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CONTINUITY AND DIFFERENTIABILITY
1
CONTINUITY
1 Definition
A function f(x) is said to be continuous at x = a , if axLim
f (a) = axLim
f (x) = f(a)
i.e. LHL = RHL = value of the function at a i.e.ax
Lim
f (x) = f (a) .
If f (x) is not continuous at x = a, we say that f (x) is discontinuous at x = a .
NOTES:
All Polynomials, Trigonometrical functions, exponential and Logarithmic functionsare continuous in their domain .
We never talk about continuity/discontinuity at a points at which we cantapproach from either side of the point. These points are called isolated points .
e.g. f(x) = x + x at x = .
2 Single point of Continuity :
There are some functions which are continuous only at one point.
e.g. f (x) =
Qxifx
Qxifxand g (x) =
Qxif0
Qxifxare both continuous only at x =
0
3 Reasons of Discontinuity
(a) One or more than one of the three quantities , LHL , RHL and f (a) is not defined . Lets
consider some examples
(i) f (x) =x
1around x = 0
LHL = , RHL = + , f (0) is not defined ,
f (x) =x
1is discontinuous at x = 0 which is
obvious from the graph .
(ii) f (x) =
1xfor1x
1x2
around x = 1
LHL = RHL = 2 but f (x) is not defined . Therefore this function graph has a hole at x
= 1 , it is discontinuous at x = 1
(b) All the three quantities are defined , but any pair of form is unequal (or all three are unequal).
(i) f (x) = [x] around any integer I
LHL = I 1 , RHL = I , f (I) = 1
LHL RHL = f (I) , so this frunction is also discontinuous at all integers
(ii) f (x) = {x} around any integer I
LHL = 1 , RHL = 0 , f (I) = 0
LHL RHL = f (I) , so this frunction is also discontinuous at all integers
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(iii) f (x) =
Zx,0
Zx,1around any integer I
From the figure , we notic that at nay integer I ,
LHL = 1 , RHL = 1 , f (I) = 0
LHL =RHL f (I) , so this frunction is again discontinuous
(iv) f (x) =
0x,0
0x,xx
around x = 0
At x = 0 we see that
LHL = 1 , RHL = 1 , f (0) = 0
LHL RHL f (0) and the function is discontinuous
To summarise , if we intend to evaluate the continuity of a function a x = a which means thatwe want to determine whether f (x) will be continuous at x = a or not , we have to evaluate all
the three quantities , LHL , RHL and f (a) . If these three quantities are finite and equal , f (x)
is continuous at x = a . In all other cases it is discontinuous at x = a
LHL (at x = a) = RHL , (at x = a) = f (a) , for continuity at x = a .
ILLUSTRATION
Discuss the continuity of the function ,
f (x) at x =2
1, where f (x) =
1x2
1
,x2
321x,1
21x0,x
21
Solution :
We have :
2
1xatLHL =
21x
Limf (x) =
21x
Lim
x2
1
2
1x0forx
2
1)x(f
=2
1
2
1= 0 [ Using direct substitution method ]
and
2
1xatRHL =
2
1x
Limf (x) =
21x
Lim
x2
3
1x2
1forx
2
3)x(f
=2
3
2
1= 1 [ Using direct substitution method ]
21x
Lim
21x
Lim
Hence f (x) is not continuous at x =2
1. Clearly f (x) is discontinuious at x =
2
1
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3
ILLUSTRATION
Discuss the continuity of the function , f (x) at x = 2 .
f (x) =
2x,x2
2x,x2
Solution :
We have : (LHL at x = 2) = 2xLim
f (x) = 2xLim
2 x . [ f (x) = 2 x for x < 2 ]
= 2 2 = 0
and (LHL at x = 2) = 2xLim
f (x) = 2xLim
2 + x . [ f (x) = 2 + x for x 2 ]
= 2 + 2 = 4
2xLim 2x
Limf (x) . Hence f (x) is not continuous at x = 2 .
Conceptual Exercise 01
Test the continuity of the following functions .
1. f (x) =
0x,10x,x
x
at x = 0
2. f (x) =
0x,0
0x,x1sinx
at x = 0
3. f (x) =
0x,0
0x,1e
1ex/1
x/1
at x = 0
4. f (x) =
2x1,x3x4
1x0,4x5
3 at x = 1
5. f (x) =
1x21,x1
21x,
21
21x0,x
at x =2
1
6. f (x) =
0x,2
0x,xcosx
xsinat x = 0
3 Continuity In An Open IntervalA function f(x) is said to be continuous in an open interval (a, b) if it is continuous at each
and every point of (a, b) i.e. y = {x} is continuous in (1, 2)
4 Continuity In A Closed Interval
A function f(x) is said to be continuous in a closed interval [a , b] if
It is continuous in (a, b)
Value of the function at b is equal to left hand limit at b i.e., f (b) = bxLim f (x)
Value of the function at a is equal to right hand limit at a i.e., f (a) = axLim
f(x)
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illustration
Show that the function , f (x) =
1x0,2x
0x2,1x
2x3,3x2
is discontinuous at x = 0 and
continuous at every point in interval [ 3 , 1 ] .
Solution :
f (x) =
1x0,2x
0x2,1x
2x3,3x2
is plotted as shown .
Here , if we observe in the graph , we could conclude that at x = 0 ,
0xLim
f (x) = 1 and 0xLim
f (x) = 2
which shows that the function is discontinuous at x = 0 and
continuous at every other point in [ 3 , 1] .
5 Geometrical Meaning of Continuity
A function f(x) will be continuous at x = a if and only if there is no break in the graph of the
function f(x) at the point (a, f(a)). In an interval function is said to be continuous if the is no
break in graph of function in the entire interval.
Examples :f (x) = sin x is continuous in its entire domain
0
y
x23 2
f (x) = tan x is discontinuous at x = (2n + 1)2
, where n I.
2
32
O
2
2
3
x
y
f(x) will be discontinuous at x = a, in any of the following cases:
(i) axLim f (x) and axLim f (x) exist but are not equal
For example y = [x] at x I is discontinuous.
1
2
1
1
2
2
1
1
2
2 3x
y
O
x
y
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CONTINUITY AND DIFFERENTIABILITY
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(ii) axLim f (x) = ax
Lim f (x) but not equal to f (a)
5
x = 2For example f (x) =
2x,5
2x,2x4x2
at x = 2
Here LHL = 2 and RHL = 2 , but at x = 2 , f (x) = 5 .
