2 objective mht-cet mathematics 2...
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2 Objective MHT-CET Mathematics
IntroductionLet • y = f (x) be a real valued function, then derivative of the function w.r.t. x is given by
f x dydx
f x h f xhh
′( =+ −
→) lim ( ) ( )or
0
The derivative f x dydx′( ) or is also known as differential
coefficient of f (x). If the derivative of a function at a point exists then
the function is called differentiable or derivable at that point.
Illustration: If f (x) = x sin x, find f ′(p), using first principle.
Soln.: f (x) = x sin x \ f (p) = (p) sin p = 0 f (p + h) = (p + h)sin(p + h) By first principle
f f h f
hh′ =
+ −→
( ) lim ( ) ( )p
p p0
= + + −→
lim ( )sin( )h
h hh0
0p p
= + +→
lim ( )(sin cos cos sin )h
h h hh0
p p p
= − +
→lim ( )sinh
h hh0
p = –(p + 0) × 1 = – p
Illustration: If f (x) = log(3x – 1), find f ′(2), using first principle.
Soln.: Given f (x) = log (3x – 1) \ f (2) = log (6 – 1) = log 5 and f (2 + h) = log [3(2 + h)–1] = log (3h + 5) By first principle
f f h f
hh′( ) =
+ −→
22 2
0lim ( ) ( )
= + −→
lim log( ) logh
hh0
3 5 5
= +
→
lim logh h
h0
1 3 55 = +
→
lim logh h
h0
11 3
5
=+
×→
limlog
h
h
h0
1 35
35
35 =
+
→
35
1 35
35
0lim
log
h
h
h
= ×35
1 h h ttt
→ → + =
→
0 35
0 1 10
, lim log( )and
= 35
Left Hand DerivativeIf • y = f (x) is a real valued function and a is any real
number, then
lim ( ) ( ) ,h
f a h f ah→ −
+ −0
if it exists, is
called the left hand derivative of f (x) at x = a and is denoted by Lf ′(a) or f ′(a) – or f ′(a–).
Right Hand Derivative If • y = f (x) is a real valued function and a is any real
number, then lim ( ) ( ) ,h
f a h f ah→ +
+ −0
if it exists, is
called the right hand derivative of f (x) at x = a and is denoted by Rf ′(a) or f ′(a)+ or f ′(a+).
Note : If f ′(a+) and f ′(a–) are equal then f (x) is said to be differentiable at x = a.
2.1 Relationship Between Continuity and Differentiability
2.2 Derivative of Composite Functions
2.3 Derivative of Inverse Functions
2.4 Logarithmic Differentiation
2.5 Derivative of Implicit Functions
2.6 Derivative of Parametric Functions
2.7 Higher Order Derivatives
2 Differentiation
3Differentiation
Illustration: Discuss the differentiability of f(x) = x|x| at x = 0.
Soln.: We have, f x x xx x
x x( )
,
,= =
≥
− <
2
2
0
0
Now, ( ) lim ( ) ( )LHD at x f x fxx
= =−−→ −
00
00
⇒ = = − −−→
( ) limLHD at x xxx
0 000
2
⇒ = = − =→
( ) limLHD at x xx
0 00
and, ( ) lim ( ) ( )RHD at x f x fxx
= =−−→ +
00
00
⇒ = = −−→
( ) limRHD at x xxx
0 000
2
⇒ = = =→
( ) limRHD at x xx
0 00
So, f (x) is differentiable at x = 0.
2.1 Relationship Between Continuity and Differentiability
Every differentiable function is continuous but the •converse need not be true.
Illustration: A function is defined by
f x
x
x x( )
,
sin , .=
− < <
+ ≤ <
1 2 0
1 0 2
for
for
p
p
What can you say about right hand derivative and left hand derivative at x = 0?
Is f continuous at x = 0 ? Soln.: Right hand derivative at x = 0 is
f f h fhh
′ =+ −+
→ +( ) lim ( ) ( )0
0 00
=−
→ +lim ( ) ( )
h
f h fh0
0
= + − +→
lim ( sin ) ( sin )h
hh0
1 1 0
[... f (x) = 1 + sinx, 0 ≤ x < p2
]
= + −→
lim sinh
hh0
1 1 = =→
lim sinh
hh0
1
Left hand derivative at x = 0 is
f f h fhh
′(0 ) = + −−
→ −lim ( ) ( )
0
0 0 =−
→ −lim ( ) ( )
h
f h fh0
0
= − +→
lim ( sin )h h0
1 1 0 = − − =
→ →
lim limh hh h0 0
1 1 0 0 = 0
(h → 0, h ≠ 0) Since f ′(0+) ≠ f ′(0 –) \ f is not differentiable at x = 0. Continuity at x = 0
f x x( ) ,= − < <12
0for p
\ = =→ →−lim ( ) lim( )
x xf x
0 01 1
f (x) = 1 + sinx, for 02
≤ <x p
\ f (0) = 1 + sin 0 = 1 and lim ( ) lim( sin )
x xf x x
→ →+= +
0 01 = 1 + sin0 = 1
\ = =→ →− +lim ( ) ( ) lim ( )
x xf x f f x
0 00
\ f is continuous at x = 0.
2.2 Derivative of Composite FunctionsIf • y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f (g(x)) is a
differentiable function of x and dydx
dydu
dudx= ⋅
This is called chain rule.If • y is a differentiable function of u1, ui is a differentiable function of ui + 1, for i = 1, 2, ..... n – 1 and un is a differentiable function of x, then y is a differentiable function of x and
dydx
dydu
dudu
dudu
dudx
n= × × × ×1
1
2
2
3......
Illustration: Differentiate : sin (x2+ 5) w.r.t. x Soln. : Let y = sin (x2 + 5) Also let u = x2
+ 5 …(i) \ y = sinu …(ii)
Then, dydu u= cos (from (ii))
dudx x= 2 (from (i))
By chain rule, dydx
dydu
dudx x x= ⋅ = + ×cos( )2 5 2
= 2x cos (x2 + 5)
Illustration: If findy x x dydx= + +( ) , .7 11 392
32
Soln.: y = (7x2 + 11x + 39)3/2
Let u = 7x2 + 11x + 39 …(i) ⇒ y = u3/2 …(ii)
\ dydu u=
−32
32 1
( ) = 32
12u (from (ii))
dudx x= +14 11 (from (i))
So, by chain rule
dydx
dydu
dudx x x x= × = + + ⋅ +3
27 11 39 14 112 1 2( ) ( )/
= + + +32 14 11 7 11 392( )x x x
4 Objective MHT-CET Mathematics
Derivative of Some Standard Composite Functions
Sr. No. y dydx
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
[ f (x)]n
f x( )
1f x( )
sin[ f (x)]
cos[ f (x)]
sec[ f (x)]
cosec[ f (x)]
tan[ f (x)]
cot[ f (x)]
a f (x)
e f (x)
log[ f (x)]
loga[ f (x)]
n [ f (x)]n–1 × f ′(x)
12 f x
f x( )
)× ′(
− × ′(12[ ( )]
)f x
f x
cos[ f (x)] × f ′(x)
– sin[ f (x)] × f ′(x)
sec[ f (x)] · tan[ f (x)] × f ′(x)
– cosec[ f (x)] · cot[ f (x)] × f ′(x)
sec2[ f (x)] ×f ′(x)
– cosec2[ f (x)] × f ′(x)
a f (x) · log a × f ′(x)
e f (x) × f ′(x)1
f x f x( ) )× ′(
1f x a f x
e( ) log )⋅
× ′(
Illustration: Differentiate : sin[cos (tan x)] w.r.t. x Soln. : Let y = sin[cos (tan x)]
\ =dydx
ddx x{sin[cos(tan )]}
= ⋅cos[cos(tan )] [cos(tan )]x ddx x
= ⋅ − ⋅cos[cos(tan )] [ sin(tan )] (tan )x x ddx x
= – cos[cos(tanx)] · sin(tanx) · sec2x = – sec2x · cos[cos(tanx)] · sin(tan x).
2.3 Derivative of Inverse Functions
If • y = f (x) is a differentiable function of x such that the inverse function x = f –1(y) exists, then x is a differentiable function of y and
dxdy dy
dx
dydx= ≠ 01 , where
2.3.1 Derivative of Inverse Trigonometric Functions
Sr. No. f (x) f ′(x)
1. sin–1x, x ∈[–1, 1] 1
11
2−<
xx,
2. cos–1x, x ∈[–1, 1] −
−<1
11
2xx,
3. tan–1x, x ∈R1
1 2+ x
4. cot–1x, x ∈R−+
11 2x
5. sec–1x,x ∈(– ∞, –1) ∪ (1, ∞)
1
11
2x xx
−>,
6. cosec–1x, x ∈(– ∞, –1) ∪ (1, ∞)
−
−>1
11
2x xx,
Illustration: Differentiate : cot−1 x w.r.t. x
Soln. : Let y x= −cot 1
dydx
ddx x
xddx x= = −
+⋅−(cot )
( )1
21
1
= −+
⋅ = −+
11
12
12 1x x x x( )
2.3.2 Derivative of Inverse Composite Functions
Function Derivative
sin-1( f (x))1
11
2−⋅ ′( <
( ( ))), ( )
f xf x f x
cos–1( f (x)) −
−⋅ ′( <1
11
2( ( ))), ( )
f xf x f x
tan–1( f (x)) 11 2+
⋅ ′(( ( ))
)f x
f x
cot–1( f (x)) −+
⋅ ′(11 2( ( ))
)f x
f x
sec–1( f (x)) 1
11
2f x f xf x f x
( ) ( ( ))), ( )
−⋅ ′( >
cosec–1( f (x)) −
−⋅ ′( >1
11
2f x f xf x f x
( ) ( ( ))), ( )
5Differentiation
Illustration: Differentiate : sin–1(cos3x) w.r.t. x Soln.: Let y = sin–1(cos3x)
\ = −dydx
ddx x[sin (cos )]1 3
=
−×1
1 33
2(cos )(cos )
x
ddx x
=−
× − ⋅1
1 33 3
2cos( sin ) ( )
xx d
dx x
= × − ×1
33 3
2sin( sin )
xx = × − × = −1
33 3 3sin ( sin )x x
Illustration: Differentiate :
tan sin sin
sin sin− + + −
+ − −
1 1 11 1
x xx x
w.r.t. x
Soln.: Let y x xx x
= + + −+ − −
−tan sin sinsin sin
1 1 11 1
Now, 12 2
22 2
2 2+ = + +sin cos sin sin cosx x x x x
= +cos sinx x2 2
Similarly, 12 2
− = −sin cos sinx x x
\ =+
+ −
+
−
−yx x x x
x xtan
cos sin cos sin
cos sin cos
1 2 2 2 2
2 2xx x2 2−
sin
=
=
− −tancos
sintan cot1 1
2 2
2 22
x
xx
= −
= −−tan tan1
2 2 2 2p px x
Diff. w.r.t. x, we get dydx = − 1
2
Derivatives of Inverse Trigonometric Functions by Substitution
With proper substitution the given inverse function •can be reduced to a simple form. Then the derivative may be easily obtained.
Standard substitutions are :Sr. No.
Expression involving the term
Substitution
1. a x2 2− x = a sin q or a cos q
2. a x2 2+ x = a tan q or a cot q
3. x a2 2− x = a sec q or a cosec q
4. a x a x− +or x = a cos q or a cos2q
5. 1 – 2x2 x = sin q6. 2x2–1 x = cos q
7. 21
212 2
xx
xx− +
, and
11
2
2−+
xx
x = tanq
Illustration: Differentiate :
tan− + − −+ + −
1 1 11 1
x xx x
w.r.t.x
Soln.: Let y x xx x
= + − −+ + −
−tan 1 1 11 1
Put x x= ⇒ = −cos cos2 12
1q q
\ = + − −+ + −
−y tan cos coscos cos
1 1 2 1 21 2 1 2
q qq q
= −
+
−tan cos sin
cos sin1
2 2
2 2
2 2
2 2
q q
q q
= −+
−tan cos sincos sin
1 q qq q
= −+
= −
− −tan tantan
tan tan1 111 4
p q
= − = − −p q p4 4
12
1cos x
On differentiating w.r.t. x, we get
dydx x x
= −−
−
=−
12
1
1
1
2 12 2
Illustration: Differentiate : cos−−
−−+
1 a aa a
x x
x x w.r.t.x
Soln.: Let y a aa a
aa
aa
x x
x x
xx
xx
= −+
=−
+
−−
−−cos cos1 1
1
1
= −+
−cos 12
211
aa
x
x
Put ax = tanq ⇒ q = tan–1(ax)
\ = −+
=− −y cos tantan
cos (cos )12
211
12q
= 2 q = 2 tan–1(ax)
\ = −dydx
ddx ax2 1[tan ( )] = ×
+⋅2 1
1 2( )( )
addx ax
x
=+
× = ⋅+
21
212 2a
a a a aax
xx
xlog log
6 Objective MHT-CET Mathematics
2.4 Logarithmic DifferentiationWhen we want to find the derivative of a function •which can be expressed as :
(i) a product of number of functions (ii) a quotient of functions (iii) a function which is of the form [f (x)]g(x),
then it is easy to find the derivative of the logarithm of the function.
This method is known as logarithmic differentiation. Note that by chain rule,
ddx y d
dy y dydx y
dydx(log ) (log ). .= = ⋅1
Illustration: If findx y xx
dydx
= −
−exp tan , .1 2
2
Soln.: x y xx
= −
−exp tan 1 2
2
Taking log on both sides, we get
log tanx y xx
= −
−1 2
2
⇒ − =y xx
x2
2 tan(log ) ⇒ y = x2 tan(logx) + x2
On differentiating w.r.t. x, we get dy
dxx x x x
xx= + +2 22
2tan(log ) sec (log )
⇒ = + +dydx
x x x x x2 22tan(log ) sec (log )
⇒ = + +dydx
x x x x2 1 2[ tan(log )] sec (log )
2.5 Derivative of Implicit Functions • If a relation between x and y is such that y can be expressed in terms of x i.e., y = f (x) then y is called explicit function of x.If a relation between • x and y is such that y cannot be expressed in terms of x, then y is called an implicit function of x. When a given relation expresses • y as an implicit
function of x and we want to find dydx , then we
differentiate every term of the relation w.r.t. x, remembering that a term in y is first differentiated w.r.t. y and then multiplied by dy
dx .
Note : While taking the derivative of implicit function, function of y is considered as composite function of x.
Illustration: If xmyn = (x + y)m + n, then find dydx Soln.: Given xmyn = (x + y)m + n
Taking log on both sides, we get m log x + n log y = (m + n) log(x + y) Differentiating both sides w.r.t. x, we get
⇒ + = ++
× +
mx
ny
dydx
m nx y
dydx1
⇒ − ++
= ++
−
mx
m nx y
m nx y
ny
dydx
⇒−+
=−+
my nxx x y
my nxy x y
dydx( ) ( )
⇒ =dydx
yx
Illustration: If sec ,x y
x y a−+
= then find dydx .
