Chapter 5 Continuity and Differentiability
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Exercise 5.1
Question 1:
Prove that the function is continuous at
Answer
Therefore, fis continuous atx= 0
Therefore, f is continuous atx= 3
Therefore, fis continuous atx= 5
Question 2:
Examine the continuity of the function .
Answer
Thus, fis continuous atx= 3
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Question 3:
Examine the following functions for continuity.
(a) (b)
(c) (d)
Answer
(a) The given function is
It is evident that fis defined at every real number kand its value at kis k 5.
It is also observed that,
Hence,f is continuous at every real number and therefore, it is a continuous function.
(b) The given function is
For any real number k 5, we obtain
Hence, fis continuous at every point in the domain of fand therefore, it is a continuous
function.
(c) The given function is
For any real number c 5, we obtain
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Hence, fis continuous at every point in the domain of fand therefore, it is a continuous
function.
(d) The given function is
This function fis defined at all points of the real line.
Let cbe a point on a real line. Then, c< 5 or c= 5 or c> 5
Case I: c< 5
Then, f (c) = 5 c
Therefore, fis continuous at all real numbers less than 5.
Case II : c= 5
Then,
Therefore, f is continuous atx= 5
Case III: c> 5
Therefore, fis continuous at all real numbers greater than 5.
Hence,f is continuous at every real number and therefore, it is a continuous function.
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Question 4:
Prove that the function is continuous atx= n, where nis a positive integer.
Answer
The given function is f(x)=xn
It is evident that fis defined at all positive integers, n, and its value at nis nn.
Therefore, f is continuous at n, where nis a positive integer.
Question 5:
Is the function fdefined by
continuous atx= 0? Atx= 1? Atx= 2?
Answer
The given function fis
Atx= 0,
It is evident thatf is defined at 0 and its value at 0 is 0.
Therefore, fis continuous atx= 0
Atx= 1,
f is defined at 1 and its value at 1 is 1.
The left hand limit off atx= 1 is,
The right hand limit of f atx= 1 is,
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Therefore,f is not continuous atx= 1
Atx = 2,
f is defined at 2 and its value at 2 is 5.
Therefore, fis continuous atx = 2
Question 6:
Find all points of discontinuity of f, where fis defined by
Answer
The given function fis
It is evident that the given function fis defined at all the points of the real line.Let cbe a point on the real line. Then, three cases arise.
(i) c< 2
(ii) c> 2
(iii) c= 2
Case (i) c< 2
Therefore, fis continuous at all pointsx, such thatx< 2
Case (ii) c> 2
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Therefore, fis continuous at all pointsx, such thatx> 2
Case (iii) c= 2
Then, the left hand limit of f atx = 2 is,
The right hand limit of fatx= 2 is,
It is observed that the left and right hand limit of fatx= 2 do not coincide.
Therefore, fis not continuous atx= 2
Hence,x= 2 is the only point of discontinuity of f.
Question 7:
Find all points of discontinuity of f, where fis defined by
Answer
The given function fis
The given function fis defined at all the points of the real line.
Let cbe a point on the real line.
Case I:
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Therefore, fis continuous at all pointsx, such thatx< 3
Case II:
Therefore, fis continuous atx= 3
Case III:
Therefore, fis continuous in (3, 3).
Case IV:
If c= 3, then the left hand limit of f atx = 3 is,
The right hand limit of f atx = 3 is,
It is observed that the left and right hand limit of fatx= 3 do not coincide.
Therefore, fis not continuous atx= 3
Case V:
Therefore, fis continuous at all pointsx, such thatx> 3
Hence,x= 3 is the only point of discontinuity of f.
Question 8:
Find all points of discontinuity of f, where fis defined by
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Answer
The given function fis
It is known that,
Therefore, the given function can be rewritten as
The given function fis defined at all the points of the real line.
Let cbe a point on the real line.
Case I:
Therefore, fis continuous at all pointsx< 0
Case II:
If c= 0, then the left hand limit of f atx = 0 is,
The right hand limit of f atx = 0 is,
It is observed that the left and right hand limit of fatx= 0 do not coincide.
Therefore, fis not continuous atx= 0
Case III:
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Therefore, fis continuous at all pointsx, such thatx> 0
Hence,x= 0 is the only point of discontinuity of f.