(iii) Atleast one of the limits
(L.H.L. or R.H.L.) does not exist .
For example :
y = sin
x
1at x = 0
illustration
If f (x) =
0x,3x
0x,0
0x,3x2
2. Discuss the continuity .
Solution :
Here 0xLim
f (x) = 3
0x
Limf (x) = 3
f (0) = 0
Thus 0xLim
f (x) = 0xLim
f (x) = 3 f (0)
Hence f (x) is discontinuous at x = 0
illustration
If f (x) =1x
1x2
. Discuss the continuity at x 1 .
Solution :
Which shows , 1xLim
f (x) = 2 but f (1) is not defined .
1x
1x2
So f (x) is discontinuous at x = 1 .
Conceptual Exercise 02
1. Find all possible values of a and b so that f (x) is continuous for all x R if
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f (x) =
x,3xcos
x0,b2xx2sinb
0x1,ax3
1x,3xa
2
2. Let f (x) =
0x,x2tan1n
1e
0x,1x1
xcosn
x4sin
24
Is it possible to define f (0) to make the function continuous at x = 0 . If yes what
is the value of f (0) , if not then indicate the nature of discontinuity .
3. The function f (x) =
xifxcos1
xif2b
x0if56
2
2
2
b
xtana
x5tan
x6tan
Determine the values of a and b , if f is continuous at x =2
4. Determine the value of a , b and c for which the function is continuous at x = 0 .
f (x) =
0xfor
0xforc
0xfor
2/3
2/12/12
xb
xxbx
x
xsinx)1a(sin
5. Find the locus of (a , b) for which the function f (x) =
2xforaxb
2x1forx3
1xforbxa
2
is continuous at x = 1 but discontinuous at x = 2 .
6 Algebra of Continuous Functions
Let f (x) and g (x) are continuous functions at x = a. Then,
(i) f (x) is continuous at x = a where c is any constant
(ii) f (x) g (x) is continuous at x = a
(iii) f (x). g (x) is continuous at x = a
(iv))x(g
)x(fis continuous at x = a, provided g (a) 0 .
illustration
f (x) =
0x,k
0x,x4
tanx/1
. For what value of k , f (x) is continuous at x = 0 ?
Solution :
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CONTINUITY AND DIFFERENTIABILITY
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Here , 0xLim
f (x) = 0xLim
x/1
x4
tan
0xLim
f (x) = 0xLim
x/1
xtan1
xtan1
( 1 form )
0xLim
f (x) = 0xLim
x/1
1xtan1
xtan11
0xLim
f (x) =x
1
xtan1
xtan2Lim
0xe
0xLim
f (x) =)xtan1(x
xtanLim2
0xe
=e2
Here , f (x) is continuous at x = 0 , when 0xLim
f (x) = f (0) k = e2
NOTES:
If f(x) is discontinuous and g(x) is also discontinuous at x = a , then the sum of thefunctions is continuous.
Example : f (x) = [x] and g (x) = {x} , then (f + g) (x) = x is a continuous function.
If f(x) is continuous and g(x) is discontinuous at x = a then the product function
)x(g).x(f)x( is not necessarily be discontinuous at x = a.
Example : f(x) = x and g (x) =
0x,0
0x,x1sin
(x) = 0 for all x R
If f(x) and g(x) both are discontinuous at x = a then the product function
)x(g).x(f)x( is not necessarily be discontinuous at x = a.
Example : f (x) =
0x,1
0x,1and g (x) =
0x,1
0x,1
Continuity of an inverse Function : If the function y = f(x) is defined, continuous andstrictly monotonic on the interval X, then there exist a single valued inverse function
x = (y) defined, continuous and also strictly monotonic in the range of the function
y = f (x).
Conceptual Exercise 03
1. Find the value of f (0) so that the function f (x) =x
x1x1 3 is continuous at x = 0.
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2. If the function f (x) =x
)xb1(n)xa1(n is undefined at x = 0. Then find the value
which should be assigned to f at x = 0 so that it is continuous at x = 0.
3. If f (x) =)x2(sin
4x2 , x 0 is continuous function at x = 0, then find the value of f(0)
4. If f(x) = x + { x} + [x] , where [x] is the integral part and {x} is the fractionalpart of x . Discuss the continuity of f in [ 2 , 2] .
5. Discuss the continuity of the function; f (x) = [[x]] [x 1] (where [.] denotes thegreatest integral function).
6. Examine the continuity or discontinuity of the function f (x) = [x] + [x].
7 Continuity of Composite Function
If f is continuous at x = c and g is continuous at x = f(c) then the composite g(f(x)) is
continuous at x = c. e.g. f(x) =2x
xsinx2
and g(x) = x are continuous at x = 0, hence
the composite (gof)(x) =2x
xsinx2
will also be continuous at x = 0 .
illustration
Find the point(s) of discontinuity of y =2uu
12
, where u =1x
1
.
Solution:
The function u = f (x) =1x
1
is discontinuous at the point x = 1. . . . (i)
The function y = g(x) =2uu
12
=)()( 1u2u
1
is discontinuous at u = 2 and u = 1.
when u = 2 ,1x
1
= 2 x =
2
1
u = 1 1x
1
= 1 x = 2
Hence, the composite function y = g (f (x)) is discontinuous at three points x =2
1, 1, 2
Conceptual Exercise 04
1. Draw the graph of the function f (x) = x x x2 , 1 x 1 and discuss itscontinuity or discontinuity at f in the interval 1 x 1 .
2. Given the function g (x) = x26 and h (x) = 2 x2 3 x + a . Then
(a) Evaluate h (g (2))
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CONTINUITY AND DIFFERENTIABILITY
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(b) If f (x) =
1x,)x(h
1x,)x(g, find a so that f is continuous .
3. Let f (x) =
3x2,x3
2x0,x1. Determine the form of g(x) = f (f(x)) and hence
find the point of discontinuity of g , if any .
4. If f (x) = 1 + x 1, 1 x 3 and g(x) = 2 x + 1, 2 x 2 .then calculate f(g(x)) and g(f(x)). Discuss the continuity of f(g(x)).
5. Let f (x) =
1x0,x
0x1,1x2 and g (x) = sin x . Further let
h(x) = f(g(x)) + f (g(x)). Discuss the continuity of h(x) in [1, 1].