Soln.: Given sec x yx y a−
+
= ⇒−+
= −x yx y asec 1
Differentiating both sides w.r.t. x, we get
( ) ( )
( )
x y dydx
x y dydx
x y
+ −
− − +
+=
1 102
⇒ + − + − − − − =x y x y dydx x y x y dy
dx( ) ( ) ( ) 0
⇒ = + + − ⇒ =2 2 2y dydx x y x y y x dy
dx( ) ⇒ =dydx
yx
2.6 Derivative of Parametric FunctionsIf • x and y are expressed as functions of the variable t, i.e., x = f (t) and y = g(t), then these equations are called parametric equations and t is called parameter.If • x = f (t) and y = g(t) are differentiable functions of parameter t, then y is a differentiable function of x and
dydx
dydtdxdt
dxdt= ≠ 0,
Illustration: If x tt
= −+
11
2
2 and y tt
=+2
1 2 , then find dydx .
Soln. : x tt
y tt
= −+
=+
11
21
2
2 2and
Put t = tanq in both the equations, we get
x = −+
=11
22
2tantan
cosqq
q …(1)
and y =+
=21
22tantan
sinqq
q …(2)
On differentiating (1) and (2) w.r.t. q, we get
dxd
dydq
q= − =2 2 2 2sin cosand
Therefore, dydx
dyddxd
xy= = − = −q
q
cossin
22
7Differentiation
dydx x x a
ddx x x a=
+ +× + +1
2 22 2( )
=+ +
++
1 1 2
22 2 2 2x x a
x
x a
=+ +
× + +
+
12 2
2 2
2 2x x ax a x
x a
=+
12 2x a
⇒
+ =dy
dx x a2
2 2 1.( )
Again, differentiating w.r.t. x, we get
dydx
ddx x a x a d
dxdydx
⋅ + + + ⋅
=
22 2 2 2
20( ) ( )
⇒
⋅ + + ×
=dydx x x a dy
dxd ydx
22 2
2
22 2 0( )
⇒ + +
=2 02
22 2dy
dxd ydx
x a x dydx( )
⇒ ( + + =x a d ydx
x dydx
2 22
2 0) [Q y1 ≠ 0]
Illustration: If y = log(1 + cos x), then prove that
d ydx
d ydx
dydx
3
3
2
2 0+ ⋅ =
Soln. : Given, y = log(1 + cos x)
⇒ =+
−dydx x x1
1 cos ( sin ) = −+sincos
xx1
⇒ = −+ ⋅ − −
+
d ydx
x x x x
x
2
2 2
1
1
( cos ) cos sin ( sin )
( cos )
= − + ++
cos cos sin
( cos )x x x
x
2 2
21
= − ++
= −+
cos( cos ) cos
xx x1
11
12
⇒ = − − + −−d ydx
x x3
321 1 1( )( )( cos ) ( sin )
= −+
sin( cos )
xx1 2
Now, d ydx
d ydx
dydx
3
3
2
2+ ⋅
= −+
+ −+
× −
+
sin( cos ) cos
sincos
xx x
xx1
11 12
= − ++
=sin sin( cos )
x xx1
02
Illustration: If x = sin–1(3t – 4t3) and
y = cos–1 ( ),1 2− t then find dydx
.
Soln. : Given that, y t t= − =− −cos sin1 2 11 ... (1) and x = sin–1(3t – 4t3) = 3 sin–1t ... (2) On differentiating (1) and (2) w.r.t. t, we get
dydt t
dxdt t
=−
=−
1
1
3
12 2and
\ = = −
−
=dydx
dydtdxdt
t
t
1
1
31
1
13
2
2
2.7 Higher Order Derivatives (Second Order Derivatives)
Derivative of • y = f (x) w.r.t. x (if it exists) is denoted by dydx
or f ′(x) and is called the first order derivative of y.
Derivative of dydx or f ′(x) w.r.t. x (if it exists) is denoted
by d ydx
2
2 or f ′′(x) and is called the second order derivative
of y.
Derivative of
d ydx
2
2 or f ′′(x) w.r.t. x (if it exists) is
denoted by d ydx
3
3 or f ′′′(x) and is called the third order
derivative of y.
In general,
d ydx
n
n is the nth order derivative of y w.r.t. x.
Note : These higher order derivatives may also be denoted by y1, y2, y3, ..... yn
Illustration: If y = x3 log x, then find d ydx
2
2 .
Soln. : Let y = x3 log x
\ = ⋅ + ⋅ = +dydx x x x x x x x3 2 2 21 3 3log log
Again differentiating w.r.t. x, we get
d ydx
x x x x x x x x2
222 3 1 6 5 6= + ⋅ + ⋅ = +log log
Illustration: If y x x a= + +log{ },2 2 then prove that
( )x a d ydx
x dydx
2 22
2 0+ + = .
Soln. : We have, y x x a= + +log{ }2 2
On differentiating w.r.t. x, we get
7Differentiation
2.1 Relationship Between Continuity and Differentiability
1. The function f xx x x
x( )
sin( / ),,
,=≠ 0=
10 0
whenwhen
is
(a) continuous but not differentiable at x = 0(b) continuous and differentiable at x = 0(c) neither continuous nor differentiable at x = 0(d) none of these
2. f(x) = x3 is (a) continuous but not differentiable at x = 3(b) continuous and differentiable at x = 3(c) neither continuous nor differentiable at x = 3(d) none of these
3. f(x) = [x] is neither continuous nor differentiable at(a) integral points (b) rational points(c) real points (d) none of these
4. f(x) = |x – 2| is(a) continuous but not differentiable at x = 2(b) continuous and differentiable at x = 2(c) neither continuous nor differentiable at x = 2(d) none of these
5. Let f xx x
x x( )
( ),,
=− ≥ 1
≤ <
20 1
whenwhen
, then f(x) is
(a) continuous but not differentiable at x = 1(b) continuous and differentiable at x = 1(c) neither continuous nor differentiable at x = 1(d) none of these
6. The function f xx x
x x( )
( ),
( ),=
− < 1
− ≥
1
1 12
when
when is
(a) continuous but not differentiable at x = 1(b) continuous and differentiable at x = 1(c) neither continuous nor differentiable at x = 1(d) none of these
7. If f xx xx x
( ),,
,=+ ≤− >
1 25 2
forfor
then
(a) f(x) is continuous and differentiable at x = 2(b) f(x) is continuous but not differentiable at x = 2(c) f(x) is everywhere differentiable(d) f(x) is not continuous at x = 2
8. If f x x x x
x( ) cos ,
,,=
≠ 0
=
1
0 0
for
forthen at x = 0, f(x) is
(a) not continuous(b) continuous and differentiable(c) continuous but not differentiable(d) none of these
9. If f xx xx x
( ),,
,=− <− ≥
1 22 3 2
forfor
then at x = 2, f(x) is
(a) continuous but not differentiable(b) continuous and differentiable(c) neither differentiable nor continuous(d) none of these
10. Let f x x x x
x
p( ) sin ,
,,=
≠
=
1 0
0 0
for
for then f(x) is
continuous but not differentiable at x = 0, if(a) 1 ≤ p < ∞ (b) 0 < p ≤ 1(c) – ∞ < p < 0 (d) p = 0
11. If f(x) = [x], – 2 ≤ x ≤ 2, then at x = 1(a) f(x) is continuous and differentiable(b) f(x) is neither continuous nor differentiable(c) f(x) is continuous but not differentiable(d) none of these
12. If f xx x
x( )
,,
,=≤>
forfor
00 0
then at x = 0
(a) f(x) is differentiable and continuous(b) f(x) is neither continuous nor differentiable(c) f(x) is continuous but not differentiable(d) none of these
13. Which of the following is not always true?(a) If f(x) is continuous at x = a, then it is
differentiable at x = a(b) If f(x) is not continuous at x = a, then it is not
differentiable at x = a(c) If f(x) and g(x) are differentiable at x = a, then
f(x) + g(x) is also differentiable at x = a(d) If f(x) is continuous at x = a, then lim ( )
x af x
→
exists.
Multiple Choice QuestionsLEVEL - 1
8 Objective MHT-CET Mathematics
14. If f(x) = | x – 1|, x ∈R, then at x = 1(a) f(x) is not continuous(b) f(x) is continuous but not differentiable(c) f(x) is continuous and differentiable(d) none of these
15. If f xx xx x
( ),
,,=
≤ ≤− <
forfor
then0 1
2 1 1(a) f(x) is discontinuous at x = 1(b) f(x) is differentiable at x = 1(c) f(x) is continuous but not differentiable at x= 1(d) none of these
16. Let f x x a x a( ) ( )cos= − −
1 for x ≠ a and let f(a) = 0.
Then f(x) at x = a is(a) continuous but not differentiable(b) continuous and differentiable(c) neither continuous nor differentiable(d) none of these
17. f xx xx x
x( ),,
=+ ≥− <
=1 01 0
0at is
(a) discontinuous(b) continuous but not differentiable(c) differentiable(d) none of these
18. Find the values of a, b respectively if
f xx x a x
bx x( )
,,
=+ + ≤
+ >
2 3 12 1
is differentiable at every x.
(a) 3, 2 (b) 2, 4(c) 3, 5 (d) 5, 3
19. Let f(x) = x + |x|. Then f at x = 0 is (a) continuous and differentiable(b) continuous but not differentiable(c) neither continuous nor differentiable(d) none of these
20. Examine the differentiability of the function f defined
by f xx xx xx x
( ),
,,
=+ − ≤ < −+ − ≤ <+ ≤ ≤
2 3 3 21 2 02 0 1
ififif
(a) differentiable at all x ∈ R –{0, – 2}.(b) continuous only(c) discontinuous everywhere(d) differentiable at all x ∈ R
21. The function f xx x
x x x( )
| |,
,=
− ≥
− + <
3 1
432
134
12 is
(a) continuous and differentiable at x = 1
(b) continuous but not differentiable at x = 1(c) discontinuous at x = 1(d) none of these
22. f(x) = e|x| is(a) discontinuous everywhere(b) differentiable at x = 0(c) continuous at x = 0 only(d) not differentiable at x = 0
23. The function f(x) = e– |x| is(a) continuous everywhere but not differentiable at
x = 0(b) continuous and differentiable everywhere(c) not continuous at x = 0(d) none of these
24. Let f(x) = |sin x|. Then,(a) f(x) is everywhere differentiable(b) f(x) is everywhere continuous but not
differentiable at x = n p, n ∈ Z(c) f(x) is everywhere continuous but not
differentiable at x n n Z= + ∈( ) ,2 1 2p
(d) none of these
25. The function f(x) = |cos x| is(a) differentiable at x = (2n + 1) p/2, n ∈ Z(b) continuous everywhere but not differentiable at
x = (2n + 1) p/2, n ∈ Z(c) neither differentiable nor continuous at x = np,
n ∈ Z(d) none of these
26. If f (x) =|3 – x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f(x) is(a) continuous and differentiable at x = 3(b) continuous but not differentiable at x = 3(c) neither differentiable nor continuous at x = 3(d) none of these
27. Let f xx
x xx
( ),
| |,,
=≤ −
− < <≥ 1
1 11 1
0. Then, f is
(a) continuous at x = –1(b) differentiable at x = – 1(c) everywhere differentiable(d) none of these
28. f(x) = |x3| at x = 0 is(a) continuous and differentiable(b) continuous but not differentiable(c) discontinuous(d) none of these
9Differentiation
29. If
f xx x xax b x
( ),
,,=
+ ≤+ >
2 2 00
then value of a and b such
that f(x) is continuous and differentiable at x = 0 are(a) a = 1, b = 0 (b) a = 2, b = 0(c) a = 0, b = 2 (d) a = 0, b = 1
30. If
f xx x
x x( )
,
,=
+ − ≤ <
− ≤ ≤
3 2 32 0
3 2 0 32
, then f(x) at x = 0 is
(a) discontinuous(b) differentiable(c) continuous but not differentiable(d) none of these
31. Values of a and b such that the function
f x
x xax b x
( ) =≤ 1
+ >
2
2 1ifif
is derivable at x = 1 are
respectively(a) 1, –1 (b) –1, 1(c) 1, 1 (d) –1, –1
32. If f(x)= [x], where [x] denotes the greatest integer less than or equal to x. Then, f(x) at x = 0 is(a) neither continuous nor differentiable(b) continuous only(c) differentiable(d) none of these
33. If f(x) = 12x – 13, if x ≤ 3 = 2x2 + 5, if x > 3is differentiable at x = 3. Then, f ′(3) is equal to (a) –12 (b) 12(c) 23 (d) –23
34. Value of a so that the functionf(x) = x2 + 3x + a, if x ≤ 1 = 6x + 2, if x > 1is differentiable at x = 1, is(a) –3 (b) 6(c) 3 (d) 4
35. f(x) =1 + x, if x ≤ 2 = 5 – x, if x > 2, is
(a) discontinuous at x = 2(b) continuous but not differentiable at x = 2(c) continuous and differentiable at x = 2(d) none of these
2.2 Derivative of Composite Functions
Direction (36 to 97) : Differentiate w.r.t. x.
