2x and g(x) = then which of the following can be a discontinuous function.
A. f(x) + g(x)
B. f(x) - g(x)
C. f(x) . g(x)
D.
Answer:
We know that if two functions f(x) and g(x) are continuous then {f(x) +g(x)},{f(x)-g(x)}, {f(x).g(x)} and are continuous.
Since, f(x) = 2x and are polynomial functions, they are continuous everywhere.
⇒ {f(x) +g(x)},{f(x)-g(x)}, {f(x).g(x)} are continuous functions.
for,
now, f(x) = 0
⇒ 4x = 0
⇒ x = 0
∴ is discontinuous at x=0.
Question:90
Let . Then
A. f is everywhere differentiable
B. f is everywhere continuous but not differentiable
C. f is everywhere continuous but not differentiable at
D. None of these
Answer:
B)
Given that,
Let g(x) = sinx and h(x) = |x|
Then, f(x) = hog(x)
We know that, modulus function and sine function are continuous everywhere.
Since, composition of two continuous functions is a continuous function.
Hence, f(x) = hog(x) is continuous everywhere.
Now, v(x)=|x| is not differentiable at x=0.
So, f(x) is not differentiable where
We know that at
Hence, f(x) is everywhere continuous but not differentiable
Question:95
The value of c in Rolle’s theorem for the function in the interval is
A. 1
B.-1
C.
D.
Answer:
A)
Question:96
For the function the value of c for mean value theorem is
A. 1
B.
C. 2
D. None of these
Answer:
B)
Mean Value Theorem states that, Let f : [a, b] → R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that
Since, f(x) is a polynomial function it is continuous on [1,3] and differentiable on (1,3).
Now, as per Mean value Theorem, there exists at least one c ∈ (1,3), such that
Question:97
Answer:
Consider, f(x) = |x-1| + |x-2|
Let’s discuss the continuity of f(x).
We have, f(x) = |x-1| + |x-2|
When x<1, we have f(x) = -2x+3, which is a polynomial function and polynomial function is continuous everywhere.
When 1≤x<2, we have f(x) = 1, which is a constant function and constant function is continuous everywhere.
When x≥2, we have f(x) = 2x-3, which is a polynomial function and polynomial function is continuous everywhere.
Hence, f(x) = |x-1| + |x-2| is continuous everywhere.
Let’s discuss the differentiability of f(x) at x=1 and x=2.
We have
⇒ f(x) is not differentiable at x=2.
Thus, f(x) = |x-1| + |X-2| is continuous everywhere but fails to be differentiable exactly at two points x=1 and x=2.
Question:102
Answer:
False
As per Rolle’s Theorem, Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers. Then there exists some c in (a, b) such that f’(c) = 0.
We have, f(x) = |x - 1| in [0, 2].
Since, polynomial and modulus functions are continuous everywhere f(x) is continuous
Now, x-1=0
⇒ x=1
We need to check if f(x) is differentiable at x=1 or not.
We have,
⇒ Lf’(1) ≠ Rf’(1)
⇒ f(x) is not differentiable at x=1.
Hence, Rolle’s theorem is not applicable on f(x) since it is not differentiable at x=1 ∈ (0,2).
Question:103
Answer:
True.
Given that, f is continuous on its domain D.
Let a be an arbitrary real number in D. Then f is continuous at a.
Now,
If |f| is continuous at x=a.
since a is an arbitrary point in D. Therefore |f| is continuous in D.
Question:104
Answer:
True.
Let f be a function defined by f(x) = |1-x + |x||.
Let g(x) = 1-x + |x| and h(x) = |x| be two functions defined on R.
Then,
hog(x) = h(g(x))
⇒ hog(x) = h(1-x + |x|)
⇒ hog(x) = |(1-x + |x|)|
⇒ hog(x) = f(x) ∀ x ∈ R.
Since (1-x) is a polynomial function and |X| is modulus function are continuous in R.