8 Intermediate Value Theorem
Suppose f (x) is continuous on an interval I, and a and b are two points of I. Then if y0
is a
number between f (a) and f (b), there exists a number c between a and b such thatf(c) = y
0.
illustration
Let f : [0 , 1] [1 , e] be a continuous function, then prove that f(c) = ec for some c [0 , 1]Solution :
Let h(x) = f(x) ex
h(0) = f(0) 1 0h (1) = f (1) e 0h(c) = 0 for at least one c [0 , 1]
illustration
Let f , g : [0 , 1] [0 , ) be continuous function satisfying 1x0.Max
f (x) = 1x0.Max
g (x) .
Prove that there exists [0 , 1] with 2)(f + 3 f () = 2)(g + 3 g () .Solution :
Let 1x0.Max
f (x) at x = , [0 , 1] and 1x0.Max
g (x) at x = , [0 , 1]
Now consider , h (x) = f (x) g(x) x [ , ]h () = f () g () 0h () = f () g () 0
h () = 0 for atleast one [0 , 1] f () g () = 0
2)(f + 3 f () = 2)(g + 3 g () for atleast one [0 , 1]
illustration
f : (0 , 1) [0 , 1] be a continuous function .(A) if it is onto then it must be many to one (B) f must be onto
(C) f must be oneone (D) f can be bijective
Solution :Let f to be one to one then range of f will be open and as it cant be onto . Hence if it is
onto then it must be many to one .
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NOTES:
If f is a continuous function in [a, b] and is any real number such thatf(a) < < f(b), then there exists at least one solution of the equation f(x) = inthe open interval (a, b).
In general odd number of solution of f(x) = in the open interval (a, b). Inparticular if f(a) and f(b) possess opposite signs, then there exists at least one
solution of the equation f(x) = 0 in the open interval (a, b). In general odd number
of solutions of f(x) = 0 in the open interval (a, b). If f is continuous at every point of a closed interval I, then f assumes both an
absolute maximum value M and an absolute minimum value m somewhere in I.
That is there are numbers x1
and x2
in I with f(x1) = m, f(x
2) = M, and
M)x(fm for every other I.
In other words if m =bxa
min f(x), M =bxa
max f(x), then for any A satisfying the inequalities
m A M there exist a point x0
[a, b] for which f(x0) = A.
A continuous function whose domain is closed must have a range also in closedinterval but it is not necessary that domain is open then range is open (range can
be closed). f(x) has the minimum and maximum values on [a, b].
Conceptual Exercise 05
1. Let f : [1 , e] [0 , 1] is a continuous but need not be differentiable function thenprove that f (x) = n x has atleast one solution in [1, e] .
2. Let f : (1 , 10) [2 , 11] be a continuous function , but need not be differentiable,
then the correct statement is :
(A) it can be an odd function (B) it cant be an invertible function
(C) it can be an invertible function (D) none of these
3. Let f be a continuous function on R and periodic with fundamental period 1 i.e.
f (x + 1) = f (x), then prove that there will be a real number x0, such thatf(x
0+ ) = f(x
0) .
4. Let f : [0 , 1] R be a continuous function such that f(0) = f(1), then prove that
there is a solution of the equation f (x) f
n
1x = 0 , in
n
1n,0 for every
natural number n
5. Let f : [ 1 , 1] [ 1 , 1] be a continuous function, then prove thatf (c) = c3 for some c [ 1 , 1]
.
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DIFFERENTIABILITY
1. Definition of The Derivative
The derivative f (x) of a function y = f (x) at a given point x is defined as
f (x) = 0xLim
x
y
= 0xLim
x
)x(f)xx(f
= finite . If this limit exists finitely then the
function f (x) is called differentiable at the point x . The number f (x)
finitex
)x(f)xx(fLim
0xis called the left hand derivative at the point x .
Similarly the number f+ (x)
finitex
)x(f)xx(fLim
0xis called the right hand
derivative at the point x . The necessary and sufficient condition for the existence of the
derivative f (x) is the existence of the finite right and left hand derivatives , and also of the
equality
f+ (x) = f
(x) = finite .
2. Geometrical Meaning of The Derivat iveLet us consider the function f(x) and the corresponding curve y = f(x). Clearly line joining
two points M0(x , y) and M
1(x + x , y + y) on the curve will be the secant to the curve
and the slope of this secant is given by tan = xy
(Where is the angle made by the
secant with the positive direction of the xaxis). In the limiting case when x 0 thepoint M
1approaches M
0and the secant joining these two points will become the tangent
at M0
whose slope will be given by tan = 0xLim
x
y
= f (x)
which means that slope of the tangent to the curve y = f (x) at
any argument is equal to the value of the derivative at that
argument.
O x x+x
y
yxy
x
y=f(x)M1
M0
Geometrically, a function is not differentiable in the following cases :
3 Differentiability on An Interval
A function y = f (x) is differentiable on an interval (finite or infinite) if it has a derivative at
each point of the interval . It is differentiable on a closed interval [a , b] if it differentiableat every point of the open interval (a, b) and if the limits
0hLim
h
)a(f)ha(f = finite (Right hand derivative at a)
0hLim h)hb(f)b(f = finite (Left hand derivative at b) exist finitely
4 Relation Between Derivability And Continuity
(a) If )a(f exists then f(x) is derivable at )x(fax is continuous at x = a .Ingeneral a function f is derivable at x then f is continuous at x. i.e. if f(x) is derivable for
every point of its domain of definition, then it is continuous in that domain. The converse
of the above
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result is need not be true e.g. the functions f (x) = |x| and g (x) =
0x,0
0x,x1sinx
both are continuous at x = 0 but not derivable at x = 0.
(b) Let f+
(a) = and f
(a) = where and are finite then :(i) = f is derivable at x = a f is continuous at x = a.(ii) f is not derivable at x = a but f is continuous at x = a. If a
function f not differentiable but is continuous at x = a, it geometricallyimplies a sharp corner or kink at x = a .
(iii) If f is not continuous at x = a then it is not differentiable at x = a.
5. Reason of non differentiability
Case I(i) a corner, where the onesided derivatives differ
(ii) a cusp, where the slope of PQ approaches from one side and from the other
(iii) a vertical tangent, where the slope of PQ approaches from both sides or
a p p r o a c h e s from both sides (here, )
Q +
Q
P
(iv) a discontinuity
(a) (b)
Remarks :
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There are functions which are continuous at every point but differentiable at no point.
e.g. f (x) =
0n
n
3
2
cos (9n x) .
A formula that expresses f as an infinite sum of cosines with increasing higher frequencies.
By adding wiggles to wiggles infinitely many times, so to speak, the formula produces a
graph that is too bumpy in the limit to have a tangent anywhere.
6. Algebra of a Differentiable Function
(i) If f(x) and g(x) are derivable at x = a then the functions f(x) + g(x), f(x) g(x),f(x).g(x) will be derivable at x = a and if g (a) 0 , then the function f(x)/g(x) will alsobe derivable at x = a.
(ii) If f(x) is differentiable at x = a and g(x) is not differentiable at x = a, then the product
function f(x) g(x) can still be differentiable at x = a. e.g., f(x) = x and g(x) = |x| at
x = 0.
(iii) If f(x) is differentiable at x = a and g(x) is not differentiable at x = a, then the sum
function f(x) + g(x) is not differentiable at x = a.
(iv) If f(x) and g(x) both are not differentiable at x = a, then the sum function may be adifferentiable function. e.g. f(x)= x and g(x) = x
(v) If f(x) and g(x) both are not differentiable at x = a, then the product function f(x) g(x)
can still be differentiable at x = a i.e., f (x) = x and g(x) = x at x = 0 .(vi) If f(x) is derivable at x = a then it need not be true that )x(f is continuous at x = a.
e.g. f (x) =
0xif0
0xifx1sinx2
.
Conceptual Exercise 06
1. If f (x) is differentiable at x = a , find axLim
ax
)x(fa)a(fx 22
.
2. Discuss the differentiability of f (x) =
0x,0
0x,exx1
x1
at x = 0 .
3. Show that the function f (x) =
0xwhen0
0xwhensinxx1
is continuous but not differ
entiable at x = 0 .
4. A function f is defined by f (x2) = x3 for all x > 0 . Show that f is differentiable at 4.
5. Find whether the function f (x) = x3 is differentiable or not .
6. Let f (x) =
1xforbxa
1xforx1
2. If f (x) is continuous and differentiable every
where, then find the value of a and b .
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Functional Equation :
If x, y are independent variables, then :
(i) f(xy) = f(x) + f(y) f (x) = k ln x or f(x) = 0 .
(ii) f(xy) = f(x) . f(y) f (x) = xn , n R
(iii) f(x + y) = f(x) . f(y) f (x) = akx .
(iv) f(x + y) = f(x) + f(y) f (x) = kx, where k is a constant .
illustration
(a) Let f be a function such that f(x + f (y)) = f (x) + y x , y R , then find f (0).
(b) Now if it is given that there exists a positive real , such that f (h) = h for 0 < h < then find f(x) and hence f(x).
Solution :(a) Let x = 0, y = 0 in f (x) + f (y) = f (x) + y
f (0) + f (0) = f (0) + 0 2 f (0) = f (0) f (0) = 0
(b) Given f (h) = h
then f (x) = 0hLim
h
)x(f)hx(f for 0 < h <
f(x) = 0hLim
h
)x(f)h(fxf )( (given f (h) = h)
f(x) = 0hLim
h
)x(fh)x(f f(x) = 0hLim
h
h
f(x) = 1 , integrating both sides we get f (x) = x + c where f (0) = 0 c = 0
So, f (x) = xThus f (x) = 1 and f (x) = x
illustration
Let f : R R is a function satisfies conditionf (x + y3) = f (x) + [f (y)]3 for all x , y R. If f(0) 0 . Find f (10) .
Solution :Given f (x + y3) = f (x) + [f (y)]3 ................. (i)
and f (0) 0 ................. (ii)Replacing x , y by 0
f (0) = f (0) + f(0)
3
f (0) = 0 ................. (iii)also f (0) = 0h
Lim h
)0(f)h0(f = 0h
Lim h
)h(f
Let I = f (0) = 0hLim
33/1
33/1
)(
)(
h
)0(fh0f
= 0hLim
33/1
33/1
)(
)(h
hf= 0h
Lim
3
3/1
3/1
)(
)(
h
hf
= I3
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CONTINUITY AND DIFFERENTIABILITY
15
I = I3
or I = 0 , 1 , 1 as f (0) 0 f(0) = 0 , 1 ................ (v)
Thus, f (x) = 0hLim
h
)x(f)hx(f = 0h
Lim 33/1
33/1
)(
)(
h
)x(fhxf
f(x) = 0hLim
33/1
33/1
)(
)(
h
)x(fhf)x(f [ using (i)]
f(x) = 0hLim
)()(
3/1
3/1
h
hf= (f (0))3
f(x) = 0 , 1 [as f (0) = 0, 1 using (v)]Integrating both sides,
f(x) = c or x + c as f(0) = 0 f(x) = 0 or xThus f(10) = 0 or 10
Conceptual Exercise 07
1. Let f
2
yx=
2
)y(f)x(f for all real x and y. If f (0) exists and equals to
1 and f (0) = 1 , find f (x) .
2. I f f (x) + f (y) = f
yx1
yxfor all x , y R (xy 1) and 0x
Lim x
)x(f= 2 .
Find f
3
1and f (1)
3. Let f (x + y) = f(x) + f(y) 2xy 1 for all x and y . If f (0) exists andf (0) = sin , then find f{ f (0)} .
4. If f
3
yx=
3
)y(f)x(f2 for all real x and y and f (2) = 2 , then determine
y = f (x) .
5. Let f
2
yx=
2
)y(f)x(ffor all x and y . If f (1) = f (1) , show that
f(x) + f(1 x) = constant for all nonzero real x.
6. If f
y
x=
)y(f
)x(f x, y R, y 0 and f(t) 0, if t 0 and f(1) = 3 then find f(x).
Maxima/Minima/Middle function :
A function f (x) = max { } where max means the expression whose values lies above of
other expressions .
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DIFFERENTIAL CALCULUS
16
A function f (x) = min { } where min means the expression whose values lies below of other
expressions .
A function f (x) = mid { } where mid means the expression whose values lies between those
of other two expressions .
illustration
Let f(x) = min.{tan x, cot x} x R. Find(i) Range of f(x) (ii) Period (if periodic)(iii) Points of discontinuity of f(x) (iv) Points of nondifferentiability of f(x)
Solution :
x = 3 /2x = - x = x = - x =
O X
We know,
f(x) = min{tan x, cot x} can be plotted in two steps
(i) We should plot the graph of tan x and cot x
(ii) We should find their points of intersection
and neglect the area above their point of
intersection
It can seen from the graph. That,
(a) range of f(x) = (, 1] [0, 1](b) period of f(x) =
(c) Points of discontinuity are , 2
, 0 ... which can be put in the form of
2
n, nI
(d) Also the points of non differentiability are ,4
3 ,
2
, 0 , ... which can be put in
form of4
n, n I.
Here darked lines of the curves represents min {tan x , cot x} and dotted lines is max. {tan
x , cot x}
illustration
A function f is given by f (x) = mid {1 x , x , 2 x + 1} where mid means the expressionwhose values lies between those of other two expressions . Find the function and test its
continuity in entire number scale .
Solution :
y = 2 x + 1
y = x
y = 1 x
11
12
13
The bold lines denotes the mid of the function
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CONTINUITY AND DIFFERENTIABILITY
17
Hence f (x) = mid {1 x , x , 2 x + 1} =
x21,x
21x0,x1
0x31,1x2
31x,x
Conceptual Exercise 08
1. On how many points the functions ,
f (x) = max { x2 , (x 1)2 , 2 x (1 x)} ,0 x 1 , is not differentiable.
2. Let f (x) = sin x and g (x) =
xfor2
xcos1x0forxt0,)t(f.max
Discuss the continuity and differentiability of g (x) in (0 , ) .
3. Let f (x) = x4 8 x3 + 22 x2 24 x and g (x) =
1x,10x
1x1,1xtx,)t(fmin
Discuss the continuity and differentiability of g (x) in [ 1 , ) .
4. Let f (x) = 1 + 4x x2 , x R , g (x) = max. { f (t) ; x t (x + 1) ; 0 x < 3 }
= min. {(x + 3) ; 3 x 5} . Verify continuity of g (x) for all x [0 , 5] .
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DIFFERENTIAL CALCULUS
18
ANSWERS
Conceptual Exercise 01
1. f (x) is not continuous 2. f (x) is continuous 3. f (x) is not continuous
4. f (x) is continuous 5. f (x) is continuous 6. f (x) is continuous
Conceptual Exercise 02
1. a = 0 , b = 1 2. f (0+) = 2 ; f (0) = 2 , hence f (0) are not possible to define
3. a = 0 , b = 1 4. a = 2
3, b 0 , c =
2
1
5. Locus (a , b) x , y is y = x 3 excluding the points where y = 3 intersects it .
Conceptual Exercise 03
1.6
12. a + b 3.
8
14. Discontinuous at all integral values in
[2, 2]
5. Continuous on R 6. f (x) is discontinuous at x I
Conceptual Exercise 041. f is continuous at 1 x 1 2. (a) 4 3 2 + a (b) a = 3
3. g (x) =
3x2,x4
2x1,x2
1x0,2x
and g(x) is discontinuous at x = 1 and x = 2
4. Continuous everywhere 5. h (x) is not continuous at x = 0
Conceptual Exercise 05
2. B
Conceptual Exercise 06
1. 2 a f (a) a2 f (a) 2. Not differentiable at x = 0 5. differentiable every-where
6. a = 2
1, b =
2
3
Conceptual Exercise 07
1. f(x) = 1 2. f(1) = 1, f1
33
3. f { f (0)} = 1
4. f (x) = 2 x + 2 6. f (x) = x3
Conceptual Exercise 08
1. 2 2. continuous and differentiable for all x (0 , )
3. g (x) is cont. in [ 1 , ) but not diff. at x = 1 4. continuous for all x [0 , 5] except x = 3
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DIFFERENTIATION
19
DIFFERENTIATION
Methods of Differentiation
1 Derivative of f(x) From The First Principle/ab Initio Method
If f (x) is a derivable function then,
0xLim
x
y
= 0xLim
x
)x(f)xx(f
= f (x) =xd
yd
or simply f (x) = 0hLim
h
)x(f)hx(f
illustration
Using the definition of the derivative, find the derivative of the function cos ax at x.
Solution :
Let y = cos a x , we have
y = cos a (x + x) cos ax = 2 sin
x2
axa sin
2
xa
x
y
=
x
2xa
sinx2axasin2
Hence y = 0xLim
x
y
= 2 0xLim
sin
x2
axa
0xLim
x
xxasin
= a sin ax .
In particular, if a = 1, then y = cos x and y = sin x
illustration
Find the derivative of tan1 x with respect to x by using first principle.
Solution :
Let tan1 x = ,
2,
2 x = tan
and tan1 (x + h) = + . . . (i)
x + h = tan ( + ) . . . (ii)
Let 0xLim
h
xtan)hx(tan 11 = L
L = 0xLim
h
= 0x
Lim h
from (i) and (ii)
= 0hLim
tan)(tan= 0
Lim
tan)(tan
= 0Lim
sin
cos)(cos= cos2 =
2sec
1= 2x1
1
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DIFFERENTIAL CALCULUS
20
2 Algebra of Derivatives
If u and v are derivable function of x , then :
(i)xd
d(K u) = K
xd
ud, where K is any constant
(ii)xd
d(u v) =
xd
ud
xd
vd
(iii) Product Rule :xd
d(u . v) = u
xd
vd v
xd
ud
(iv) Quotient Rule :xd
d
v
u=
2v
xdvd
uxdud
v
where v 0
(v) Chain Rule : If y = f (u) and u = g (x) thenxd
yd=
ud
yd.
xd
ud
illustration
If y = x1/2 + log5x + xcos
xsin+ 2x , find xd
yd?
Solution :Here y = x1/2 + log
5x + tanx + 2x on differentiating w.r.t. x we get,
xd
yd=
xd
d(x)1/2 +
xd
d(log
5x) +
xd
dtan x +
xd
d(2x)
= 2
1(x)1/2 1 +
5logx
1
e+ sec2 x + 2x n 2 =
2
1(x)3/2 +
5logx
1
e
+ sec2 x + 2x n 2
I llustration 43 :
Differentiate: y =12x1tan
esinn2
.
Solution :
y =12x1tan
esinn2
xd
yd=
12x1tanesinn2
n 2 .
1xtan 21esin
1cos
1xtan 21e
. 1xtan21
e .
1x112 1x2
1
2 . 2 x
=12x1tan
esinn2
.
n 2 .
1xtan 21esin
1cos
1xtan 21e . 1xtan
21
e .
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DIFFERENTIATION
21
1xx
1
2
3 Some Standard Formulae of Differentiat ion
xd
d(constant) = 0
xd
dxn = nxn1
xdd ax = ax n a
xdd ex = e x
xd
d(log
a|x| ) =
x
1log
ae
xd
d n x =
x
1
xd
dsinx = cosx
xd
dcos x = sin x
xd
dtan x = sec2x
xd
dsec x = sec x tan x
xd
dcot x = cosec2x
xd
dcosec x = cosec x cot x
Conceptual Exercise 01
Differentiate the following functions w.r.t. x .
1. ex log x tan x 2. log sin x2
3. (x2 + x + 1)4 4. log (sec x + tan x)
5. If y = log
x
1x , prove that
xd
yd=
)1x(x2
1x
6. If y = x
x
e1
e1
show that
xd
yd=
x2xx
e1e1
e
4 Inverse Function And Their Derivat ives
Theorem :
If the inverse functions f and g are defined by y = f (x) and x = g (y) and if f (x) exists and
f (x) 0 then g (y) = )x(f1
. This result can also be written as , if xdyd
exists and xd
yd
0
,
thenxd
yd= xdyd
1or
xd
yd.
yd
xd= 1 or
xd
yd= xdyd
1where
0
yd
xd
xd
dsin1 x =
2x1
1
, 1 < x < 1
xd
dcos1 x =
2x1
1
, 1 < x < 1
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DIFFERENTIAL CALCULUS
22
xd
dtan1 x = 2x1
1
, x R
xd
dcot1 x = 2x1
1
, x R
xd
dsec1 x =
1xx
1
2 , x > 1
xd
dcosec1 x =
1xx
1
2 , x > 1
5 Differentiation of A Function Defined Parametrically
Let x and y be the functions of parameter t , i.e., x = f (t) , y = (t) , then
xd
yd=
tdxd
tdyd
=)t(f
)t(
illustration
If x =2te
and y = tan1 (2 t + 1), find
xd
yd?
Solution :
Here x =2te so ,
tdxd = 2 . t
2te and y = tanan1 (2 t + 1)
On differentiating both sides, we get td
yd=
2)1t2(1
1
(2)
xd
yd=
tdxd
tdyd
=
2t
2
e
t2
1t4t412
Hence ,
xd
yd= 1t2t2t2
e2
t2
Conceptual Exercise 02
Findxd
ydin the following cases .
1. x = a
2
ttanlog
2
1tcos 2 and y = a sin t .
2. x = a ( sin ) and y = a (1 cos )
3. x = a e (sin cos ) , y = a e (sin + cos )
4. x =2
ee tt and y =
2
ee tt
5. x = e
1and y = e
1
6. x = cos12t1
1
and y = sin1
2t1
t
, t R
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DIFFERENTIATION
23
6 Logarithmic Differentiation : To find the derivative of :
(i) a function which is the product or quotient of a number of functions
OR
(ii) a function of the form [f(x)]g(x) where f and g are both derivable, it will be found
convenient to take the logarithm of the function first and then differentiate. This is
called Logarithmic Differentiation.
illustration
If xy
. yx
= 1, find xd
yd
?
Solution :
Taking n on both sides;
y n x + x n y = n 1
Differentiating both sides, we get
y .xd
d(n x) +
yxd
d. n x +
xxd
d. ny + x
ynxd
d = 0
or y .x
1+ n x .
yd
xd+ 1 . n y + x .
y
1.
xd
yd= 0
y
xxn
xd
yd=
ynx
y
xd
yd=
xnyx
ynxy
.x
y
Conceptual Exercise 03
Differentiatiate the following functions w.r.t. x .
1. xx (x > 0) 2. (sin x)log x 0 x2
3.xxx (x > 0)
4. (sin x)tan x + (cos x)sec x 0 x2
5. xy = ex y (x > 0)
6. ex + ey = ex + y . Prove thatxd
yd+ ey x = 0
7 Differentiation with substitution
Following are some substitutions useful in finding derivatives .
Expression Substitution
a2 + x2 x = a tan or a cot a2 x2 x = a tan or a cot x2 a2 x = a sec or a cosec
xa
xa
orxa
xa
x = a cos 2
22
22
xa
xa
or22
22
xa
xa
a2 = a2 cos 2
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DIFFERENTIAL CALCULUS
24
illustration
Differentiate y = sin1 (3 x 4x 3)
Solution :
Let y = sin1 (3x 4x3) . Putting x = sin , we get
y = sin1 (3 sin 4 sin3 ) = sin1 (sin 3 ) = 3
or y = 3 sin1 x [ x = sin = sin1 x ]
xdyd = xd
d(3 sin1 x) = 3 xd
d(sin1 x) = 3 . 2x1
1
=2x1
3
illustration
Differentiate y = sin1
2x1
x2
Solution :
Let y = sin1
2x1
x2. Putting x = tan , we get
y = sin1
2tan1
tan2 = sin1 (sin 2 ) = 2 .
or y = 2 tan1 x [ x = tan = tan1 x ]
xd
yd=
xd
d(2 tanan1 x) = 2
xd
d(tan1 x) = 2 . 2x1
1
= 2x1
2
Conceptual Exercise 04
1. Differentiate the following function w.r.t. x .
(i) sin1 (sin x) , x [0 , 2 ] (ii) cos1 (cos x) , x [0 , 2 ]
(iii) tan1
(tan x) , x [0 , ] 2
2. Differentiate cos1 (2 x2 1) with respect to x , if :
(i) 0 < x < 1 (ii) 1 < x < 0
3. Differentiate the following functions w.r.t. x .
(i) tan1
xcos1
xcos1, 0 < x < (ii) tan1
xsin1
xsin1, 2
< x < 2
4. If y = sin1
2x1xx1x and 0 < x < 1 , then find
xd
yd.
5. Differentiate sin1
2x1x2 + cos1
2
2
x1x1 w.r.t. x , if :
(i) x (0 , 1) (ii) x ( , 1)
6. Differentiate tan1
2x1
x2+ cos1
2
2
x1
x1w.r.t. x , when :
(i) x (0 , 1) (ii) x ( , 1)
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DIFFERENTIATION
25
8 Differentiation of Implicit Function
If the relation between the variables x and y is given by an equation containing both, and
this equation is not immediately solvable for y, then y is called an implicit function of x.
Implicit functions are given by (x , y) = 0.
(i) In order to find xdyd
, in the case of implicit functions, we differentiate each term
w.r.t. x regarding y as a functions of x and then collect terms in xdyd
together on
one side to finally find xdyd
.
(ii) In answers of xdyd
in the case of implicit functions, both x and y are present.
illustration
If x2 + y2 + xy = 2, findxd
yd?
Solution :x2 + y2 + xy = 2, Differentiating both sides w.r.t. x we get,
xd
d(x2) +
xd
d(y2) +
xd
d(x y) =
xd
d(2)
or 2 x + 2 yxd
yd+
xd
xdy + x
yxd
d= 0 or 2 x + 2 y
xd
yd+ 1 . y + x .
xd
yd= 0
(2 y + x)xd
yd= (2 x + y)
xd
yd=
)xy2(
)yx2(
illustration
If 6x1 + 6y1 = a3 (x3 y3) , prove thatxd
yd=
2
2
y
x
6
6
x1
y1
Solution :
Here , 6x1 + 6y1 = a3 (x3 y3)
Let x3 = sin , y3 = sin then we get , 2sin1 + 2sin1 = = a3 (sin sin )
cos + cos = a3 (sin sin )
2 cos
2. cos
2= a3
2sin
2cos2
cos
2 = a3 sin
2 = cot
2 = a3
= 2 cot1 (a3)
or sin1 x3 sin1 y3 = 2 cot1 (a3), Differentiating both sides w.r.t. x we get,
6x1
1
. 3 x2
6y1
1
. 3 y2 .
xd
yd= 0 Hence ,
xd
yd=
62
62
x1y
y1x
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DIFFERENTIAL CALCULUS
26
Conceptual Exercise 05
Findxd
ydfor the following .
1. a x2 + 2 h x y + b y2 + 2g x + 2 f y + c = 0
2. log (x2 + y2) = 2 tan1
x
y3. x y = c2
4. (x2 + y2)2 = x y 5. ex y = log
y
x
6. If sin y = x sin (a + y) , prove thatxd
yd=
asin
)ya(sin2
9 Differentiation of A Functions With Respect To Another Function
)x(d
)x(fd
=
)x(xd
d
)x(fxd
d
=
)x(
)x(f
illustration
Differentiate : sin1
2x1
x2with respect to tan1 x
Solution :
xtanxd
d
x1x2sinxdd
1
12
= xtan
xdd
xtan2xd
d
1
1
= xtanxd
d2
xtanxd
d2
1
1
= 2
illustration
Differentiate ln tan x with respect to sin1 (ex) .
Solution :
)( x1 esindxtannd )(
=
)( x1 esin
xtann
xdd
xdd
=
x2e1
1.e
xsecxcotx
2
=xcos.xsin
e1e x2x
Conceptual Exercise 06
Differentiate
1. tan1
x
1x1 2w.r.t. tan1x , x 0 .
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DIFFERENTIATION
27
2. sin1
2x1
x2w.r.t. tan1 x , 1 < x < 1 . 3. xx w.r.t. x log x .
4. x2 w.r.t. x3 . 5. log (1 + x2) w.r.t. tan1 x .
6. sin1
2
x1
x2w.r.t. cos1
2
2
x1
x1, if 0 < x < 1 .
7. tan1
xa1xa1
w.r.t. 22 xa1
8. cos1 (4 x3 3 x) w.r.t. tan1
x
x1 2
, if2
1< x < 1
9. tan1
2x1x
w.r.t. sin1
2x1x2 , if
2
1< x 0 x R .
Solution :Putting x = 0 = y in the given functional equation, we get f (0) = 1
f (0) = 0hLim h
)x(f)hx(f = 0h
Lim h
)x(f1hx2)h(f)x(f
= 2x 0hLim h
)0(f)h(f = 2 x f (0) = 2 x 2aa3
f (x) = x2 2aa3 x + C
Putting x = 0 we have C = 1. Hence f (x) = x2 2aa3 x 1
The discriminant = 3 + a a2 + 4 = a2 + a + 1
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EXAMPLES
35
=
4
3
2
1a
2
< 0. Hence f (x) > 0 for all x R .
Example 5.
Draw the graph of the function f defined by f (x) =
4x1,x4
1x1,3x. Discuss the
continuity and differentiability of f at x = 1 .
Solution :The graph of the function is shown in the adjacent figure.
It is clear from the graph that it is continuous for
all x [ 1, 4] and not differentiable at x = 1,because at x = 1 ; LHD > 0 while RHD = 1 < 0.
Y
x4
1
1
3
01
Example 6.The function f is defined by y = f (x) where x = 2 t t, y = t2 + t t, t R . Draw the graphoff for the interval 1 x 1 . Also discuss its continuity and differentiability at x = 0.
Solution :When t 0; |t| = t x = 2 t t = t x 0and y = t2 + t2 = 2t2
i.e. y = 2 x2, when x 0W hen t < 0 ; t= t x = 2 t ( t) = 3t x < 0and y = t2 + t( t) = 0
i.e. y = 0, when x < 0
x
(1, 2)
x1 1O
y
yThus f (x) =
0x10x,0
1x00x,x2 2
Now the student can prove easily that the function is continuous and differentiable in [ 1,
1].
Example 7.
Let f (x) = x3 x2 + x + 1 and g (x) =
2x1,x3
1x0,xt0;)t(fmax
Discuss the continuity and differentiability of the function g (x) in the interval (0 , 2) .
Solution :f (t) = t3 t2 + t + 1 f (t) = 3 t2 2 t + 1its disc = ( 2)2 4.3.1 = 8 < 0
and coefficient of t2 = 3 > 0
Hence f (t) > 0 for all real t. f (t) is always increasingThus f(t) is maximum when t is maximum and t
max= x
max f(t) = f(x) g(x) =
2x1,x3
1x0,1xxx 23
Now it can be easily seen that f(x) is continuous in (0, 2) and differentiable in (0, 2) except
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DIFFERENTIAL CALCULUS
36
at x = 1. because at x = 1 LHD > 0 while RHD = 1 < 0.
Example 8.If 2 x = y1/5 + y1/5 , then express y as an explicit function of x and prove that
(x2 1) +2
2
xd
yd+ x
xd
yd= 25 y .
Solution :
25/15/1 yy = 25/15/1 yy 4 = 4 x2 4 { y1/5 y1/5 = 2 x }
2 y1/5 = 5/15/1 yy + 5/15/1 yy = 2 x + 2 1x2 y =
52 1xx
Differentiating w.r.t. x ,
xd
yd
= 5
42
1xx
1x
x2
.2
1
1 2 = 5 1x
1xx
2
52
= 1x
y5
2 { Using
(i) }
(x2 1)2
xd
yd
= 25 y2
Differentiating w.r.t. x , 2 x
2
xd
yd
+ (x2 1) . 2
xd
yd.
2
2
xd
yd= 50 y
xd
yd
or ,xd
yd+ (x2 1) .
2
2
xd
yd= 25 y
ttanconsaisythen,0xd
yd
Example 9.
If f (x) = 3x
xcosBxsinAx2sin is continuous at x = 0 . Find the values of A and B . Also
find f(0).
Solution :As f (x) is continuous at x = 0,
0xLim f (x) = f (0) and both f (0) and 0x
Lim f (x) are finite .
f (0) = 0xLim 3xxcosBxsinAx2sin
As denominator 0 , when x 0 .Numerator should also 0 when x 0 which is possible only if sin 2 (0) + A sin (0) + B cos (0) = 0 B = 0
f (0) = 0xLim 3x
xsinAx2sin
f(0) = 0xLim
x
xsin
2x
Axcos2= 0x
Lim
2x
Axcos2
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EXAMPLES
37
Again we can see that denominator 0 as x 0 Numerator should also approach 0 as x 0 2 + A = 0 A = 2
f (0) = 0xLim
2x
2xcos2= 0x
Lim
2
2
x
2/xsin4= 0x
Lim
4x
2xsin2
2
= 1
So, we get A = 2 , B = 0 and f ( 0) = 1
Example 10.
Show that 2xLim
2x
xtan
+ xLim
x
2x
11
> 3 .
Solution :
2xLim
2x
xtan
= 0yLim y
)2y(tan =
and xLim
)x/1(x
2
2
x
11
=
x
1
xLim2x
2x x
11Lim
e
= e0 = 1
xLim
2x
xtan
+ xLim
x
2x
11
= + 1 > 3 .
CONCEPT BUILDING OBJECTIVE EXAMPLES
Example 1.
0xLim
xtanx2
xsinx21
1
is equal to :
(A)31 (B)
21 (C) 0 (D) None of these
Solution :
0xLim
....3xxx2
....6xxx2
3
3
=3
1
Hence (A) is the correct answer.
Example 2.
1xLim 1x
}x{sinx
, where { x } denotes the fractional part of x, is equal to
(a) 1 (B) 0 (C) 1 (D) does not exist
Solution :
01xLim
{x} = 01xLim
(x [ x ]) = 1 0 = 1
01xLim
{x} = (x [ x ]) = 1 1 = 0
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DIFFERENTIAL CALCULUS
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01xLim
1x
}x{sinx
= 01xLim
1x
x
sin {x} =
01xLim
}x{
}x{sinx
1x
1x
= 1 1 1 = 1
Since, L.H. limit R. H. limitHence (D) is the correct answer.
Example 3.
Let f (x) = xLim sin2n x , then number of point(s) where f (x) is discontinuous is :
(A) 0 (B) 1 (C) 2 (D) infinitely many
Solution :
f (x) = nLim sin2 n x = n
Lim (sin2 x)n =
In,2
)1n2(x,0
In,2
)1n2(x,1
Clearly , f (x) is discontinuous at x = (2 n + 1) 2
, n I .Hence (D) is the correct answer .
Example 4.
Let f (x) =
0x,0
0x,x1sinxp
Then f (x) is continuous but not differentiable at x = 0 if :
(A) p < 0 (B) p = 0 (C) 0 < p 1 (D) p 1Solution :
f(0) = 0
For 0xLim f (x) = 0 0x
Lim x
p sinx1 = 0 .
This is possible only when p > 0 ... (i)
f (0) = 0hLim h
)0(f)h0(f = 0h
Lim h
0h1sinhp
= 0hLim h
p 1 sinh
1
f (0) will exist only when p > 1 f (x) will not be differentiable if p 1 ... (ii)From (i) and (ii), for f(x) to be not differentiable but continuous at x = 0 , possible values
of p are given by 0 < p 1 .Hence (C) is the correct answer.
Example 5.
nn22n 2tan
2
1.....
2tan
2
1
2tan
2
1tanlim
(A)1
(B)1
2 cot 2 (C) 2 cot 2 (D) None of these
Solution :tan = cot 2 cot 2
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EXAMPLES
39
2
1tan
2
=
2
1cot
2
cot
n2
1tan n2
= n2
1cot n2
1n2
1 cot 1n2
Required limit = nLim
Sn = nLim
2cot22
.2
2tan2
1
nn
nn =
1
2 cot 2
Hence (B) is correct answer.
Example 6.In order that the function f(x) = (x + 1)cotx is continuous at x = 0, f(0) must be defined as
(A) 0 (B) e (C)e
1(D) None of these
Solution :
0xLim f (x) = 0xLim xtan/x
x/1)x1( = e1
So , f (0) = e
Hence (B) is the correct answer .
Example 7.
For m , n I+ , 0xLim m
n
)x(sin
xsinis equal to :
(A) 1 , if n < m (B) 0 , if n > m (C) n/m (D) 0 , if n = m
Solution :Writing the given expression in the form
n
n
x
xsin
m
n
x
x m
xsin
x
and noting that the 0
Lim
sin= 1 , we see that the required
limit equals to 1 , if n = m , and 0 if n > m.
Hence (B) is the correct answer .
Example 8.
If f (x) is continuous and f (0) = 2 , then 0xLim x
ud)u(fx
0
is :
(A) 0 (B) 2 (C) f (2) (D) f (1)
Solution :
0xLim x
ud)u(fx
0
= 0xLim 1
)x(f
0xatcontinuousis)x(f
Rules'Hospital'LgsinU
= f (0) = 2.
Hence (b) is the correct answer.
Example 9.
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DIFFERENTIAL CALCULUS
Let g (x) be the inverse of the function f (x) and f (x) =3x1
1
. Then g (x) is equal to :
(A) 3)x(g1
1
(B)
3)x(f1
1
(C) 1 + (g(x))3 (D) 1 + (f (x))3
Solution :
Since g (x) is the inverse of f (x) , therefore g (x) = f1 (x) f {g (x)} = xDifferentiate both sides w.r.t. x
f { g(x) } g (x) = 1 g (x) =))x(g(f
1
= 1 + (g(x))3
Hence (C) is the correct answer.
Example 10.
1xLim 1x
1xlogxlogxx2
23
is equal to :
(A) 2/3 (B) 3/2 (C) 0 (D) None of these
Solution :
The given l imit = 1xLim 1x
xlog1x1x
2
23 )()(
= 1x
Lim
)()( 1x1x
xlog1x1x1xx)1x( )()()( 3
= 1xLim )1x()1x(
xlog)1x(1xx)1x( ][ 2
=)11(
1log)11(1112
=2
3
Hence (B) is the correct answer .