36. (5x3 – 4x2 – 8x)9
(a) 9(5x3 – 4x2 – 8x)8 · (15x2 – 8x – 8)(b) 9(5x3 – 4x2 – 8x)8 · (15x2 + 8x + 8)(c) 9(5x3 + 4x2 + 8x)8 · (15x2 – 8x – 8)(d) 9(5x3 + 4x2 + 8x)8 · (15x2 + 8x + 8)
37. log (3x2 + 2x + 1)
(a) 2 3 1
3 2 12( )x
x x+
− − (b) 2 3 1
3 2 12( )x
x x+
+ +
(c) 2 3 13 2 12
( )xx x
−+ +
(d) ( )3 1
3 2 12x
x x+
+ +38. sin(x2
+ x) (a) (2x + 1)cos(x2 + x) (b) (2x –1)cos(x2 + x)(c) –(2x + 1)cos(x2 + x) (d) (2x –1)cos(x2 – x)
39. ( )( )log ( )10 10 10 10cosec x
(a) 10 cosec(10x) · cot (10x) · log10 · (10)cosec (10x)
(b) –10 cosec(10x) · cot (10x) · log10 · (10)cosec (10x)
(c) –10 cosec(10x) · cot (10x) · log10(d) 10 cosec(10x) · cot (10x) · log10
40. ( ) ( )log sin log cos4 92 3x x+ (a) 0 (b) 1(c) sinx + cosx (d) –1
41. a a x2 2 2+ +
(a) x
a x a x2 2 2 2+ ⋅ +
(b) x
a x a a x2 2 2 2 2 2+ ⋅ + +(c) x
a x a x2 2 2+ ⋅ +
(d) x
a x a a x2 2 2 2 2+ ⋅ + +
42. cos x
(a) − sin xx2
(b) sin xx2
(c) sin x (d) − sin x
43. cos x3 · sin2(x5)(a) 5x4 · sin(2x5) + 3x2 sin x3
(b) 5x4 · sin(2x5) · cosx3 + 3x2sinx3 · sin2(x5)(c) 5x4 · sin(2x5) · cosx3 – 3x2sinx3 · sin2(x5)(d) –5x4 · sin(2x5) · cosx3 – 3x2sinx3 · sin2(x5)
44. sin23x · tan32x(a) 6 sin23x · sec22x + 3tan32x(b) –6 sin23x · tan22x · sec22x + 3tan32x · sin 6x(c) 6 sin23x · tan22x · sec22x – 3tan32x · sin 6x(d) 6 sin23x · tan22x · sec22x + 3tan32x · sin 6x
10 Objective MHT-CET Mathematics
45. cos(sinx)(a) – cosx · sin(sinx) (b) cosx · sin(sinx)(c) cos (sinx) · sinx (d) – cos (sinx) · sinx
46. tan x
(a) − sin
tan
2
4
x
x x (b) sec
tan
2
4
x
x x
(c) − sec
tan
2 x
x x (d) sec
tan
2 x
x x47. sin(ax2 + bx + c)
(a) (2ax – b) cos(ax2 + bx + c)(b) –(2ax + b) cos(ax2 + bx + c)(c) (2ax + b) cos(ax2 + bx + c)(d) (2ax + b) cos (2ax + b)
48. sin(x2 + x + 5) (a) (2x + 1) cos(2x +1) (b) (2x –1) cos(x2 +x + 5)(c) –(2x + 1) cos(x2 + x + 5) (d) (2x +1) cos(x2 + x + 5)
49. sin x
(a) cossin
xx2
(b) − cossin
xx2
(c) cossin
xx
(d) − cossin
xx
50. 1cot( tan cot )a x b x+
(a) sec2(a tan x + b cotx) · (a sec2x – b cosec2x)(b) sec2(a tan x – b cotx) · (a sec2x – b cosec2x)(c) sec2(a tan x + b cotx) · (a sec2x + b cosec2x)(d) sec(a tan x + b cotx) · (a sec2x + b cosec2x)
51. tan2 (logx3)
(a) 6 3 2x
x xtan(log ) sec (log )⋅
(b) − ⋅6 3 2x
x xtan(log ) sec (log )
(c) 6 3 2 3x
x xtan(log ) sec (log )⋅
(d) tan(logx3) · sec2(log x3)
52. sec(tan )x
(a) sec(tan ) tan(tan ) secx x xx
⋅ ⋅ 2
2
(b) − ⋅ ⋅sec(tan ) tan(tan ) secx x xx
2
2
(c) sec(tan ) tan(tan ) secx x xx
⋅ ⋅ 2
(d) − ⋅ ⋅sec(tan ) tan(tan ) secx x xx
2
53. ( )sina x x
(a) − +
⋅a x x xx
x xsin cos sin2
(b) a x x xx
x x⋅ +
sin cos sin2
(c) a a x x xx
x x⋅ ⋅ +
sin (log ) cos sin2
(d) − ⋅ +
⋅a a x x xx
x xsin (log ) cos sin2
54. sin sin x
(a) cos sin cos
sin
x x
x x
⋅
⋅
(b) cos sin cos
sin
x x
x x
⋅
⋅4
(c) cos sin
sin
x
x x
(d) − ⋅
⋅
cos sin cos
sin
x x
x x4
55. exsin2x + cos2x
(a) ( cos sin ) sin cos2 2 2 2 2x x x ex x x+ +
(b) ( cos sin ) sin cos2 2 2 2 2x x x ex x x+ −
(c) ( cos sin ) sin cos2 2 2 2 2x x x ex x x− −
(d) ( cos sin ) sin cos2 2 2 2 2x x x ex x x− +
56. e ex x5⋅
(a) e xx x55 14+ ⋅ −( ) (b) e xx x5
5 14− ⋅ +( )
(c) e xx x55 14− ⋅ −( ) (d) e xx x5
5 14+ ⋅ +( )
57. ex + 2 logx
(a) (x + 2x2)ex (b) (x – 2x2)ex
(c) (x2 + 2x)ex (d) (x2 – 2x)ex
58. ax5sin
(a) a xx x5 5 5
sin sin cos log⋅ ⋅ ⋅
(b) 5 5sin cos logx x⋅ ⋅
(c) a x ax x5 5 5
sin sin cos log log⋅ ⋅ ⋅ ⋅
(d) − ⋅ ⋅ ⋅ ⋅a x ax x5 5 5
sin sin cos log log
59. log (log )e
x2
(a) 12x xlog
(b) − 12x xlog
(c) 12 log x
(d) − 12 log x
11Differentiation
60. 71x x+
(a) xx
xx
2
2
11 7 7−
⋅ ⋅+
log
(b) xx
xx
2
2
11 7 7+ ⋅ ⋅
+log
(c) xx
xx
2
2
11 7 7− ⋅ ⋅
−log
(d) xx
xx
2
2
11 7 7+ ⋅ ⋅
−log
61. eax bx c2+ +
(a) ( )22
ax b eax bx c− ⋅ + +
(b) ( )22
ax b eax bx c+ ⋅ + +
(c) ( )22
ax b eax bx c+ ⋅ − −
(d) − + ⋅ + +( )22
ax b eax bx c
62. ( )3 44 352x x− +
(a) 152
4 1 3 42 4 332x x x x( )( )− − −
(b) − − − +152
4 1 3 42 4 332x x x x( )( )
(c) 152
4 1 3 42 4 332x x x x( )( )− − +
(d) − + − −152
4 1 3 42 4 332x x x x( )( )
63. sin sinx x+
(a) cossin
cossin
xx
xx x2 4
+
(b) cossin
cossin
xx
xx x2 4
−
(c) − +cossin
cossin
xx
xx x2 4
(d) − −cossin
cossin
xx
xx x2 4
64. 11
−+
coscos
xx
(a) 12 2
2sec x (b) − 12 2
2sec x
(c) sec22x (d) − sec2
2x
65. 84x e x⋅
(a) − ⋅ ⋅ ⋅ +⋅8 8 4 14 4x e xx
e x(log ) ( )
(b) 8 8 4 14 4x e xx
e x⋅ ⋅ ⋅ ⋅ +(log ) ( )
(c) 8 8 4x xe⋅ ⋅(log )(d) − ⋅ ⋅8 8 4x xe(log )
66. log (ex· sin5x)(a) 1 – 5cotx (b) 1 + cotx(c) 1 – 5cotx5 (d) 1 + 5cotx
67. sin2[log(2x + 3)]
(a) −+
⋅ +22 3
2 2 3x
xsin[ log( )]
(b) 22 3
2 2 3x
x+
⋅ +sin[ log( )]
(c) 22 3
2 3x
x+
⋅ +sin[log( )]
(d) −+
⋅ +22 3
2 3x
xsin[log( )]
68. log(emx – e–mx)
(a) e ee e
mx mx
mx mx+−
−
− (b) m e ee e
mx mx
mx mx( )+
−
−
−
(c) e ee e
mx mx
mx mx−+
−
− (d) e ee e
mx mx
mx mx
−
−−
+
69. log sinsin
11
+−
xx
(a) tanx (b) cotx(c) secx (d) –secx
70. 1x a x+ −
(a) −+
+
12
1 1a x a x
(b) 12
1 1a x a x+
−
(c) 1 1x a x+
+ (d) 12
1 1a x a x+
+
71. ( )2 1 52x x+ +
(a) 4 10
5
2
2
x x
x
+ +
+ (b) 4 10
5
2
2x x
x+ +
+
(c) 4 10
5
2
2
x x
x
− −
+ (d) 4 10
5
2
2
x x
x
− +
+72. elog tan x
(a) sec2x (b) – sec2x(c) secx tanx (d) – secx tanx
73. x2 log sin x(a) x2cot x – 2x log sin x (b) xcot x + 2x log sin x(c) xcot x + 2 log sin x (d) x2cot x + 2x log sin x
74. ex log(sin 2x)(a) ex{2 cotx + log(sin 2x)} (b) ex{2 cot2x + log(sin 2x)}(c) ex{2 cotx – log(sin 2x)} (d) –ex{2 cot2x + log(sin 2x)}
12 Objective MHT-CET Mathematics
75. e x
(a) ex
x12
4 (b) − e
x
x12
4
(c) ex
x
4 (d) − e
x
x
4
76. log (log x), x > 1
(a) 1x xlog
(b) − 1x xlog
(c) 1log x
(d) 1x
77. log tan x2
(a) –cosec x (b) –cot2x(c) cosec x (d) cot2x
78. log(log(log x3))
(a) 1x xlog
(b) 13x x x(log )(log )
(c) − 33x x x(log )(log log )
(d) 33 3x x x(log )(log log )
79. ecot x
(a) ecotx(cosec2x) (b) ecotx(cosecx cotx)(c) –ecotx(cosec2x) (d) –ecotx(cosecx cotx)
80. tan(log x)
(a) sec (log )2 x
x (b) −
sec (log )2 xx
(c) sec tan (log )x x xx
(d) −sec tan (log )x x x
x
81. log sin (ex + 5x + 8)(a) –(ex + 5) · cot(ex + 5x + 8)(b) (ex + 5) · cot(ex + 5x + 8)(c) (ex + 5) · cosec2(ex + 5x + 8)(d) –(ex + 5) · cosec2(ex + 5x + 8)
82. e–5xcot 4x(a) –e–5x(5 cot 4x + 4 cosec24x)(b) e–5x(5 cot 4x + 4 cot24x) (c) e–5x(5 cot 4x + 4 cosec24x) (d) –e5x(5 cosec 4x + 4 cosec24x)
83. 3x + 2
(a) (9 × 3x log 3) (b) –(9 x log 3)(c) –(9 × 3x log 3) (d) 9x log 3
84. log( )x x+ +1 2
(a) −+
1
1 2x (b) 1
1 2x x+ +(c) 1
1 2+ x (d) −
+ +
1
1 2x x
85. log sin x2 1+
(a) x
xx
22 2
11
++cot (b) x
xx
22
11
++cot
(c) −+
+x
xx
22
11cot (d) x
xx
22
11
++tan
86. ex cos x
(a) excosx(cosx – x sinx) (b) –excosx(cosx – x sinx)(c) e–xcosx(xcosx – sinx) (d) –e–xcosx(cosx – sinx)
87. 11
+−
ee
x
x
(a) −− −
e
e e
x
x x( )1 1 2 (b)
e
e e
x
x x( )1 1 2+ −
(c) e
e e
x
x x( )1 1 2− − (d) −
+ −
e
e e
x
x x( )1 1 2
88. x3excos x(a) –exx2(x sin x – x cos x + 3 cos x)(b) –exx2(x cos x– x sin x + 3 cos x)(c) exx2(x sin x – x cos x + 3 sin x)(d) exx2(x cos x – x sin x + 3 cos x)
89. esinxsin(ex)(a) esinx(ex sin(ex) + sin x cos(ex))(b) esinx(ex cos(ex) – cos x cos(ex))(c) esinx(ex cos(ex) + cos x sin(ex))(d) –esinx(ex cos(ex) + cos x sin(ex))
90. e x−2
(a) − −1 2
xe x (b) 1 2
xe x−
(c) 1 2
xe x (d) −1 2
xe x
91. tan(x + 45)(a) sec2(x + 45) (b) –sec2(x + 45)(c) sec(x + 45) tan(x + 45)(d) –sec(x+ 45) tan(x + 45)
92. 32 2x x+
(a) − ⋅ ++( log ) ( )3 3 2 12 2x x x
(b) ( log ) ( )3 3 2 12 2x x x+ ⋅ −
(c) 2 3 12 2( ) ( )x x x+ ⋅ +
(d) 2 1 3 32 2( ) ( log )x x x+ ⋅ +
13Differentiation
93. cos(log x)2
(a) 2 2log sin(log )x x
x (b)
log sin(log )x xx
(c) − log sin(log )x x
x
2
(d) −2 2log sin(log )x x
x94. eex
(a) e ee xx× (b) − ×e ee xx
(c) e ex e x× − (d) ee x
95. (x2 + x + 1)4
(a) –4(x2 + x + 1)3 · (2x + 1) (b) –4(x2 + 2x + 1)3 · (2x + 1)(c) 4(x2 + x + 1)3 · (2x + 1) (d) 4(x2 + x + 1)· (2x + 1)
96. 1
2 2a x−(a) −
−x
a x( ) /2 2 3 2 (b) 12 2 3 2( ) /a x−
(c) xa x( ) /2 2 3 2−
(d) −−
12 2 3 2( ) /a x
97. elog(logx)· log3x
(a) log( )x
x
2 (b)
log( )3 2xx
(c) −log( )3 2x
x (d)
− log( )xx
2
98. Find dydx when y = 2u3 + 1 and u
x= 1
2 3/ .
(a) 43x
(b) −4
3x
(c) 4x3 (d) –4x3
99. Find the derivative of the following w.r.t. x : (2x2 + 3)5/3(x + 5)–1/3
(a) − + + −+
( ) ( )( )
/
/2 3 18 100 3
3 5
2 2 3 2
4 3x x x
x
(b) ( ) ( )( )
/
/2 3 18 100 3
3 5
2 2 3 2
4 3x x x
x+ − −
+
(c) ( ) ( )
( )
/
/2 3 18 100 3
3 5
2 2 3 2
4 3x x x
x+ + −
+
(d) − + + ++
( ) ( )( )
/
/2 3 18 100 3
3 5
2 2 3 2
4 3x x x
x
100. Differentiate |2x2 – 1| w.r.t. x.
(a) 4 2 1
2 1
2
2x x
x|( ) |−
− (b) −
−−
4 2 12 1
2
2x x
x| |
(c) 4 2 12 1
2
2x x
x( )
| |−
− (d) 2 1
2 1
2
2xx
−+
2.3 Derivative of Inverse Functions
2.3.2 Derivative of Inverse Composite Functions
Direction(101 to 160) : Differentiate w.r.t.x.