⇒ g(x) = 1-x + |x| is everywhere continuous.
⇒ h(x) = |x| is everywhere continuous.
Hence, f = hog is everywhere continuous.
Question:106
Answer:
False
Let and
which is a constant function and continuous everywhere.
But, is discontinuous at
Students can use NCERT Exemplar Class 12 Math solutions chapter 5 PDF download, prepared by experts, for better understanding of concepts and topics of probability. The topics and subtopics are mentioned below.
The sub-topics covered in this chapter are:
In Class 12 Math NCERT exemplar solutions chapter 5, students will learn about continuity, differential, limits and their properties in detail. Other than learning about continuity and discontinuity in detail, one will also cover Heine's definition, Cauchy's theorem, and algebra of functions. In a separate subtopic of differentiability, one will learn about differentiability concept, fundamentals, derivatives and its types, relations between differentiability and continuity. NCERT exemplar solutions for Class 12 Math chapter 5 will also cover implicit functions, composite functions, inverse trigonometric functions and its derivatives.
NCERT exemplar class 12 Math solutions chapter 5 also includes logarithmic and exponential functions, their concept and their differentiability. The topics also cover logarithmic differentiation in detail and all its fundamentals. One will also get a closer look at parametric forms of functions and their derivatives and how to solve the questions. Other than all this, one will learn two of the most major advanced calculus theorems, Mean Value Theorem and Rolle's Theorem.
We have a team of highly experienced teachers, who have solved the NCERT exemplar Class 12 Math solutions chapter 5 continuity and differentiability. Our teachers have solved the questions in the simplest way; so that every student can understand the answer in a single get-go. The answers are exhaustive, meaning every step is mentioned as per the exam requirement. No shortcuts and no tricks are used in solving the questions, as per the NCERT syllabus. One will learn some of the most crucial theorems of calculus mathematics. Better understanding will help in higher education and also in getting a clear idea about advanced calculus.
Continuity and differentiability is a chapter, which every student should be well prepared for. NCERT exemplar Class 12 Math chapter 5 solutions not only helps in understanding calculus better but also helps in other subjects like physics. This chapter explains the base of the continuity function and differentiability function in detail.
In NCERT exemplar Class 12 Math solutions chapter 5, students will learn that Continuity is characteristic of the function. This characteristic shows that the function can have an unbroken and continuous wave. Differentiability, on the other hand, is a function, which shows that the derivative of the domain exists at every single point. Differentiability shows that there can be a slope to the graph at each point.
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | Continuity and Differentiability |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 |
Yes, you can download the NCERT exemplar Class 12 Maths chapter 5 solutions from the link given in the page.
Yes, the NCERT exemplar solutions for Class 12 Maths chapter 5 are prepared to help students prepare well for board exams.
You can prepare notes for this chapter by highlighting or underlining the important points which will make it easier for you to read for quick revision.
The chapter has only 1 exercise with 107 questions in total that are all problem based with variable weightage.
Application Date:09 September,2024 - 14 November,2024
Application Date:09 September,2024 - 14 November,2024
Admit Card Date:04 October,2024 - 29 November,2024
Admit Card Date:04 October,2024 - 29 November,2024
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:
Re-evaluate Your Study Strategies:
Consider Professional Help:
Explore Alternative Options:
Focus on NEET 2025 Preparation:
Seek Support:
Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.
I hope this information helps you.
Hi,
Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.
Scholarship Details:
Type A: For candidates scoring 60% or above in the exam.
Type B: For candidates scoring between 50% and 60%.
Type C: For candidates scoring between 40% and 50%.
Cash Scholarship:
Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).
Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.
Hope you find this useful!
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
Hello Akash,
If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.
You can get the Previous Year Questions (PYQs) on the official website of the respective board.
I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.
Thank you and wishing you all the best for your bright future.
Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide
Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 15th NOV'24! Trusted by 3,500+ universities globally
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE