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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:
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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.
hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
Hello dear,
If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.
As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.
Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.
Believe in Yourself! You can make anything happen
All the very best.
Hello Student,
I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.
You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.
All the best.
If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.
Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.
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The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.
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A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.
An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.
Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.
Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities.
Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.
Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.
Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.
A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability.
Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.
A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.
Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.
An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.
The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.
The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.
For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.
Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.
Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.
Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.
Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.
A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.
The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.
A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.
A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.
In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.
Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.
For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.
In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.
Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.
Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.
Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).
Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian.
The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.
Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.
A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.
A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.
A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment.
An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.
An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.
Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack
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