101. cos ( )−1 4 x
(a) 21 16( )− x
(b) −−2
1 16x x( )
(c) 21 16x x( )+
(d) 216 1( )x −
102. 5x.sec–12x
(a) 5
4 15 5 2
21
xx
x xx
−+ −(log ) sec
(b) 5
4 15 5
21
xx
xx
−+ −(log ) sec
(c) 54 1
5 5 21x
x
x xx
( )(log ) sec
−+ −
(d) 5
4 15 2
21
xx
x xx
−+ −sec
103. cos cos− +1 12
x
(a) −12
(b) 12
(c) 1 (d) –1
104. cosec(3 tan–1x)
(a) −
+
− −3
1
1 1
2cosec (tan ) cot(tan )x x
x
(b) 3 3 3
1
1 1
2cosec ( tan ) cot( tan )− −
−
x x
x
(c) 3 3
1
1 1
2cosec (tan ) cot( tan )− −
+
x x
x
(d) −
+
− −3 3 3
1
1 1
2cosec ( tan ) cot( tan )x x
x105. sin–1(2 cos2x –1)
(a) –2 (b) –1
(c) 1 (d) 12
106. cos sin cos− +
1 3 45
x x
(a) 12
(b) 2(c) –1 (d) 0
14 Objective MHT-CET Mathematics
107. sincos sin− +
1
2x x
(a) 2 (b) 1(c) 0 (d) –2
108. tan− −−
13
231 3x x
x(a) 3
1 2+ x (b) −
+3
1 2x
(c) 11 2+ x
(d) −+
11 2x
109. sin−
+
12
21
xx
(a) −+
21 2x
(b) 21 2+ x
(c) 11 2+ x
(d) −+
11 2x
110. tan tantan
− +−
1 a b xb a x
(a) 11 2+ x
(b) –1
(c) 1 (d) 0
111. sin–1(1 –x2)
(a) −
+ −
2
1 1 2 2( )x (b)
−
− −
2
1 1 2
x
x( )
(c) 2
1 1 2
x
x+ −( ) (d)
−
− −
2
1 1 2 2
x
x( )
112. sin–1(2x)
(a) 2 2
1 4
x
x
log
− (b)
2
1 4
x
x−
(c) log 2
1 4+ x (d)
2
1 2
x
x
xlog
−
113. sin− +121
2x
(a) −
−
x
x1 4 (b)
x
x1 2−
(c) x
x1 4− (d)
x
x
2
41 −
114. cosec–1(2x +1)
(a) 2
2 1 12( )x x+ − (b)
2
2 1 2 12( )x x+ +
(c) −
+ + −
2
2 1 2 1 12| | ( )x x
(d) 1
2 1 2 1 12| | ( )x x+ + −
115. cos–1(log 2 x)
(a) −
−
1
2 2 2x x(log ) (log )
(b) 12x xlog log−
(c) 1
2 2 2(log ) (log )− x
(d) 12x log
116. tan sincos
−+
11
xx
(a) −12
(b) 12
(c) 1 (d) 0
117. tan sincos
− −
1 1 xx
(a) 1 (b) 0
(c) −12
(d) –1
118. cot–1(cosec x + cotx)
(a) 1 (b) 12
(c) −12 (d) –1
119. tan coscos
− −+
1 11
xx
(a) 1 (b) 12
(c) −12
(d) –1
120. sincos sin− +
1 4 541
x x
(a) 1 (b) –1
(c) 0 (d) 12
121. sin ( )− −1 22 1x x
(a) 2
1 2− x (b)
−
−
2
1 2x
(c) 1
1 2− x (d) −
−
1
1 2x
15Differentiation
122. sin− + + −
1 1 12
x x
(a) 1
2 1 2− x (b)
−
−
1
2 1 2x
(c) −
+
1
2 1 2x (d)
1
1 2− x
123. tan− −−
12 3
3 23
3a x x
a ax
(a) 3
2 2a x+ (b) 3
2 2a
a x−(c) 3
2 2a
a x+ (d) 3
2 2
a
a x+
124. sin− −+
12
21 251 25
xx
(a) 10
1 25+ x (b)
101 25 2− x
(c) 1
1 25 2+ x (d)
−+
101 25 2x
125. tan−
−
12
71 12
xx
(a) 41 16
31 92 2+
++x x
(b) 41 16
31 9+
++x x
(c) 1
1 161
1 92 2++
+x x (d)
121 16 1 92 2( )( )+ +x x
126. tan− −+
1 x ax a
(a) a
x a
2
2 2+ (b)
−+a
x a2 2
(c) a
x a2 2+ (d)
12 2x a+
127. tan−+
−
112
1 4
x
x
(a) 2 2
1 4
1x
x
+
+log
(b) 2 2
1 4
x
xlog
+
(c) −
+
+2 2
1 4
1x
xlog
(d) 2
1 4
x
x+
128. sin−
1 xa
(a) −
−
12 2a x
(b) 1
a x−
(c) 12 2a x+
(d) 1
2 2a x−
129. cot−
1 xa
(a) a
a x2 2+ (b)
12 2a x+
(c) −+a
a x2 2 (d) −+1
2 2a x
130. 1
+
− −b
xb a
xa
tan tan1 11
(a) −+
++
1 12 2 2 2x b x a
(b) 1 1
2 2 2 2x b x a++
+
(c) 1 12 2 2 2x b x a+
−+
(d) none of these
131. sin(2cos–1x)
(a) 2 2
1
1
2
sin( cos )−
−
x
x (b)
−
+
−2 2
1
1
2
sin( cos )x
x
(c) −
−
−2 2
1
1
2
cos( cos )x
x (d) 2 2
1
1
2
cos( cos )−
−
x
x132. 5x · sec–1x
(a) 5 11
51x
x xx
−+
−( )(log )cosec
(b) 1
15
21
x xx
−+ −( )(log )sec
(c) 5 1
15
21x
xx
−+
−( )(log )sec
(d) 5 1
15
21x
x xx
−+
−( )(log )sec
133. cos–1(2 – x)
(a) 1
1 2 2− −( )x (b) −
− −
1
1 2 2( )x
(c) −
+ −
1
1 2 2( )x (d)
1
2 2− x
134. cosec −
1 14x
(a) −
−
4
1 16 2x (b)
4
1 16 2+ x
(c) 4
1 16 2− x (d)
1
1 16 2− x
135. e xsin(cos )ec −1
(a) e
x
x−1
2
/ (b)
−ex
x1
2
/
(c) ex
x1
2
/ (d) e
x
x−1/
136. sin cosec−
17x
(a) −72x
(b) 72x
(c) 12x
(d) −7x
16 Objective MHT-CET Mathematics
137. sin–1(1 – 2sin2x )(a) 0 (b) 1(c) –1 (d) –2
138. sin sin cos− +
1 4 35
x x
(a) 1 (b) – 12
(c) 0 (d) 12
139. tan− −+
12
211
xx
(a) 2
1 4xx+
(b) −+2
1 4xx
(c) 1
1 2+ x (d) 2
1 2xx+
140. tan− −
1 212
xx
(a) 21 2+ x
(b) 1
1 2+ x
(c) −+
21 2x
(d) −+
11 2x
141. sec co− −+−
1 12 1
x x
xsec
(a) 2
12x − (b)
−
−
2
12x
(c) 1
12x x − (d)
2
12x x −
142. cos ( )− − − ⋅ −1 2 21 1ax a x
(a) −
−
1
1 2x (b)
1
1 2− x
(c) 1
1 2x x− (d)
x
x1 2−143. cos–1(1 – 2x2)
(a) −
−
2
1 2x (b)
2
1 2− x(c) 1
1 2− x (d) 2
12x −144. cot–1(tan 3x)
(a) 3 (b) –3(c) –3x (d) 3x
145. tan− + −
1 21 1xx
(a) 1
1 2+ x (b)
12 1 2( )+ x
(c) −+
12 1 2( )x
(d) −+
11 2x
146. tan sin coscos sin
− +−
1 a x b xa x b x
(a) 1 (b) –1
(c) 0 (d) 1
1 2+ x
147. tan− +−
1 4 33 4
xx
(a) −+
11 2x
(b) 2
1 2+ x
(c) 1
1 2+ x (d)
−+
21 2x
148. tan− ( )1 5x x
(a) 3
1 2+ x (b) −
+3
1 2x
(c) 13 1 2( )+ x
(d) none of these
149. cossin cos
ec−+
1 54 3x x
(a) 0 (b) 1
(c) 54
(d) 53
150. tan cossin
− −
1 1 33
xx
(a) 1
1 2+ x (b)
11 9 2+ x
(c) 32
(d) −32
151. tan− + +
+ −
1 2
2
1
1
x x
x x
(a) 1
1 2+ x (b)
−+
12 1 2( )x
(c) −+
11 2x
(d) 1
2 1 2( )+ x152. sin–1[cos(2x – 3)]
(a) 0 (b) 1(c) 3 (d) none of these
153. tan− +− −
12
5 13 6
xx x
(a) 1
9 12 51
2 2 12 2x x x x+ ++
− +
(b) 39 12 5
12 2 12 2x x x x+ +
+− +
(c) 3
9 12 51
2 2 12 2x x x x+ +−
− +
(d) 3
9 12 51
2 2 12 2x x x x− −+
+ −
17Differentiation
154. cot log
log log− −
+
12
2
x x
e x
x
x x
(a) 11
112 2+
++x x x[ (log ) ]
(b) 1
11
12 2++
+x x[ (log ) ]
(c) 11
112 2+
−+x x[ (log ) ]
(d) 11
12 2+
+x x x(log )
155. cos cos( / ) sin( / )− +
1 3 2 22
x x
(a) –2 (b) −12
(c) 32
(d) − 32
156. cos− −+
1 11
xx
(a) 11 + x
(b) 1
1x x( )+
(c) 1
1( )+ x x (d) −
+1
1( )x x
157. sin sin , .− −+ − − ≤ ≤1 1 21 1 1x x x
(a) 1
11
2−+
x
xx| |
(b) −
−−
1
11
2x
xx| |
(c) 11
1−
−
x
xx| |
(d) none of these
158. 4 1 21 2
1tan coscos
− +−
xx
(a) 0 (b) 3(c) –4 (d) –2
159. tan− −+
1
1x a
ax
(a) 11x x( )+
(b) 1
2 1x x( )+
(c) −
+1
2 1x x( ) (d) 1
2 x
160. tan−
−
12 2
x
a x
(a) 12 2a x−
(b) −
−
12 2a x
(c) 1
2 2a x+ (d)
−
+
12 2a x
2.4 Logarithmic Differentiation
Direction (161 to 183) : Differentiate each of the following w.r.t. x.
161. xx + (tanx)x
(a) xx (1 – log x) + (tan x)x [2cosec 2x + log sin x](b) xx (1 + log x) + (tan x)x [2x cosec 2x + log tan x](c) xx (1 + log x) + (tan x)x [2cosec x + log tan x](d) xx (1 + log x) + 2cosec x + log tan x
162. (x tan x)secx
(a) sec x (x tan x)x [ 1x
+ tan x log (x tan x) + 2cosec x]
(b) sec x (x tan x) [ 1x
+ tan (log (tan x)) + 2cosec 2x]
(c) sec x (x tan x) [x + tan x + 2cosec 2x]
(d) sec x (x tan x)sec x [ 1x
+ tan x · log (x tan x) + 2cosec 2x]
163. (sin )logxx
x
1 2+
(a) sin (log ) log sin( )
x x xx
xx
+ −+2
1 2 2
(b) loglog sin
xx
xxx
+ −+2
1 2
(c) (sin ) cot log
log sinlogxx
x xx
xxx
x
12
12 2++ −
+
(d) cot (log ) log sinx x xx
xx
+ −+2
1 2
164. (sec )21
x x
(a) 2 2 1
2(sec ) [ tan log sec ]
/xx
x x xx
−
(b) 2 sec [ tan log sec ]xx
x x x−
(c) 2 2
2tan [ tan log cos ]xx
x x x−
(d) − −2 2 1
2(sec ) [ tan log sec ]
/xx
x x xx
165. ( ) ( ) ( )x x x+ ⋅ + ⋅ −1 2 3 7 332
52
72
(a) ( ) ( ) ( )
( ) ( )
/ / /x x x
x x x
+ + −
++
++
−
1 2 3 7 33
2 15
2 321
2 7 3
1 2 3 2 7 2
(b) ( ) ( ) ( )
( ) ( )
/ / /x x x
x x x
+ + −
++
+−
−
1 2 3 7 33
15
2 321
2 7 3
5 2 3 2 9 2
18 Objective MHT-CET Mathematics
(c) ( ) ( ) ( )
( ) ( )
/ / /x x x
x x x
+ + −
++
+−
−
1 2 3 7 33
2 15
2 321
2 7 3
3 2 5 2 7 2
(d) ( )( )
/7 3 32 1
52 3
7 2−+
++
xx x
166. (tanx)sinx
(a) (sin x)tan x [cosec x + (cos x) (log tan x)](b) (tan x)sin x [sec x + (cos x) log tan x](c) (tan x)cos x [sec x + (sin x) (log tan x)](d) (tan x)sin x [tan x + (cos x) (log sec x)]
167. xx + ax + xa + aa
(a) (1 + log x) + (ax log a + axa–1)(b) xx(1 + log x) + log a + axa–1
(c) xx(1 + log x) + xa log x + axa–1
(d) xx(1 + log x) + ax log a + axa–1
168. (logx)x – xlogx
(a) (log )log
log(log ) loglogxx
x x xx
x x1 2+
−
(b) (log )log
log(log )loglogx
xx x
xx
x x1 2−
+
(c) (log )log
log(log )log
xx
xx
xx 1 2
+
+
(d) (log ) [log(log )]loglogx x x
xx
x x−
2
169. xx
(a) xx(1 + log x) (b) xx(1 – log x)(c) –xx(1 + log x) (d) xx log x
170. 22 42+
− ⋅ +xx x( )
(a) 22
4 12
12 4
22
+−
++
+−
++
xx
xx x
xx
( )
(b) 22
4 12 2
12 2
24
22
+−
++
+−
++
xx
xx x
xx
( )( ) ( )
(c) 22
4 12 2
12 2
24
22
+−
++
−−
−+
xx
xx x
xx
( )( ) ( )
(d) 22
4 12
12 4
22
+−
++
−−
−+
xx
xx x
xx
( )
171. (1 + x)x
(a) ( ) log( )11
1++
− +
x xx
xx
(b) ( ) log( )11
1 2++
+ +
x xx
xx
(c) ( ) log( )11
1++
+ +
x xx
xx
(d) ( ) log( )11
1 2++
− +
x xx
xx
172. x
x x
xcos
2 4 5+ +
(a) x
x xx
xx x x
x x
xcos cos (sin )(log )2 24 52 4
4 5+ +− − +
+ +
(b) x
x x
xx
x x x
x x
xsin sin (cos )(log )2 24 5
2 4
4 5+ +− − +
+ +
(c) x
x x
xx
x x x
x x
xcos cos (sin )(log )2 24 5
2 4
4 5+ ++ + +
+ +
(d) x
x xx
xx x x
x x
xsin sin (cos )(log )2 24 52 4
4 5+ ++ + +
+ +
173. ex sin x+ xlog x
(a) e x x x xx
xx x xsin log( sin cos )
log+ +
2
(b) e x x x xx
xx x xsin log( sin cos )
log− +
2
(c) e x x x xx
xx x xsin log( cos sin )
log− +
2
(d) e x x x xx
xx x xsin log( cos sin )
log+ +
2
174. x xx xsin (sin )−
+ −1 1
(a) x xx
x
x
xsin sin log− −+
−
1 1
21
+− ⋅
+
−−
−(sin )sin
log(sin )12 1
1
1x x
x xxx
(b) x xx
x
xx
x
xx
x xsin sin log(sin )
log(sin )
− −−
−
−−
+
−−
1 1
21
21
1
1
(c) x xx
x
xx
x
xx
x xsin sin log(sin )
log(sin
− −−
−
+−
−
−+
1 1
21
21
1
2 1))
(d) x xx
x
xx
xx
x xsin cos log(sin )
sinlog(sin
− −−
−−
+−
−
+
1 1
21
11
1
xx)
19Differentiation
175. 1010x
x
.
(a) − + ⋅
10 10 10 1 1010 10x x
x xx
xxlog log log
(b) 10 10 1 1010x xx
xx
log log log+ ⋅
(c) 10 10 10 1 1010 10x xxx
xx x
log log log+ ⋅
(d) 10 10 10 1 1010x x
x
xxlog log log+ ⋅
176. (sin )cosx x−1
(a) (sin ) (sin )(cot )log sincosx x x
x
x
x− − −−
1 121
(b) (sin ) (cos )(cot ) log sincosx x x xx− − −{ }1 1
(c) (sin ) (sin )( )log sincosx x x
x
x
x− − −−
1 121
cosec
(d) (sin ) (cos )(cot )log sincosx x x
x
x
x− − −−
1 121
177. (xx)x
(a) x(x x)x (1 + 2 logx) (b) (x x)x (1 – 2 logx)(c) (x x)x (1 + 2 logx) (d) x(x x)x (1 – 2 logx)
178. xx + x1/x
(a) x x xx
xx x( log )
( log )/111
2+ ++
(b) x x xx
xx x( log )
( log )/111
2+ +−
(c) x x xx
xx x( log )
( log )/111
2+ −−
(d) − + −−
x x xx
xx x( log )
( log )/111
2
179. (x cos x)x + (x sin x)1/x
(a) ( cos ) cos sincos
( sin ) sin cossin
/
x x x x xx
x x x x xx x
x
x
−
+ +
12
(b) ( cos ) {log( cos )} ( sin )log( sin )
/x x x x x xx x
x
x x+
1
2
(c) ( cos ) cos sincos
( sin )
sin cossin
/x x x x xx
x x
x x xx x
x x+
+
−
1
2
(d) ( cos ) cos sincos
log( cos )x x x x xx
x xx − +
+ + −
( sin ) sin cossin
log( sin )x x x x x
x x
x x
xx1
2 2
180. y e e ee x eex ex xx
= + + .
(a) e e e e x ex
x
e e x x
x e e x e x
e x x
x e x x ex
xx x
+ +
+ +
1
1
log
( log )
(b) e e e e x ex
xx e e x e xx e x x ex
− +
1 log
(c) e e e e e x xx e e e x xx e x xx x− +( )1 log
(d) e e xx
x e e x xx x e e x xe x x xx x1 1+
+ +log ( log )
181. (sin x)cosx – x1/x
(a) (sin ) (cos cot sin log(sin ))
log
cosx x x x x
xx
x
x
x
−
−−
1
21
(b) (sin ) (cos sin )logcosx x x x
x
xx x− −
−
1
21
(c) (sin ) (sin )(log(sin ))logcosx x x x
x
xx x−
−
1
21
(d) none of these
182. x xx( )2 32
1+
+
(a) x xx x
xx x
( )2
22
11
26
21
1+
++
+−
+
(b) x x
x xx
x x( )2 3
22
11
26
21
1+
++
+−
+
(c) x xx x
xx x
( ) /2 3 2
22
11
26
21
1++
++
++
(d) ( )x
x xx
x x
2 3
221
12
62
11
++
++
−+
183. (2x + 3)x–5
(a) ( ) ( ) log( )2 3 2 52 3
2 35x xx
xx+ −+
+ +
−
(b) (2x + 3)x – 5 [2(x – 5) + log (2x + 3)]
(c) (2x + 3)x – 5[(x – 5) + log (2x + 3)]
(d) ( ) log( )2 3 52 3
2 35x xx
xx+ −+
+ +
−
20 Objective MHT-CET Mathematics
184. If x = ex/y, then find dydx
.
(a) −+( )
logx y
x x (b) −−( )
logx y
x x
(c) ( )
logx y
x x+
(d) x y
x x−
log185. Differentiate w.r.t. x : x2ex sinx
(a) x e xx
xx2 2 1sin cot− −
(b) x e xx
xx2 2 1sin cot+ +
(c) xe xx
xx sin cot1 +
(d) xe xx
xx sin cot1 1+ +
186. If findy x xx
dydx= +
−( )
( ), .
/
/4
4 3
3 2
4 3
(a) x xx x x x( )
( ) ( ) ( )
/
/+
−⋅ +
+−
−
44 3
12
32 4
163 4 3
3 2
4 3
(b) – x xx x x x( )
( ) ( ) ( )
/
/+
−⋅ +
+−
−
44 3
12
32 4
163 4 3
3 2
4 3
(c) x x
x x x x( )
( )
/
/+
−−
+−
−
44 3
12
34
164 3
3 2
4 3
(d) none of these
187. Differentiate x xx sin−1 w.r.t. x.
(a) ( sin )( log )x x x x
x x
xx
− − +−
12
1
(b) ( sin )( log )( )
x x x x
x x
xx
− − +−
12
12
(c) ( sin )( log )( )
x x x x
x x
xx
− + +−
12
12
(d) ( sin )( log )( )
− + −−
−x x x x
x x
xx
12
12
Direction (188 to 200) : Find dydx
.
188. If y = (x)cosx + (cosx)sinx
(a) ( ) cos (log )sincosx xx
x xx ⋅ −
+ ⋅ − + ⋅(cos ) { sin tan cos log(cos )}sinx x x x xx
(b) ( ) cos (log ) cos
(cos ) {cos (log cos )}
cos
sin
x xx
x x
x x x
x
x
⋅ −
+
(c) ( ) cos (log )sin
(cos ) { sin tan }
cos
sin
x xx
x x
x x x
x
x
⋅ −
+ −(d) none of these
189. y = xx · e(2x+5)
(a) xx e2x + 5 (b) xx e2x + 5(3 – logx)
(c) xx e2x + 5(1 – logx) (d) xx e2x + 5·(3 + logx)
190. y x xex=
3 sin
(a) x xe x
xx
3 3 1sin cot− +
(b) x x
e xxx
sin cot3 1+ −
(c) x xe x
xx
3 3 1sin cot+ −
(d) x x
e xxx
3 1 1sin cot+ −
191. y = (2 – x)3(3 + 2x)5
(a) ( ) ( )2 3 2 153 2
82
3 5− ++
−−
x xx x
(b) ( ) ( )2 3 2 153 2
32
3 5− ++
+−
x xx x
(c) ( ) ( )2 3 2 103 2
32
3 5− ++
−−
x xx x
(d) ( ) ( )2 3 2 103 2
32
3 5− + ⋅+
+−
x xx x
192. y = x sin2x
(a) x xx
x xxsin sin ( cos ) log2 2 2 2⋅ +
(b) x xx
x xxsin cos ( sin ) log2 2 2 2⋅ +
(c) x xx
x xxsin sin ( cos ) log2 2 2 2⋅ −
(d) x xx
x xxsin cos ( sin ) log2 2 2 2⋅ −
193. y x x x= ( )
(a) x x x x x xx x x xx( ) ( )[ (log ) (log ) ]⋅ + +−1 2
(b) x x x xx xx( ) ( )[ log (log ) ]− + +1 2
(c) x x x x xx x xx( ) ( )[ log (log ) ]− + +1 2
(d) x x x x x xx x x xx( ) ( )[ log (log ) ]− + −1 2
21Differentiation
194. y = (tan x)1/x
(a) (tan )sin log(tan )
xx x x
xx1
22 2 −
(b) (tan )cot log(tan )
xx x x
xx1
22 2 −
(c) (tan )log(tan )
xx x x
xx1
22 2cosec +
(d) (tan )log(tan )
xx x x
xx1
22 2cosec −
195. y x xx
= ++
( )3 51
2
(a) x x
x x x x( )3 5
11
26
3 51
1
2++
−+
−+
(b) x x
x x x x( )
( ) ( )3 5
11
26
3 51
2 1
2++
⋅ ++
−+
(c) x xx x x x
( )( )
3 51
1 63 5
12 1
2++
−+
−+
(d) x x
x x x x( )3 5
11 6
3 51
1
2++
++
−+
196. y = esinx + (tan x)x
(a) esinxcosx + (tanx)x[2x cosec2x + log tanx](b) esinxcosx – (tanx)x[x cosec2x + log cotx](c) esinxcosx + (tanx)x[x cosec2x + log sinx](d) esinxcosx – (tanx)x[2x cosec2x + log secx]
197. y = (sinx)logx
(a) (sin ) (log )log sinlogx x x
xx
x ⋅ +
cosec2
(b) (sin ) (log ) cot(log sin )logx x x
xx
x ⋅ −
(c) (sin ) (log ) cot(log sin )logx x x
xx
x ⋅ +
(d) (sin ) (log ) coslog sinlogx x x
xx
x ⋅ −
ec2
198. yx
x= 5
5
(a) 5 55
x
xlog (b)
5 5 55
x
x xlog −
(c) 5 5
5
x
x xlog
(d) − −
5 5 55
x
x xlog
199. y x xx
x x= + +−
cos2
211
(a) xxcosx ·{cosx – xsinx (logx) + cosx(logx)}
−−
412 2
xx( )
(b) xxcosx ·{sinx – xcosx (logx) + sinx(logx)}
−−
412 2
xx( )
(c) xxcosx ·{cosx + xsinx (logx) – cosx(logx)}
−−
412 2
xx( )
(d) xxcosx·{sinx + xcosx logx – sinx(logx)}
200. y x x=
(a) x x
x
x ( log )22
− (b) x
xx
x ( log )12−
(c) x x
x
x ( log )22
+ (d) x
xx
x 12+
log
2.5 Derivative of Implicit Functions
Direction (201 to 235) : Find dydx in each of the
following.
201. x2 + siny = y2 + log (x + y)
(a) 1 2
2 1− +
+ − + −x x y
x y y y x y( )
( ) cos ( )
(b) 1 2
2 1+ +
+ − + +x x y
x y y y x y( )
( ) cos ( )
(c) 1
2 1− +
+ − + −x x y
x y y y x y( )
( )sin ( )
(d) 1
1+ +
+ − + −x x y
x y y y x y( )
( ) cos ( )
202. cos− −+
=1
2 2
2 2x yx y
a
(a) −yx
(b) yx
(c) xy
(d) − xy
203. 1 12 2− + − = −x y a x y( )
(a) 11
2
2−−
xy
(b) 11
−−
yx
(c) 11
−−
xy
(d) 1
1
2
2−−
y
x
204. x3 + y3 + 4x3y = 0
(a) 3 1 4
3 4
2
2 3x y
y x
( )++
(b) −−
−3 1 4
3 4
2
2 3x y
y x
( )
(c) −+
+3 1 4
3 4
2
2 3x y
y x
( ) (d) −
++
3 1
3 4
2
2 3x y
y x
( )
22 Objective MHT-CET Mathematics
205. x3 + x2y + xy2 + y2 = 81
(a) ( )3 2
2 3
2 2
2 2x xy y
x xy y
+ ++ +
(b) − + +
+ +( )3 2
2 3
2 2
2 2x xy y
x xy y
(c) ( )3 2
2 3
2 2
2 2x xy y
x xy y
+ −− +
(d) 3
3
2 2
2 2x xy y
x xy y
+ ++ +
206. xy = yx
(a) log
log
y yx
x xy
−
− (b)
log
log
x yx
y xy
−
−
(c) log
log
y xy
x yx
−
− (d)
log
log
y yx
x xy
+
+
207. sec x yx y a+
−
= 2
(a) −yx (b)
yx
(c) − xy (d)
xy
208. sin tan ( )− −−+
=1
2 2
2 21x y
x ya
(a) yx
(b) −yx
(c) xy
(d) − xy
209. log10
3 3
3 3 2x yx y
−+
=
(a) 99101
2
2xy
(b) 10199
2
2xy
(c) − 99101
2
2xy
(d) − 10199
2
2yx
210. xy = ex – y
(a) x y
x−
+( log )1 (b)
x yx
++( log )1
(c) x y
x x−
+( log )1 (d)
x yx x
++( log )1
211. x3 + y3 = 3ax2y
(a) − −
−x ay x
y ax
( )22 2 (b)
x ay x
y ax
( )22 2
++
(c) x ay x
y x
( )−−2 2 (d)
x ay x
y ax
( )22 2
−−
212. y3 – 3y2x = x3 + 3x2y
(a) x xy yx xy y
2 2
2 222
+ ++ −
(b) − + ++ −
x xy yx xy y
2 2
2 222
(c) x xy yx xy y
2 2
2 2+ ++ −
(d) − + ++ −
x xy yx xy y
2 2
2 2
213. x = cos(xy)
(a) −+
1 y xyx xy
sinsin
(b) 1 +
y xyx xy
sinsin
(c) 1 − y xy
xysin
sin (d) 1 + y xy
xysin
sin214. esiny = xy
(a) −
−y
x y y( cos )1 (b) y
y ycos − 1
(c) y
y ycos + 1 (d) y
x y y( cos )− 1
215. tan− −+
=1
2 2
2 2 2x yx y
k
(a) − xy (b)
yx
(c) xy (d) −
yx
216. y = kx + y
(a) −−
y ky klog
log1 (b) log
logk
y k1 −
(c) −−
loglogk
y k1 (d) y k
y klog
log1 −217. x + y = sin(xy)
(a) y xy
x xycos( )
cos( )−
−1
1 (b) −
−−
y xyx xy
(cos )cos( )
11
(c) −−
y xyx xycoscos( )1
(d) y xyx xycoscos( )1 −
218. x2 + y2 = ex + y
(a) −+ −− −
( )x y x
y x y
2 2
2 22
2 (b) x y x
y x y
2 2
2 22
2
+ −− −
(c) x y x
y x y
2 2
2 22
2
+ ++ +
(d) x y x
y x y
2 2
2 22
2
+ +− +
219. y x xx
=∞
( )...
(a) −
−y
x y x
2
2( log ) (b) yy x
2
2 + log
(c) y
x y x
2
2( log )+ (d) y
x y x
2
2( log )−
220. x y axy3 3 2+ =
(a) − −−
8 33 8
2 2 2
2 2 2a xy xy a x y
(b) 8 3
3 8
2 2 2 2
2 2 2a x y x
y a x
−−
(c) 8 3
3 8
2 2 2
2 2 2a xy x
y a x y
−−
(d) −−
+( )8 3
3 8
2 2 2 2
2 2 2a x y x
y a x y
23Differentiation
221. xy = ex + y
(a) log
( log )
x
x
−− +
2
1 2 (b) −+
log
( log )
x
x1 2
(c) log
( log )
y
x1 2+ (d)
−+
log
( log )
y
x1 2
222. (cosx)y = (cosy)x
(a) − +
+(log cos tan )log cos tan
y y xx x y
(b) log cos tanlog cos tan
x x yy y x
++
(c) log cos tanlog cos tan
y y xx x y
++
(d) − ++
log cos tanlog cos tan
x x yy y x
223. y x y= +
(a) −
−1
2 1y (b) 1
2 1y −
(c) −+1
2 1y (d) 1
2 1y +
224. xy + yx = ab
(a) − +
+
−
−{ . (log )}{ (log ) }
( )
( )y x y y
x x x y
y x
y x
1
1
(b) y x y yx x x y
y x
y x. (log )
(log )
( )
( )
−
−+
+
1
1
(c) y x y y
x x x y
y x
y x+
+
(log )
(log )
(d) − +
+
{ (log )}
(log )
y x y y
x x x y
y x
y x
225. (x2 + y2)2 = xy
(a) − −
−y
xx y
x y
( )
( )
3
3
2 2
2 2 (b) xy
x y
x y
( )
( )
3
3
2 2
2 2−
−
(c) y x y
x x y
( )
( )
3
3
2 2
2 2−
− (d)
− −−
xy
x y
x y
( )3
3
2 2
2 2
226. xy = yx2
(a) − −
−y
xx y yy x x
( log )( log )
(b) xy
x y yy x x
( log )( log )
−−
(c) − −
−x
yx y yy x x
( log )( log )
(d) yx
x y yy x x
( log )( log )2 2
2−
−
227. sin( )xy xy x y+ = −2
(a) 2 2 3
2 2xy y y xy
xy xy x y
− −− +
cos( )
cos( )
(b) −− −
− +2 2 3
2 2xy y y xy
xy xy x y
cos( )
cos( )
(c) 2 2 3
2xy y y xy
xy xy x y
− +− +
cos( )
cos
(d) − − +
− +( cos )
cos2 2 3
2xy y y xy
xy xy x y228. xy2 – x2y = 4
(a) −−−
2
2
2
2xy y
xy x (b) −
−−
2
2
2
2xy x
xy y
(c) 2
2
2
2xy x
xy y
−−
(d) 2
2
2
2xy y
xy x
−−
229. 2x2 –3xy + 4y2 = 8
(a) 4 33 8
x yx y
−−
(b) 3 84 3
x yx y
−−
(c) −−−
4 33 8
x yx y
(d) 3 48 3
x yx y
−−
230. sec(x + y) = xy
(a) −− + +
+ + −
y x y x yx y x y xsec( )tan( )
sec( )tan( )
(b) y x y x y
x y x y x− + +
+ + −sec( )tan( )
sec( )tan( )
(c) y x y x yx x y x y
+ + +− + +
sec( )tan( )sec( )tan( )
(d) y x y
x y x
+ ++ −
tan ( )
tan ( )
2
2
231. x3 + y3 = 3 axy
(a) −−−
ay x
y ax
2
2 (b)
y ax
ay x
2
2−−
(c) ay x
y ax
−−
2
2 (d) −−−
y ax
ay x
2
2
232. sin2 x + cos2y = 1
(a) sinsin
22
yx
(b) − sinsin
22
xy
(c) −sinsin
22
yx
(d) sinsin
22
xy
233. log( ) tanx y xy
2 2 12+ =
−
(a) y xy x
−+
(b) −−+
y xy x
(c) y xy x
+−
(d) −+−
y xy x
24 Objective MHT-CET Mathematics
234. y = x sin y
(a) −−
sin( cos )
yx y1
(b) sin
( cos )x
x y1−
(c) sin
( cos )y
x y1− (d)
−−
sin( cos )
xx y1
235. y xx
xx
=+
+ + ∞
sincos
sincos ...
11 1 to
(a) −+ +
+ + −( )cos sin
cos sin1
1 2y x y xy x x
(b) ( )cos sin
cos sin1
1 2+ +
+ + −y x y xy x x
(c) sin
( cos )y
x y1−
(d) −−
sin( cos )
xx y1
2.6 Derivative of Parametric Functions
236. Find dydx , if x = sin(log t), y = log (sin t)
(a) t t
tcot
cos(log ) (b) − t t
tcot
cos(log )
(c) −t t
tcosec2
cos(log ) (d) −t t t
tcos cotcos(log )
ec
237. If x = a sec3q and y = a tan3q, find dydx at q p= 3
(a) 12 (b) 3
2
(c) −12
(d) − 3
2238. Differentiate log (1 + x2) w.r.t. tan–1 x
(a) – 2x (b) 1
1 2+ x
(c) 2x (d) 2
1 2xx+
239. Find dydx , if x e e y e em m m m
= − = +− −
2 2,
(a) xy
(b) − xy
(c) yx
(d) −yx
240. If xt
ty
t=
+
=+
− −sin , cos ,12
121
1
1
then find dydx
(a) – 1 (b) 0
(c) 1 (d) 12
241. Find dydx , if x = a(q – sinq), y = a(1 – cosq)
(a) − cot q2
(b) cosec2 q2
(c) −cosec2 q2
(d) cot q2
242. Find dydx , if x u y u= + = +1 12 2, log( )
(a) 2
1 2+ u (b)
−
+
2
1 2u
(c) 1
1 2+ u (d)
21 2
uu+
243. Find dydx , if x = q – sinq, y = 1 – cosq, at q p= 2 .
(a) 1 (b) 0
(c) – 1 (d) 12
244. Find dydx , if x = 2 cos t + cos 2t, y = 2 sin t – sin 2t, at
t = p4 .
(a) 2 1− (b) 1 2+
(c) 1 2− (d) 12
245. Find dydx , if y = log(secq + tan q), x = secq, at q p= 4 .
(a) 1 (b) – 1
(c) 12
(d) 2
246. If x a t ty a t t
= −
= +
1 1, , then find dydx
(a) − xy
(b) xy
(c) yx
(d) −yx
247. If x = esin3t, y = ecos 3t, then find dydx
(a) y xx y
loglog
(b) −y xx y
loglog
(c) x xy y
loglog
(d) −x xy y
loglog
248. Differentiate 5x w.r.t. log5x.(a) 5x ⋅ x ⋅ (log 5)2 (b) 5x log 5(c) – 5x ⋅ x ⋅ (log 5)2 (d) 5x ⋅ x ⋅ log 5
249. Differentiate cos–1(sinx) w.r.t. tan–1x.(a) 1 + x2 (b) x2 – 1(c) –(1 + x2) (d) 1 – x2
25Differentiation
250. Find dydx , if x = asec3t, y = b tan2t
(a) − 23
ba tsec
(b) 3
2a
b tsec
(c) − 32
ab tsec
(d) 2
3b
a tsec
251. Find dydx , if x y=
−
= −−
− −tan , tan12
13
22
131 3
q qq
(a) 32
(b) − 32
(c) 23
(d) − 23
252. Find dudv , if u x
xv x
x= −
+
=−
− −cos , tan12
21
211
21
(a) – 1 (b) 1(c) 0 (d) 2
253. Differentiate cos− −+
12
211
xx
w.r.t. tan–1 x
(a) – 2 (b) 1(c) 2 (d) 0
254. Differentiate tanx w.r.t. sin x.(a) sec2 x (b) sec x tan x(c) – sec3 x (d) sec3 x
255. If x = t log t, y = tt, then find dydx
(a) ex (b) – ex
(c) e–x (d) ex + 1
256. Find dydx at q p= 4 , if x = sin2q, y = tan q.
(a) 2 (b) 12
(c) 2 (d) – 2257. Differentiate x5 w.r.t. 5x
(a) −⋅
55 5
4xx log
(b) 5 5
5 4
x
x
⋅ log
(c) −⋅5 5
5 4
x
x
log (d)
55 5
4xx ⋅ log
258. Find dydx , if x = cos–1(4t3 – 3t), y t
t= −
−tan 1 21
(a) 13
(b) − 13
(c) 1 (d) – 1
259. Find the derivative of sec−
−
121
2 1x w.r.t. 1 2− x
at x = 12 .
(a) – 4 (b) 14
(c) 4 (d) − 14
260. If x = ecos2t and y = esin2t, then find dydx .
(a) y xx y
loglog
(b) x xy y
loglog
(c) −y xx y
loglog (d) −
x xy y
loglog
261. If x tt
= −+
11
2
2 and y tt
=+2
1 2 , then find dydx .
(a) − xy
(b) xy
(c) yx
(d) −yx
262. Find dydx when x = a (q – sin q), y = a (1+ cos q).
(a) tan q2 (b) cot q
2
(c) − tan q2
(d) − cot q2
263. Differentiate sin2(q2 + 1) w.r.t. q2.
(a) – sin(2q2 + 2) (b) sin(2q2 + 2)(c) cos(2q2 + 2) (d) – cos(2q2 + 2)
264. Find dydx at t = 2 when x bt
t=
+2
1 2 and y a tt
= −+
( ) .11
2
2
(a) 43
ab
(b) − 43
ab
(c) 34ba
(d) − 34ba
265. Find dydx
when x e y e= +
= −
−q qq q q q1 1,
(a) e− − − −
+ + −
2 3 2
3 21
1
q q q qq q q
( )(b)
e2 3 2
3 21
1
q q q qq q q( )− + + +
+ + −
(c) e− − + + +
+ + −
2 3 2
3 21
1
q q q qq q q( )
(d) e2 3 2
3 21
1
q q q qq q q( )− − −
+ + −
266. Find dydx
when x = a tan q, y = a cot q
(a) −yx
(b) yx
(c) xy
(d) − xy
267. Differentiate sinx3 w.r.t. x3
(a) – cos x3 (b) cos x3
(c) 3x cos x3 (d) – 3x2 cos x3
26 Objective MHT-CET Mathematics
268. If x = 3 cos t – 2 cos3 t and y = 3 sin t – 2 sin3t, then
find dydx .
(a) cot t (b) tan t(c) – cot t (d) – tan t
269. Differentiate xx xsin sinw.r.t. .
(a) x x
x− tansin2 (b)
x xx
+ tansin2
(c) x xx
− cotsin2 (d) tan
sinx x
x−
2
270. Differentiate sin–1 x w.r.t. tan–1x.
(a) 1
1
2
2
+
−
x
x (b) − +
−
( )1
1
2
2
x
x
(c) x
x
2
2
1
1
−
− (d)
1
1 2
+
−
x
x
271. If x a tt
= +−
11
2
2 and y t
t=
−2
1 2( ), find dy
dx .
(a) 12
2+ tat
(b) − +( )12
2tat
(c) tat
2 12
− (d) 2
1 2att+
272. If a > 0, x t t
a= +
1 and y at t=+( )1
, find dydx .
(a) a at
t+ −
⋅
1 1log (b) −
⋅
+
+ −
−a a
tt
tt
a
1 1
11
log
(c) a a
tt
tt
a
+ −
−⋅
+
1 1
11
log( ) (d)
log a
tt
a+
−1 1
273. Differentiate ex w.r.t. x .
(a) −2e xx (b) −e xx
(c) e xx (d) 2e xx
274. If x tt
y tt= + = +1 3 2
2log log ,and then find dy
dx .
(a) t (b) – t
(c) 1t
(d) − 1t
275. Find dydx , when x = at2, y = 2at
(a) − 1t
(b) 1t
(c) t (d) –t
276. If x tt
y tt
dydx= =sin
cos, cos
cos, .
3 3
2 2find
(a) cot 3t (b) tan 3t(c) – cot 3t (d) – tan 3t
277. Find dydx , when x
ty t
t=
+=
+− −cos , sin1
21
2
1
1 1(a) – 1 (b) 1
(c) 0 (d) 12
278. Find dydx , when x = a log t, y = b sin t
(a) bt tacos (b)
−bt tacos
(c) abt tcos
(d) −a
bt tcos
279. Find dydx , when x = eq (sin q + cos q), y = eq (sin q – cos q)
(a) – tan q (b) cot q(c) – cot q (d) tan q
280. Find dydx , when x = a cos q and y = b sin q
(a) − ba
cot q (b) ba
cot q
(c) ba
tanq (d) ab
cot q
281. If x = 2 cos q – cos 2 q and y = 2 sin q – sin 2 q, then
find dydx
(a) cot 32q
(b) −
tan 3
2q
(c) tan 32q
(d) −
cot 3
2q
282. If x a t t= +
1 and y a t t= −
1 , then find dydx
(a) − xy
(b) xy
(c) yx
(d) −yx
283. Find dydx , when x a t t= +{ }cos log tan1
2 22
and y = a sin t(a) cot t (b) – tan t(c) – cot t (d) tan t
27Differentiation
284. Find dydx , when x e y e= −
= +
−q qq q q q1 1and
(a) e23 2
2 31
1q q q q
q q q
+ + −
− + +
(b) e2
2 3
3 211
q q q qq q q
+ − −+ + +
(c) e− − − −+ + −
23 2
3 211
q q q qq q q
( ) (d) e− − + +
+ + −2
2 3
3 211
q q q qq q q
( )( )
285. Find dydx , when x = b sin2q and y = a cos2q
(a) − ab
(b) ab
(c) ba
(d) − ba
286. Find dydx when x = a (q – sin q) and y = a(1 – cos q)
(a) − cot q2
(b) cot q2
(c) tan q2
(d) − tan q2
287. Find dydx , when x t y t= =4 4,
(a) 12t
(b) t2
(c) − 12t
(d) – t2
288. If x = e–cos2t and y = e–sin2t, then dydx =
(a) y xx y
loglog (b) −
y xx y
loglog
(c) y yx x
loglog
(d) −y yx x
loglog
289. Find dydx , when x = log t, y = sin t
(a) t cost (b) cost
t(c)
1t tcos (d) – t cost
290. Find dydx , when x = sin t, y = cos 2t
(a) 4 sint (b) 4sint
(c) – 4 sint (d) sint4
2.7 Higher Order Derivatives (Second Order Derivatives)
Direction (291 to 294) : Find d ydx
2
2 .
291. x3 + 5x2 – 3x + 10(a) 6x + 10 (b) – 6x – 15(c) 6x2 + 10x (d) 3x2 + 10x –3
292. e4x · cos 5x(a) e4x (9 cos 5x + 40 sin 5x) (b) – e4x (40 cos 5x + 9 sin 5x)(c) – e4x (9 cos 5x + 40 sin 5x)(d) e4x (40 cos 5x + 9 sin 5x)
293. y = log xex
2
(a) 22x
(b) − 22x
(c) ex
x
2 (d) x2
2294. y = sec x – tanx
(a) − −
⋅ −
12 4 2 4 2
2sec tanp px x
(b) 12 4 2
2sec p −
x
(c) 12 4 2 4 2
sec tanp p−
⋅ −
x x
(d) 12 4 2 4 2
2sec tanp p−
⋅ −
x x
295. Find d ydx
2
2 , if x = at2, y = 2at
(a) 1
2 3at (b) − 1
2 3at
(c) 1t (d) log t
296. If y = x2ex, then d ydx
2
2
(a) –ex(x2 + 4x) (b) e–x(x2 + 4x)(c) ex(x2 + 4x + 2) (d) ex(2x + 4)
297. If ax2 + 2hxy + by2 = 0, then d ydx
2
2 =
(a) 0 (b) 1(c) – 1 (d) 2
298. Find d ydx
2
2 , if y = cos(log x)
(a) cos(log ) sin(log )x x
x
−2
(b) sin(log ) cos(log )x x
x
−2
(c) cos(log ) sin(log )x xx−
2
(d) sin(log ) cos(log )x xx
−2
28 Objective MHT-CET Mathematics
299. Find d ydx
2
2 , if y = log (1 – cos x)
(a) 12 2
2cosec x
(b) − 1
2 2 2cosec x xcot
(c) −
12 2
2cosec x (d)
12 2 2
cosec x xcot
300. Find d ydx
2
2 , if x = acos3q, y = asin3q
(a) secsin
4
3qqa
(b) sec
sin
2
3qqa
(c) sec
sin
4
3qq
(d) secsin
4 qq
301. If y = emx + e–mx, then d ydx
2
2 =
(a) my (b) my2
(c) m2y (d) m2y2
302. Find d ydx
2
2 , if : x = a (q – sin q), y = a (1 – cosq)
(a) 14 2
4cosec q
(b) −
14 2
4cosec q
(c) −
14 2
2sec q (d) none of these
303. Find d ydx
2
2 , if y = x3 + 4x + 3
(a) – 6x (b) 2x(c) – 2x (d) 6x
304. If x = a(1 + cosq), y = a(q + sinq), then d ydx
2
2 2at q p= is
(a) 0 (b) 1
(c) 1a (d) − 1
a
305. Find d ydx
2
2 , if y = e3x + x3 + 5
(a) 9e3x + 6x (b) 9e3x + x(c) 3e3x + x (d) 3e3x + 6x
306. If y = sin (logx), find d ydx
2
2 .
(a) − +{sin(log ) cos(log )}x x
x2
(b) {sin(log ) cos(log )}x x
x
+2
(c) sin(log )x
x2
(d) − +{sin(log ) cos(log )}x x
x4
307. If y = e4x sin 3x, find d ydx
2
2 .
(a) ex(7 sin 3x + cos 3x)(b) e2x(7 sin 3x + 24 cos 3x)(c) e4x(7 sin 3x + 24 cos 3x)(d) e4x(7 sin x + 24 cos 3x)
308. If y x d ydx
= −tan , find .12
2
(a) 2
1 2xx+
(b) −+2
1 2 2xx( )
(c) 2
1 2 2xx( )+
(d) 2
1 2 2( )+ x
309. If x = a (q + sinq) and y = a (1 – cosq), findd ydx
2
2 at
q p= 2 .
(a) 1a
(b) −1a
(c) – a (d) a
Direction (310-314) : Find d ydx
2
2 .
310. y = x11.(a) – 110 x9 (b) 110 x9
(c) 110 x2 (d) – 110 x8
311. y = (x4 + cot x)(a) 12x + 2 cosec x + cot x(b) 12x2 + 2 cosec2 x cot x(c) 12x2 + cosec2 x cot x(d) x2 + cosec2 x cot x
312. y = x3 log x(a) 5 + 6 log x (b) 5x + 6 log x(c) 5x + 6x log x (d) 5 + 6x log x
313. y = 2 sinx + 3 cos x(a) 2 sin x + 3 cos x (b) – 2 cos x – 3 sin x (c) 2 cos x + 3 sin x (d) – 2 sin x – 3 cos x
314. y = (cosec x + cot x)(a) y2 cosec x (b) y cosec x(c) y cosec2 x (d) y2 cosec2 x
315. If x = a(q – sinq) and y = a(1 – cos q), find d ydx
2
2 at q = p.
(a) 1
4a (b)
a4
(c) − 14a
(d) − a4
29Differentiation
Direction (316-327) : Find d ydx
2
2.
316. y = sin 3x cos 5x.(a) 2 sin x – 32 sin 8x (b) 2 sin 2x + 32 sin x(c) sin 2x – 32 sin 8x (d) 2 sin 2x – 32 sin 8x
317. y = x sin x (a) x sin x – 2 cos x (b) x sin x + 2x cos x(c) sin x + 2 cos x (d) – x sin x + 2 cos x
318. y xx= log
(a) ( log )2 3
3x
x
− (b)
( log )2 32x
x+
(c) 2 3
2log x
x
− (d)
2 3log xx
−
319. y = e3x sin 4x(a) e3x(24 cos x – 7sin x)(b) e2x(24 cos 4x – 7sin 4x)(c) ex(24 cos 4x – 7sin 4x)(d) e3x(24 cos 4x – 7sin 4x)
320. y = 2x3 + 3x2 + 6(a) 12x – 6 (b) 12x + 6(c) 6x + 12 (d) 12 + 3x
321. xa
yb
2
2
2
2 1− =
(a) ba y
2
2 3 (b) b
a y
4
2 3
(c) −ba y
4
2 3 (d) −b
a y
4
3 2
322. y = |x|3
(a) 6 x (b) 6x(c) 2 x (d) does not exist
323. y2 = ax2 + b
(a) aby2
(b) aby3
(c) a by
3 (d) ab
y
324. y = x
x x2 3 2− +
(a) 21
223 3( ) ( )x x−
+−
(b) 11
113 2( ) ( )x x−
+−
(c) 21
223 3( ) ( )x x−
−+
(d) −−
+−
21
423 3( ) ( )x x
325. y = ax
(a) ax (log a) (b) a2x (log a)2
(c) ax (log a)2 (d) xa log a
326. y = emsin–1x
(a) e mx
mxx
m xsin/( )
−
−+
−
1 2
2 2 3 21 1
(b) e mx
mxx
m xsin( )
−
−+
−
1
1 12 2 2
(c) e mx
mxx
xsin/( )
−
−+
−
1 2
2 2 3 21 1
(d) e mx
mxx
m xsin
1 12−+
−
327. y cos–1x(a) cosec2y cot y (b) cosec y cot2 y(c) cosec y cot y (d) –cosec2y cot y
328. If x = a(cos t + t sin t) and y = a (sin t – t cos t), findd ydx
2
2 .
(a) −1 2at
t(sec ) (b) 1 3a
t(sec )
(c) 1 3at
t(sec ) (d) 1 3at
ec t(cos )
329. If x = t2 and y = t3, find d ydx
2
2 .
(a) 34t
(b) 34t
(c) 32t
(d) 32t
330. If y = | logex|, find d ydx
2
2 .
(a) 1 0 1 1 1x
xx
xif and if< < − >
(b) 1 0 1 1 12 2x
xx
xif and if< < − >
(c) x, if 0 < x < 1(d) x2, if 0 < x < 1
331. If x = 2 cos t – cos 2t, y = 2 sin t – sin 2t, find d ydx
2
2 at
t = p2
(a) 32
(b) 1
(c) −32
(d) – 1
30 Objective MHT-CET Mathematics
332. If x = a cos q, y = b sin q, then d ydx
2
2 =
(a) −ba y
4
2 3 (b) ba y
2
2 2
(c) b
ay
3
3 (d) ba y
4
2 3
333. If y = x3 log x, then d ydx
4
4 =
(a) −6x
(b) 6x
(c) 1x
(d) −1x
Direction (334-340) : Find the second derivative in each of the following.
334. y = x3 + tan x(a) 6x – 2 sec x – tan2x (b) 6 + 2x sec2x tan x (c) 6x + 2 sec2x (d) 6x + 2 sec2x tan x
335. y = x4 log x(a) x2(5 + 6 log x) (b) 5x2 + log x(c) x2(7 + 12 log x) (d) 7x2 + 3 log x
336. y = log(log x)
(a) − +( log )( log )
12x
x x (b) 1+ log
logx
x x
(c) 12
+ log( log )
xx x
(d) 1− loglog
xx x
337. y = e–x cosx(a) 2ex sin x (b) e–x sin x(c) 2e–x sin x (d) ex sin x
338. y = e6x cos 3x(a) 9ex (3 cos 3x + 4 sin 3x)(b) 9e6x (3 cos 3x – 4 sin 3x)(c) 9e6x (3 cos 3x + 4 sin 3x)(d) 9e2x (3 cos x – 4 sin x)
339. y = xx
(a) yx
x1 1 2+ +
( log ) (b) − + +
yx
x1 1 2( log )
(c) yx
x1 2+
(log ) (d) − +
yx
x1 2(log )
340. ey (x + 1) = 1
(a) −+1
1 2( )x (b) 1
1( )x +
(c) 11 2( )x +
(d) −+11x
341. If y = axn+1 + bx–n, then x d ydx
22
2 =
(a) n(n – 1)y (b) n(n + 1)y(c) ny (d) n2y
342. If x = a cos nt – b sin nt, then d xdt
2
2 =
(a) n2x (b) –n2x(c) –nx (d) nx
343. If y xa bxe
x=
+
log , then x3 y2 =
(a) a xa bx
2 2
2( )+ (b) (1 + x)2
(c) a xbx
−
2
(d) (1 – y)2
344. If y2 = ax2 + bx + c, then y d ydx
32
2 is(a) a constant (b) a function of x only(c) a function of y only (d) a function of x and y
345. If y = etanx, then (cos2x)y2 = (a) (1 – sin 2x)y1 (b) –(1 + sin 2x)y1
(c) (1 + sin 2x)y1 (d) none of these
346. If x = a sin t – b cos t, y = a cos t + b sin t, then d ydx
2
2 =
(a) x y
y
2 2
3+
(b) yx y
3
2 2+
(c) −+y
x y
3
2 2 (d) − +x yy
2 2
3
347. If log y = tan–1 x, then (1 + x2)y2 =(a) (1 – 2x) y1 (b) (1 + 2x) y1
(c) (2x – 1) y1 (d) 2xy1
348. If y = (sin–1x)2, then (1 – x2)y2 =(a) 2 – xy1 (b) – (2 + xy1)(c) 2 + xy1 (d) xy1
349. If y = 3e2x + 2e3x, then d ydx
2
2 =
(a) 2e2x + 6e3x (b) 12e2x – 6e3x
(c) 12e2x – 18e3x (d) 12e2x + 18e3x
350. If y = x–x, find d ydx
2
2
(a) x xx
x (log )2 1+{ } (b) x xx
x ( log )1 12+ −{ }(c) x x
xx (log )2 1−{ } (d) none of these
31Differentiation
LEVEL - 2
1. If y = log5(log7x), find dydx
(a) 15x xlog . log
(b) −15x xlog . log
(c) 1x xlog
(d) 17x xlog . log
2. If y x x
x x= + +
+ −
log2
2
25
25, find dy
dx
(a) −
+
2
252x (b) 1
252x +
(c) 2
252x + (d) −
+
1
252x
3. If yx x
=+
13 cos cot
,ec
find dydx
(a) −+
coseccosec cot
xx x3
(b) cosec
cosec cotx
x x3 3 +
(c) cosec
cosec cot
2
3
xx x+
(d) −
+cosec
cosec cot
2
3
xx x
4. If y e xx
x= ⋅ −+
log ,3234
3 find
dydx
(a) 3 143 4 3
−− +( )( )x x
(b) 3 143 4 3
+− +( )( )x x
(c) 3 144 3
+− +( )( )x x
(d) 3 144 3
−− +( )( )x x
5. If y = sin2(log(2x + 3)), find dydx
(a) 22 3
2 2 3x
x+
+. sin( log( ))
(b) −+
+22 3
2 2 3x
x. sin( log( ))
(c) 12 3
2 2 3x
x+
+. sin( log( ))
(d) −+
+12 3
2 2 3x
x. sin( log( ))
6. Differentiate ( )2 7 42 53 x x− − w.r.t. x.
(a) − − − −53
2 7 4 4 72 2 3( ) .( )/x x x
(b) 53
2 7 4 4 72 2 3( ) .( )/x x x− − −
(c) 35
2 7 4 4 72 2 3( ) .( )/x x x− − +
(d) − − − −35
2 7 4 4 72 2 5( ) .( )/x x x
7. Differentiate 13 7
17 3x x+
−−
w.r.t. x.
(a) 32
13 7
17 33 2 3 2( ) ( )/ /x x+
+−
(b) −+
−−
12
13 7
17 33 2 3 2( ) ( )/ /x x
(c) 12
13 7
17 33 2 3 2( ) ( )/ /x x+
+−
(d) −+
+−
32
13 7
17 33 2 3 2( ) ( )/ /x x
8. Differentiate sec tansec tan
x xx x
+− w.r.t. x.