Question 9:
Find all points of discontinuity of f, where fis defined by
Answer
The given function fis
It is known that,
Therefore, the given function can be rewritten as
Let cbe any real number. Then,
Also,
Therefore, the given function is a continuous function.
Hence, the given function has no point of discontinuity.
Question 10:
Find all points of discontinuity of f, where fis defined by
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Answer
The given function fis
The given function fis defined at all the points of the real line.
Let cbe a point on the real line.
Case I:
Therefore, fis continuous at all pointsx, such thatx< 1
Case II:
The left hand limit of f atx = 1 is,
The right hand limit of f atx = 1 is,
Therefore, fis continuous atx= 1
Case III:
Therefore, fis continuous at all pointsx, such thatx> 1Hence, the given function f has no point of discontinuity.
Question 11:
Find all points of discontinuity of f, where fis defined by
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Answer
The given function fis
The given function fis defined at all the points of the real line.
Let cbe a point on the real line.
Case I:
Therefore, fis continuous at all pointsx, such thatx< 2
Case II:
Therefore, fis continuous atx= 2
Case III:
Therefore, fis continuous at all pointsx, such thatx> 2
Thus, the given function fis continuous at every point on the real line.
Hence, f has no point of discontinuity.
Question 12:
Find all points of discontinuity of f, where fis defined by
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Answer
The given function fis
The given function fis defined at all the points of the real line.
Let cbe a point on the real line.
Case I:
Therefore, fis continuous at all pointsx, such thatx< 1
Case II:
If c= 1, then the left hand limit of fatx= 1 is,
The right hand limit of fatx = 1 is,
It is observed that the left and right hand limit of fatx= 1 do not coincide.
Therefore, fis not continuous atx= 1
Case III:
Therefore, fis continuous at all pointsx, such thatx> 1
Thus, from the above observation, it can be concluded thatx= 1 is the only point ofdiscontinuity of f.
Question 13:
Is the function defined by
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a continuous function?
Answer
The given function is
The given function fis defined at all the points of the real line.
Let cbe a point on the real line.
Case I:
Therefore, fis continuous at all pointsx, such thatx< 1
Case II:
The left hand limit of f atx= 1 is,
The right hand limit of fatx = 1 is,
It is observed that the left and right hand limit of fatx= 1 do not coincide.
Therefore, fis not continuous atx= 1
Case III:
Therefore, fis continuous at all pointsx, such thatx> 1Thus, from the above observation, it can be concluded thatx= 1 is the only point of
discontinuity of f.
Question 14:
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Discuss the continuity of the function f, where fis defined by
Answer
The given function is
The given function is defined at all points of the interval [0, 10].
Let cbe a point in the interval [0, 10].Case I:
Therefore, fis continuous in the interval [0, 1).
Case II:
The left hand limit of f atx= 1 is,
The right hand limit of fatx = 1 is,
It is observed that the left and right hand limits of fatx= 1 do not coincide.
Therefore, fis not continuous atx= 1
Case III:
Therefore, fis continuous at all points of the interval (1, 3).
Case IV:
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The left hand limit of f atx= 3 is,
The right hand limit of fatx = 3 is,
It is observed that the left and right hand limits of fatx= 3 do not coincide.
Therefore, fis not continuous atx= 3
Case V:
Therefore, fis continuous at all points of the interval (3, 10].
Hence, f is not continuous atx = 1 andx = 3
Question 15:
Discuss the continuity of the function f, where fis defined by
Answer
The given function is
The given function is defined at all points of the real line.
Let cbe a point on the real line.
Case I:
Therefore, fis continuous at all pointsx, such thatx < 0
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Case II:
The left hand limit of f atx= 0 is,
The right hand limit of fatx = 0 is,
Therefore, fis continuous atx= 0
Case III:
Therefore, fis continuous at all points of the interval (0, 1).
Case IV:
The left hand limit of f atx= 1 is,
The right hand limit of fatx = 1 is,
It is observed that the left and right hand limits of fatx= 1 do not coincide.
Therefore, fis not continuous atx= 1
Case V:
Therefore, fis continuous at all pointsx, such thatx> 1Hence, f is not continuous only atx = 1
Question 16:
Discuss the continuity of the function f, where fis defined by
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Answer
The given function fis
The given function is defined at all points of the real line.
Let cbe a point on the real line.
Case I:
Therefore, fis continuous at all pointsx, such thatx < 1
Case II:
The left hand limit of f atx= 1 is,
The right hand limit of fatx = 1 is,
Therefore, fis continuous atx= 1
Case III:
Therefore, fis continuous at all points of the interval (1, 1).
Case IV:
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The left hand limit of f atx= 1 is,
The right hand limit of fatx = 1 is,
Therefore, fis continuous atx= 2
Case V:
Therefore, fis continuous at all pointsx, such thatx> 1
Thus, from the above observations, it can be concluded that fis continuous at all points
of the real line.
Question 17:
Find the relationship between aand bso that the function fdefined by
is continuous atx = 3.
Answer
The given function fis
If fis continuous atx= 3, then
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Therefore, from (1), we obtain
Therefore, the required relationship is given by,
Question 18:
For what value of is the function defined by
continuous atx= 0? What about continuity atx= 1?
Answer
The given function fis
If fis continuous atx= 0, then
Therefore, there is no value of for which fis continuous atx= 0
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Atx= 1,
f(1) = 4x+ 1 = 4 1 + 1 = 5
Therefore, for any values of , fis continuous atx= 1
Question 19:
Show that the function defined by is discontinuous at all integral point.
Here denotes the greatest integer less than or equal tox.Answer
The given function is
It is evident that gis defined at all integral points.
Let nbe an integer.
Then,
The left hand limit of f atx= nis,
The right hand limit of fatx = nis,
It is observed that the left and right hand limits of fatx= ndo not coincide.
Therefore, fis not continuous atx=n
Hence, gis discontinuous at all integral points.
Question 20:
Is the function defined by continuous atx = p?
Answer
The given function is
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It is evident that fis defined atx = p
Therefore, the given function fis continuous atx =
Question 21:
Discuss the continuity of the following functions.
(a) f(x) = sinx+ cosx
(b) f(x) = sinx cosx
(c) f(x) = sinx cos x
Answer
It is known that if g and h are two continuous functions, then
are also continuous.
It has to proved first that g(x) = sinx and h(x) = cosxare continuous functions.Let g (x) = sinx
It is evident that g(x) = sinxis defined for every real number.
Let c be a real number. Putx= c+ h
Ifxc, then h0
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Therefore, gis a continuous function.
Let h(x) = cosx
It is evident that h(x) = cosxis defined for every real number.
Let c be a real number. Putx= c+ h
Ifxc, then h0
h (c) = cos c
Therefore, his a continuous function.
Therefore, it can be concluded that
(a) f(x) = g(x) + h(x) = sinx+ cosxis a continuous function
(b) f(x) = g(x) h(x) = sinx cosxis a continuous function
(c) f(x) = g(x) h(x) = sinx cosxis a continuous function
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Question 22:
Discuss the continuity of the cosine, cosecant, secant and cotangent functions,
Answer
It is known that if g and h are two continuous functions, then
It has to be proved first that g(x) = sinx and h(x) = cosxare continuous functions.Let g (x) = sinx
It is evident that g(x) = sinxis defined for every real number.
Let c be a real number. Putx= c+ h
Ifx c, then h 0
Therefore, gis a continuous function.
Let h(x) = cosx
It is evident that h(x) = cosxis defined for every real number.Let c be a real number. Putx= c+ h
Ifx c, then h 0
h (c) = cos c
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Therefore, h(x) = cosxis continuous function.
It can be concluded that,
Therefore, cosecant is continuous except atx = np, n Z
Therefore, secant is continuous except at
Therefore, cotangent is continuous except atx = np, n Z
Question 23:
Find the points of discontinuity of f, where
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Answer
The given function fis
It is evident that fis defined at all points of the real line.
Let cbe a real number.
Case I:
Therefore, fis continuous at all pointsx, such thatx < 0
Case II:
Therefore, fis continuous at all pointsx, such thatx> 0
Case III:
The left hand limit of fatx= 0 is,
The right hand limit of fatx= 0 is,
Therefore, fis continuous atx= 0
From the above observations, it can be concluded that fis continuous at all points of the
real line.Thus, fhas no point of discontinuity.
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Question 24:
Determine if fdefined by
is a continuous function?
Answer
The given function fis
It is evident that fis defined at all points of the real line.
Let cbe a real number.
Case I:
Therefore, fis continuous at all pointsx 0
Case II:
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Therefore, fis continuous atx= 0
From the above observations, it can be concluded thatfis continuous at every point of
the real line.
Thus, fis a continuous function.
Question 25:
Examine the continuity of f, where fis defined by
Answer
The given function fis
It is evident that fis defined at all points of the real line.
Let cbe a real number.
Case I:
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Therefore, fis continuous at all pointsx, such thatx 0
Case II:
Therefore, fis continuous atx= 0
From the above observations, it can be concluded thatfis continuous at every point of
the real line.
Thus, fis a continuous function.
Question 26:
Find the values of k so that the function fis continuous at the indicated point.
Answer
The given function fis
The given function fis continuous at , if fis defined at and if the value of the f
at equals the limit of fat .
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It is evident that f is defined at and
Therefore, the required value of kis 6.
Question 27:
Find the values of k so that the function fis continuous at the indicated point.
Answer
The given function is
The given function fis continuous atx= 2, if fis defined atx= 2 and if the value of fat
x= 2 equals the limit of fatx= 2
It is evident that f is defined atx= 2 and
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Therefore, the required value of .
Question 28:
Find the values of k so that the function fis continuous at the indicated point.
Answer
The given function is
The given function fis continuous atx= p, if fis defined atx= p and if the value of fat
x= p equals the limit of fatx= p
It is evident that f is defined atx= p and
Therefore, the required value of
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Question 29:
Find the values of k so that the function fis continuous at the indicated point.
Answer
The given function f is
The given function fis continuous atx= 5, if fis defined atx= 5 and if the value of fat
x= 5 equals the limit of fatx= 5
It is evident that f is defined atx= 5 and
Therefore, the required value of
Question 30:
Find the values of aand bsuch that the function defined by
is a continuous function.
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Answer
The given function f is
It is evident that the given function fis defined at all points of the real line.
If fis a continuous function, then fis continuous at all real numbers.
In particular, fis continuous atx = 2 andx = 10
Since fis continuous atx = 2, we obtain
Since fis continuous atx = 10, we obtain
On subtracting equation (1) from equation (2), we obtain
8a= 16
a= 2
By putting a= 2 in equation (1), we obtain
2 2 + b= 5
4 + b= 5
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b= 1
Therefore, the values of aand bfor whichfis a continuous function are 2 and 1
respectively.
Question 31:
Show that the function defined byf (x) = cos (x2) is a continuous function.
Answer
The given function is f (x) = cos (x2)
This function fis defined for every real number and fcan be written as the composition
of two functions as,
f= g o h, where g(x) = cosxand h(x) =x2
It has to be first proved that g (x) = cosxand h(x) =x2are continuous functions.
It is evident that gis defined for every real number.
Let cbe a real number.
Then, g(c) = cos c
Therefore, g(x) = cosxis continuous function.
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h(x) =x2
Clearly, his defined for every real number.
Let kbe a real number, thenh (k) = k2
Therefore, his a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is
continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, is a continuous function.
Question 32:
Show that the function defined by is a continuous function.
Answer
The given function is
This function fis defined for every real number and fcan be written as the composition
of two functions as,
f= g o h, where
It has to be first proved that are continuous functions.
Clearly, gis defined for all real numbers.
Let cbe a real number.Case I:
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Therefore, gis continuous at all pointsx, such thatx < 0
Case II:
Therefore, gis continuous at all pointsx, such thatx> 0
Case III:
Therefore, gis continuous atx= 0
From the above three observations, it can be concluded that gis continuous at all points.
h (x) = cosx
It is evident that h(x) = cosxis defined for every real number.
Let c be a real number. Putx= c+ h
Ifxc, then h0
h (c) = cos c
Therefore, h(x) = cosxis a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is
continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
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Therefore, is a continuous function.
Question 33:
Examine that is a continuous function.
Answer
This function fis defined for every real number and fcan be written as the composition
of two functions as,
f= g o h, where
It has to be proved first that are continuous functions.
Clearly, gis defined for all real numbers.
Let cbe a real number.Case I:
Therefore, gis continuous at all pointsx, such thatx < 0
Case II:
Therefore, gis continuous at all pointsx, such thatx> 0
Case III:
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Therefore, gis continuous atx= 0
From the above three observations, it can be concluded that gis continuous at all points.
h (x) = sinx
It is evident that h(x) = sinxis defined for every real number.
Let c be a real number. Putx= c+ k
Ifxc, then k0
h (c) = sin c
Therefore, his a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is
continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, is a continuous function.
Question 34:
Find all the points of discontinuity of f defined by .
Answer
The given function is
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The two functions, gand h, are defined as
Then, f= g h
The continuity of gand h is examined first.
Clearly, gis defined for all real numbers.
Let cbe a real number.
Case I:
Therefore, gis continuous at all pointsx, such thatx < 0
Case II:
Therefore, gis continuous at all pointsx, such thatx> 0
Case III:
Therefore, gis continuous atx= 0
From the above three observations, it can be concluded that gis continuous at all points.
Clearly, his defined for every real number.
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Let c be a real number.
Case I:
Therefore, his continuous at all pointsx, such thatx < 1
Case II:
Therefore, his continuous at all pointsx, such thatx> 1
Case III:
Therefore, his continuous atx= 1
From the above three observations, it can be concluded thathis continuous at all points
of the real line.
gand hare continuous functions. Therefore, f = g h is also a continuous function.
Therefore, f has no point of discontinuity.
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Exercise 5.2
Question 1:
Differentiate the functions with respect tox.
Answer
Thus, fis a composite of two functions.
Alternate method
Question 2:
Differentiate the functions with respect tox.
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Answer
Thus, f is a composite function of two functions.
Put t= u(x) = sinx
By chain rule,
Alternate method
Question 3:
Differentiate the functions with respect tox.
Answer
Thus, f is a composite function of two functions, uand v.
Put t= u(x) = ax+ b
Hence, by chain rule, we obtain
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Alternate method
Question 4:
Differentiate the functions with respect tox.
Answer
Thus, f is a composite function of three functions, u, v, and w.
Hence, by chain rule, we obtain
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Alternate method
Question 5:
Differentiate the functions with respect tox.
Answer
The given function is , where g(x) = sin (ax+ b) and
h(x) = cos (cx + d)
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g is a composite function of two functions, uand v.
Therefore, by chain rule, we obtain
his a composite function of two functions,pand q.
Put y=p(x) = cx + d
Therefore, by chain rule, we obtain
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Question 6:
Differentiate the functions with respect tox.
Answer
The given function is .
Question 7:
Differentiate the functions with respect tox.
Answer
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Question 8:
Differentiate the functions with respect tox.
Answer
Clearly, f is a composite function of two functions, u andv, such that
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By using chain rule, we obtain
Alternate method
Question 9:
Prove that the function f given by
is notdifferentiable atx= 1.
Answer
The given function is
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It is known that a function fis differentiable at a pointx= cin its domain if both
are finite and equal.
To check the differentiability of the given function atx= 1,
consider the left hand limit of fatx= 1
Since the left and right hand limits of fatx= 1 are not equal, fis not differentiable atx
= 1
Question 10:
Prove that the greatest integer function defined by is not
differentiable atx= 1 andx= 2.
Answer
The given function fis
It is known that a function fis differentiable at a pointx= cin its domain if both
are finite and equal.
To check the differentiability of the given function atx= 1, consider the left hand limit of
fatx= 1
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Since the left and right hand limits of fatx= 1 are not equal, fis not differentiable at
x= 1
To check the differentiability of the given function atx= 2, consider the left hand limit
of fatx= 2
Since the left and right hand limits of fatx= 2 are not equal, fis not differentiable atx
= 2
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Exercise 5.3
Question 1:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
Question 2:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
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Question 3:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
Using chain rule, we obtain and
From (1) and (2), we obtain
Question 4:
Find :
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Answer
The given relationship isDifferentiating this relationship with respect tox, we obtain
Question 5:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
[Derivative of constant function is 0]
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Question 6:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
Question 7:
Find :
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Answer
The given relationship isDifferentiating this relationship with respect tox, we obtain
Using chain rule, we obtain
From (1), (2), and (3), we obtain
Question 8:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
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Question 9:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
The function, , is of the form of .
Therefore, by quotient rule, we obtain
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Also,
From (1), (2), and (3), we obtain
Question 10:
Find :
Answer
The given relationship is
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It is known that,
Comparing equations (1) and (2), we obtain
Differentiating this relationship with respect tox, we obtain
Question 11:
Find :
x Answer
The given relationship is,
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On comparing L.H.S. and R.H.S. of the above relationship, we obtain
Differentiating this relationship with respect tox, we obtain
Question 12:
Find :
Answer
The given relationship is
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Differentiating this relationship with respect tox, we obtain
Using chain rule, we obtain
From (1), (2), and (3), we obtain
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Alternate method
Differentiating this relationship with respect tox, we obtain
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Question 13:
Find :
Answer
The given relationship is
Differentiating this relationship with respect tox, we obtain
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Question 14:
Find :
Answer
lationship is
Differentiating this relationship with respect tox, we obtain
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Question 15:
Find :
Answer
The given relationship is
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Differentiating this relationship with respect tox, we obtain
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Exercise 5.4
Question 1:
Differentiate the following w.r.t.x:
Answer
Let
By using the quotient rule, we obtain
Question 2:
Differentiate the following w.r.t.x:
Answer
Let
By using the chain rule, we obtain
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Question 2:
Show that the function given by f(x) = e2xis strictly increasing on R.
Answer
Let be any two numbers in R.
Then, we have:
Hence, fis strictly increasing on R.
Question 3:
Differentiate the following w.r.t.x:
Answer
Let
By using the chain rule, we obtain
Question 4:
Differentiate the following w.r.t.x:
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Answer
Let
By using the chain rule, we obtain
Question 5:
Differentiate the following w.r.t.x:
Answer
Let
By using the chain rule, we obtain
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Question 6:
Differentiate the following w.r.t.x:
Answer
Question 7:
Differentiate the following w.r.t.x:
Answer
Let
Then,
By differentiating this relationship with respect tox, we obtain
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Question 8:
Differentiate the following w.r.t.x:
Answer
Let
By using the chain rule, we obtain
,x> 1
Question 9:
Differentiate the following w.r.t.x:
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Answer
Let
By using the quotient rule, we obtain
Question 10:
Differentiate the following w.r.t.x:
Answer
Let
By using the chain rule, we obtain
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Exercise 5.5
Question 1:
Differentiate the function with respect tox.
Answer
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
Question 2:
Differentiate the function with respect tox.
Answer
Taking logarithm on both the sides, we obtain
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Differentiating both sides with respect tox, we obtain
Question 3:
Differentiate the function with respect tox.
Answer
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
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Question 4:
Differentiate the function with respect tox.
Answer
u =xx
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
v= 2sinx
Taking logarithm on both the sides with respect tox, we obtain
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Differentiating both sides with respect tox, we obtain
Question 5:
Differentiate the function with respect tox.
Answer
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
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Question 6:
Differentiate the function with respect tox.
Answer
Differentiating both sides with respect tox, we obtain
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Differentiating both sides with respect tox, we obtain
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Therefore, from (1), (2), and (3), we obtain
Question 7:
Differentiate the function with respect tox.
Answer
u = (logx)x
Differentiating both sides with respect tox, we obtain
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Differentiating both sides with respect tox, we obtain
Therefore, from (1), (2), and (3), we obtain
Question 8:
Differentiate the function with respect tox.
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Answer
Differentiating both sides with respect tox, we obtain
Therefore, from (1), (2), and (3), we obtain
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Question 9:
Differentiate the function with respect tox.
Answer
Differentiating both sides with respect tox, we obtain
Differentiating both sides with respect tox, we obtain
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From (1), (2), and (3), we obtain
Question 10:
Differentiate the function with respect tox.
Answer
Differentiating both sides with respect tox, we obtain
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Differentiating both sides with respect tox, we obtain
From (1), (2), and (3), we obtain
Question 11:
Differentiate the function with respect tox.
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Answer
Differentiating both sides with respect tox, we obtain
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Differentiating both sides with respect tox, we obtain
From (1), (2), and (3), we obtain
Question 12:
Find of function.
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Answer
The given function is
Letxy= uand yx= v
Then, the function becomesu + v= 1
Differentiating both sides with respect tox, we obtain
Differentiating both sides with respect tox, we obtain
From (1), (2), and (3), we obtain
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Question 13:
Find of function.
Answer
The given function is
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
Question 14:
Find of function.
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Answer
The given function is
Taking logarithm on both the sides, we obtain
Differentiating both sides, we obtain
Question 15:
Find of function.
Answer
The given function is
Taking logarithm on both the sides, we obtain
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Differentiating both sides with respect tox, we obtain
Question 16:
Find the derivative of the function given by and hence
find .
Answer
The given relationship is
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
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Question 17:
Differentiate in three ways mentioned below
(i) By using product rule.
(ii) By expanding the product to obtain a single polynomial.
(iii By logarithmic differentiation.
Do they all give the same answer?
Answer
(i)
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(ii)
(iii)
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
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By taking logarithm on both sides of the equation , we obtain
Differentiating both sides with respect tox, we obtain
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Exercise 5.6
Question 1:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
Answer
The given equations are
Question 2:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
x= acos , y= bcos
Answer
The given equations arex= acos and y= bcos
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Question 3:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
x= sin t, y= cos 2t
Answer
The given equations arex= sin tand y= cos 2t
Question 4:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
Answer
The given equations are
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Question 5:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
Answer
The given equations are
Question 6:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
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Answer
The given equations are
Question 7:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
Answer
The given equations are
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Question 8:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
Answer
The given equations are
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Question 9:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
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Answer
The given equations are
Question 10:
Ifxand yare connected parametrically by the equation, without eliminating the
parameter, find .
Answer
The given equations are
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Question 11:
If
Answer
The given equations are
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Hence, proved.
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Exercise 5.7
Question 1:
Find the second order derivatives of the function.
Answer
Let
Then,
Question 2:
Find the second order derivatives of the function.
Answer
LetThen,
Question 3:
Find the second order derivatives of the function.
Answer
Let
Then,
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Question 4:
Find the second order derivatives of the function.
Answer
Let
Then,
Question 5:
Find the second order derivatives of the function.
Answer
Let
Then,
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Question 6:
Find the second order derivatives of the function.
Answer
Let
Then,
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Question 7:
Find the second order derivatives of the function.
Answer
Let
Then,
Question 8:
Find the second order derivatives of the function.
Answer
Let
Then,
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Question 9:
Find the second order derivatives of the function.
Answer
Let
Then,
Question 10:
Find the second order derivatives of the function.
Answer
Let
Then,
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Question 11:
If , prove that
Answer
It is given that,
Then,
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Hence, proved.
Question 12:
If find in terms of yalone.
Answer
It is given that,
Then,
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Question 13:
If , show that
Answer
It is given that,
Then,
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Hence, proved.
Question 14:
If show that
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Answer
It is given that,Then,
Hence, proved.
Question 15:
If , show that
Answer
It is given that,
Then,
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Hence, proved.
Question 16:
If , show that
Answer
The given relationship is
Taking logarithm on both the sides, we obtain
Differentiating this relationship with respect tox, we obtain
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Hence, proved.
Question 17:
If , show that
Answer
The given relationship is
Then,
Hence, proved.
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Exercise 5.8
Question 1:
Verify Rolles Theorem for the function
Answer
The given function, , being a polynomial function, is continuous in [4,
2] and is differentiable in (4, 2).
f(4) = f(2) = 0
The value of f(x) at 4 and 2 coincides.
Rolles Theorem states that there is a point c(4, 2) such that
Hence, Rolles Theorem is verified for the given function.
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Question 2:
Examine if Rolles Theorem is applicable to any of the following functions. Can you say
some thing about the converse of Rolles Theorem from these examples?
(i)
(ii)
(iii)
Answer
By Rolles Theorem, for a function , if
(a) fis continuous on [a, b](b) fis differentiable on (a, b)
(c) f (a) = f(b)
then, there exists some c(a, b) such that
Therefore, Rolles Theorem is not applicable to those functions that do not satisfy any of
the three conditions of the hypothesis.
(i)
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous atx = 5 andx = 9
f(x) is not continuous in [5, 9].
The differentiability of fin (5, 9) is checked as follows.
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Let n be an integer such that n(5, 9).
Since the left and right hand limits of fatx= nare not equal, fis not differentiable atx
= n
f is not differentiable in (5, 9).
It is observed that fdoes not satisfy all the conditions of the hypothesis of Rolles
Theorem.
Hence, Rolles Theorem is not applicable for .
(ii)
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous atx = 2 andx = 2
f(x) is not continuous in [2, 2].
The differentiability of fin (2, 2) is checked as follows.
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Let n be an integer such that n(2, 2).
Since the left and right hand limits of fatx= nare not equal, fis not differentiable atx
= n
f is not differentiable in (2, 2).
It is observed that fdoes not satisfy all the conditions of the hypothesis of Rolles
Theorem.
Hence, Rolles Theorem is not applicable for .
(iii)
It is evident that f, being a polynomial function, is continuous in [1, 2] and is
differentiable in (1, 2).
f (1) f(2)
It is observed that fdoes not satisfy a condition of the hypothesis of Rolles Theorem.
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Hence, Rolles Theorem is not applicable for .
Question 3:
If is a differentiable function and if does not vanish anywhere, then
prove that .
Answer
It is given that is a differentiable function.
Since every differentiable function is a continuous function, we obtain
(a) fis continuous on [5, 5].(b) f is differentiable on (5, 5).
Therefore, by the Mean Value Theorem, there exists c(5, 5) such that
It is also given that does not vanish anywhere.
Hence, proved.
Question 4:
Verify Mean Value Theorem, if in the interval , where
and .
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Answer
The given function is
f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4)
whose derivative is 2x 4.
Mean Value Theorem states that there is a point c(1, 4) such that
Hence, Mean Value Theorem is verified for the given function.
Question 5:
Verify Mean Value Theorem, if in the interval [a, b], where a= 1 and
b= 3. Find all for which
Answer
The given function fis
f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3)
whose derivative is 3x2 10x 3.
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Mean Value Theorem states that there exist a point c(1, 3) such that
Hence, Mean Value Theorem is verified for the given function and is the
only point for which
Question 6:
Examine the applicability of Mean Value Theorem for all three functions given in theabove exercise 2.
Answer
Mean Value Theorem states that for a function , if
(a) fis continuous on [a, b]
(b) fis differentiable on (a, b)
then, there exists some c(a, b) such that
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy
any of the two conditions of the hypothesis.
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(i)
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous atx = 5 andx = 9
f(x) is not continuous in [5, 9].
The differentiability of fin (5, 9) is checked as follows.
Let n be an integer such that n(5, 9).
Since the left and right hand limits of fatx= nare not equal, fis not differentiable atx
= n
f is not differentiable in (5, 9).
It is observed that fdoes not satisfy all the conditions of the hypothesis of Mean Value
Theorem.
Hence, Mean Value Theorem is not applicable for .
(ii)
It is evident that the given function f(x) is not continuous at every integral point.
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In particular, f(x) is not continuous atx = 2 andx = 2
f(x) is not continuous in [2, 2].
The differentiability of fin (2, 2) is checked as follows.
Let n be an integer such that n(2, 2).
Since the left and right hand limits of fatx= nare not equal, fis not differentiable atx
= n
f is not differentiable in (2, 2).
It is observed that fdoes not satisfy all the conditions of the hypothesis of Mean Value
Theorem.
Hence, Mean Value Theorem is not applicable for .
(iii)It is evident that f, being a polynomial function, is continuous in [1, 2] and is
differentiable in (1, 2).
It is observed that fsatisfies all the conditions of the hypothesis of Mean Value Theorem.
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Hence, Mean Value Theorem is applicable for .
It can be proved as follows.
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Miscellaneous Solutions
Question 1:
Answer
Using chain rule, we obtain
Question 2:
Answer
Question 3:
Answer
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Class XII Chapter 5 Continuity and Differentiability Maths
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Taking logarithm on both the sides, we obtain
Differentiating both sides with respect tox, we obtain
Question 4:
Answer
Using chain rule, we obtain
8/13/2019 Chapter 5 Continuity and Differentiability
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Class XII Chapter 5 Continuity and Differentiability Maths
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Question 5:
Answer
8/13/2019 Chapter 5 Continuity and Differentiability
128/144
Class XII Chapter 5 Continuity and Differentiability Maths
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Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from Welcome Metro Station)
Question 6:
Answer
8/13/2019 Chapter 5 Contin
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