(a) − +
+
tan .secp p
4 2 4 22x x
(b) sec24 2p +
x
(c) tan .secp p4 2 4 2
2+
+
x x
(d) sec . tanp p4 2 4 2
+
+
x x
9. If theny x xx x
dydx= −
+ =sec tansec tan ,
(a) − −
−
tan secp p
4 2 4 22x x
(b) − +( ) +( )tan secp p4 2 4 2
2x x
(c) tan secp p4 2 4 2
2−( ) +( )x x
(d) tan secp p4 2 4 2
2+( ) −( )x x
10. If y = exsin2x + cos2x, then dydx =
(a) (2xcos 2x – sin 2x)exsin2x + cos2x
(b) (2xcos 2x + sin 2x)exsin2x + cos2x
(c) (2cos 2x – xsin 2x)exsin2x + cos2x
(d) (2cos 2x + xsin 2x)exsin2x + cos2x
11. If y e ee e
x x
x x= +−
−
− , then dydx =
(a) 4
1
2
2 2e
e
x
x( )− (b)
−−
41
2
2 2e
e
x
x( )
(c) 21
2
2 2e
e
x
x( )− (d)
−−
21
2
2 2e
e
x
x( )
32 Objective MHT-CET Mathematics
12. If y a
x
x=
+
cos,
1 2 then dy
dx =
(a) a x a x x
x
xcos [( )(log )(sin ) ]1 2
1
2
23
+ −
+( )(b)
− + −
+( )a x a x x
x
xcos [( )(log )(sin ) ]1 2
1
2
23
(c) a x a x x
x
xcos [( )(log )(sin ) ]1 2
1
2
23
+ +
+( )(d)
− + +
+( )a x a x x
x
xcos [( )(log )(sin ) ]1
1
2
23
13. If f(x) = logx (logx), then f ′(x) at x = e is
(a) 1e (b) 1
(c) e (d) none of these
14. If y
x
x=
+ ( )− ( )
−tancos
cos,1
12
12
then dydx =
(a) 12 (b)
−12
(c) 14 (d)
−14
15. If y x x x x= − + −( )−sin ,1 21 1 then dydx =
(a) −
−+
−
2
1
1
22 2
x
x x x (b)
−
−−
−
1
1
1
22 2x x x
(c) 1
1
1
22 2−+
−x x x (d)
1
1
1
22 2−−
−x x x
16. If y xx
=+
−cos log(log )
,12
21 then
dydx =
(a) −
+2
1 2x x( (log ) ) (b) 2
1 2x x( (log ) )+
(c) −
+1
1 2x x( (log ) ) (d)
11 2x x( (log ) )+
17. If f x xx
( ) cos (log )(log )
= −+
−12
211
, then f ′(e) =
(a) 1 (b) 22e
(c) 2e (d) 1
e
18. If y x a x
x a x= + +
+ −−cot ,1
2 2
2 2 then dy
dx =
(a) −+a
a x2 2 2( ) (b)
aa x2 2 2( )+
(c) −+a
a x
2
2 22( ) (d)
aa x
2
2 22( )+
19. If y xx
xx
=+
+ −+
− −tan cot ,12
141 5
3 22 3
then dydx =
(a) 5
1 252
12 2++
+x x (b)
51 25
212 2+
−+x x
(c) −+
51 25 2x
(d) 51 25 2+ x
20. If y xx x
dydx
x=
+ −=(cos ) ,
1 2 then
(a) y x x x xx x
log(cos ) tan+ + −+ −
2 11 2
(b) y x x x xx x
log(cos ) tan− − −+ −
2 11 2
(c) y x x x xx x
log(cos ) tan+ − −+ −
2 11 2
(d) y x x x xx x
log(cos ) tan− + −+ −
2 11 2
21. If y x xx x= +3 3( ) , then dy
dx =
(a) x x x x xx x3 2 31 3 1( log ) ( log )+ + +
(b) x x x x xx x3 2 31 3 1( log ) ( log )+ + +
(c) x x x x xx x3 2 31 1( log ) ( log )+ + +
(d) x x x x xx x3 2 31 3 3 1( log ) ( log )+ + +
22. If y x xx x= +− −(sin ) ( ) ,cos1 1
then dydx =
(a) (sin ) log sinsin
− −−
−−
1 12 11
x x x
x xx
+ +−
− −x
xx
x
xxcos cos log1 1
21
(b) (sin ) log sinsin
− −−
+−
1 12 11
x x x
x xx
+ +−
− −x
xx
x
xxcos cos log1 1
21
(c) (sin ) log sinsin
− −−
+−
1 12 11
x x x
x xx
+ −−
− −x
xx
x
xxcos cos log1 1
21
(d) (sin ) log sinsin
− −−
−−
1 12 11
x x x
x xx
+ −−
− −x
xx
x
xxcos cos log1 1
21
33Differentiation
23. Differentiate ( )( )x xx x− +
+ +3 4
3 4 5
2
2 w.r.t. x.
(a) 12
3 4
3 4 5
2
2( )( )x x
x x
− +
+ +
1
32
46 4
3 4 52 2xx
xx
x x−+
+− +
+ +
(b) −− +
+ +12
3 4
3 4 5
2
2( )( )x x
x x1
3 46 4
3 4 52 2xx
xx
x x−+
++ +
− −
(c) 12
3 4
3 4 5
2
2( )( )x x
x x
− +
+ +
1
32
46 4
3 4 52 2xx
xx
x x−+
+− +
− −
(d) 12
3 4
3 4 5
2
2( )x x
x x
− +
+ +1
32
46 4
3 4 52 2xx
xx
x x−−
++ +
+ +
24. If x a y a dydx
t t= = =− −sin cos,
1 1 then
(a) yx
(b) xy
(c) − yx (d) − x
y
25. Differentiate tan− + −
1 21 1xx
w.r.t.
cos .− + +
+
1 2
2
1 1
2 1
x
x(a) 1 (b) tan x
2
(c) –1 (d) 0
26. If y = Aemx + Benx, then y2 =(a) mAemx + nBenx (b) m2Aemx + n2Benx
(c) m2emx + n2enx (d) –m2Aemx – n2Benx
27. Find dydx if x t
ty t
t= =sin
cos, cos
cos
3 3
2 2.
(a) cot 3t (b) tan 3t(c) –cot 3t (d) –tan 3t
28. If x = f(t), y = g(t) and x, y are differential functions of
t, then d ydx
2
2 =
(a) g t f t f t g tf t
″( ⋅ ′( − ″( ⋅ ′′(
) ) ) ( )[ )]2
(b) g t f t g t f tf t
″( ⋅ ′( − ′( ⋅ ″′(
) ) ) ( )[ )]3
(c) g t f t f t g tf t
′( ⋅ ″( − ′( ⋅ ″(′(
) ) ) ))
(d) f t g t g t f tf t
″( ⋅ ′( − ″( ⋅ ′(′(
) ) ) )[ )]3
29. The derivative of (logx)x w.r.t. log x is
(a) x x x xx(log ) log log(log )1 +
(b) (logx)x (log x + x)(c) x[1 + log (log x)] (d) (logx)x[1 + log (logx)]
30. The derivative of log10x w.r.t. logx 10 is
(a) − (log )(log )
x 2
210 (b)
(log )(log )
x 1010
2
2
(c) 1 (d) (log )log
x 2
210
31. Differentiate ( cos ) ( sin )x x x xx x+1
w.r.t. x.(a) (x cosx)x [sec x (cos x – x sin x) + log (x cos x)]
+
( sin ) /x x
x
x
2
1 [cosec x (x cos x + sin x)
– log (x sin x)](b) (x cosx)x [sec x (cos x + x sin x) + log (x cos x)]
+( sin ) /x x
x
x1
2 [cosec x (x cos x + sin x) – log (x sin x)](c) (x cosx)x [cos x + x sin x) + log (x cos x)]
+( sin ) /x x
x
x1
2 [(x cos x – sin x) – log (x sin x)]
(d) ( cos ) [log( cos )] ( sin ) /x x x x x x
xx
x+ ×
1
2[log (x sin x)]
32. If y = e2cos–1(3x), then ( )1 9 22
2− =x d ydx
(a) 36 9y x dydx
− (b) 9 36x dydx
y+
(c) 36 9x dydx
y+ (d) 36 9x dydx
y−
33. If (x – a)2 + (y – b)2 = c2, then 1
2 3 2
2
2
+
dy
dxd ydx
/
is
(a) a constant independent of a only.(b) a constant independent of b only.(c) a constant independent of a and b.(d) none of these
34 Objective MHT-CET Mathematics
34. If y x xm
= + +{ }2 1 , then (x2 + 1) y2 =
(a) xy1 – m2y (b) m2y1 – xy(c) xy – m2y1 (d) m2y – xy1
35. If y x x x=
∞(cos ) ,(cos )(cos )...
then dydx =
(a) y xy x
2
1tan
( log cos )− (b) −
−y xy x
2
1tan
( log cos )
(c) y xy x
tan( log cos )1 − (d)
−−
y xy x
tan( log cos )1
36. If y x x x= + + + ∞log log log ... ,to then dydx =
(a) −
−1
2 1x y( ) (b) 1x
(c) 1
2 1x y( )− (d) −1x
37. If xy log(x + y) = 1, then dydx =
(a) y x y x yx xy x y
( )( )
2
2+ ++ +
(b) − + ++ +
y x y x yx xy x y
( )( )
2
2
(c) x xy x yy x y x y
( )( )
2
2+ ++ +
(d) − + −+ −
y x y x yx x y xy
( )( )
2
2
38. If xy + yx = (x + y)x + y, find dydx
(a) ( ) { log( )} log
log ( ) { l
x y x y yx y y
x x xy x y
x y y x
y x x y+ + + + −
+ − + +
+ −
− +1
1
1
1 oog( )}x y+
(b) ( ) { log( )} log
log ( ) { l
x y x y yx y y
x x xy x y
x y y x
y x x y+ + + + +
− + + −
+ −
− +1
1
1
1 oog( )}x y+
(c) ( ) { log( )} log
log ( ) { log(
x y x y yx y y
x x x y x y
x y y x
y x y+ − + + −
− + + +
+ −
+1
1
1
))}(d) none of these
39. If y axx a x b x c
bxx b x c
cx c= − − − + − − + − +
21( )( )( ) ( )( ) ,
then dydx =
(a) −−
+−
+−
yx
aa x
bb x
cc x
(b) xy
aa x
bb x
cc x−
+−
+−
(c) yx
aa x
bb x
cc x−
+−
+−
(d) yx
aba x
bcb x
cac x−
+−
+−
40. Find dydx , when y
x a x a
x a= + − −
−
( ) ,2
2 2 where x > a > 0.
(a) −−2 2
2 2 3 2a
x a( ) / (b) 2 2
2 2 3 2a
x a( ) /−
(c) 2
2 2 3 2a
x a( ) /− (d)
−−
22 2 3 2
ax a( ) /
41. If y e ee e
x x
x x= −+
−
− , then dydx =
(a) y2 – 1 (b) y – 1(c) –y2 (d) 1 – y2
42. Find dydx , when x a t t= +
cos log tan 2
and
y = a sin t, 0 2< <t p .
(a) –tan t (b) tan t(c) cot t (d) – cot t
43. Differentiate log sin x2
3 1−
w.r.t. x
(a) cot
log sin
x
x
2
2
31
31
−
−
(b) x x
x
cot
log sin
2
2
31
31
−
−
(c) x x
x
cot
log sin
2
2
31
33
1
−
−
(d) x x
x
tan
log sin
2
2
31
33
1
−
−
44. Differentiate x x x x1 12 2+ + + +log( ) w.r.t. x
(a) − +2 12x (b) 2 12x +
(c) 1
1 2x x+ (d)
2
12x +
45. If x = secq – cosq and y = secn q – cosnq, then dydx =
(a) sec cos
sec cos
n nq qq q
−+
(b) n n n(sec cos )
sec cosq q
q q−
+
(c) sec cos
cos sec
n nq qq q
++
(d) n n n(sec cos )
sec cosq q
q q+
+
35Differentiation
46. If ya b
a ba b
x=−
+−
−222 2
1tan tan , then dydx =
(a) 1a b x− cos
(b) 1
a b x+ cos
(c) −
−1
a b xcos (d) 1
a x bcos −
47. Find dydx when
y x xx x
x= + + −+ − −
< <−cot sin sin
sin sin, .1 1 1
1 10 2
p
(a) − 12
(b) 1
(c) 12
(d) 0
48. If y xx
xx
=−
+ +−
− −tan sec12
12
22
111
find d ydx
2
2 .
(a) 1
1 12 2−+
+xx
x x| | ( )
(b) 21 2x
x x| | ( )+
(c) 1
1212 2+
+−xx
x x| | ( )
(d) none of these
49. If y x x
xx=
−+ −
−sin log ,1
22
11 then dy
dx =
(a) sin
( ) /
−
−
1
2 1 21x
x (b) −
−
−sin( ) /
1
2 3 21x
x
(c) 1
1 2 3 2( ) /− x (d) sin
( ) /
−
−
1
2 3 21x
x
50. Differentiate x xx
xx+
+
+1 1 1 w.r.t. x.
(a) xx
xx
xx
x+
−+
+ +
1 11
12
2
2log
+ + −
+x x
xx
xx
1 1
2 21 log
(b) − +
−+
+ +
+
+x
xxx
xx
xx
x1 11
12
2
2 1 1
logxx+
12
(c) xx
xx
x xx
xx+
−+
+ +
+1 11
12
2
1 1
2
(d) xx
xx
x xx
xx+
−+
+
+1 11
2
2
1 1
2log
51. If y = (logcosxsinx)(logsinx cosx)–1 + sin–1 21 2
xx+
,
find dydx at x = p
4 .
(a) 8
232
162log−
+p (b) 32
168
22p +−
log
(c) 816
3222p +
−log
(d) 82
32162log
++p
52. If x = 2 cos q – cos2q and y = 2 sin q – sin2q, find d ydx
2
2 at q p= 2 .
(a) 32
(b) 1
(c) − 32
(d) –1
53. If y x a x a xa
dydx= − +
=−2 2
2 22
1sin then
(a) x a2 2− (b) a x2 2+
(c) a2 – x2 (d) a x2 2−
54. If y = { tan log( )},2 11 2x x x dydx
− − + =then
(a) – 2 tan–1x (b) 2 cot–1x(c) 2 tan–1x (d) – 2 cot–1x
55. If y = xcosx + (cosx)x, then dydx =
(a) x xx
x x x xx xcos cos (sin )(log ) (cos ) {log cos }−{ } +
(b) x xx x xxcos cos (sin )(log )⋅ −{ }
+ (cosx)x[(log cos x)– x tan x]
(c) x xx
x xxcos cos (sin )(log )+{ } + (cosx)x {log cos x + x tan x}
(d) x xx
x xxcos cos (sin )(log )+{ } + ( cos x)x {log cos x – tan x}
56. If y x xx
x
= ++
+ + ∞
11
1...
, then dydx =
(a) y
y x( )2 − (b)
−−y
y x2
(c) xy x2 −
(d) −−x
y x2
36 Objective MHT-CET Mathematics
57. If x = a sin 2t(1 + cos2t) and y = bcos 2t (1 – cos 2t),
then dydx t
=
=at p4
(a) − ba
(b) ab
(c) ba
(d) − ab
58. Differentiate 2 4 11
2 1
2 3 2
x x xx
+ + ⋅ −−( ) / w.r.t. x.
(a) y xx
( ) log2 1 2 24 1
+ +−
(b) y x xx
( ) log2 1 2 312+ −
−
(c) y xx
xx
( )2 1 24 1
312+ +
−−
−
(d) y xx
xx
( ) log2 1 2 24 1
312+ +
−−
−
59. For the function f given by f(x) = x2 – 6x + 8, f ′(5) =(a) 4 (b) 10(c) 3 (d) 6
60. Let f xx xx x
( )( ),( ),
,=+ ≥− <
2 02 0
ifif
then f (x) is
(a) everywhere differentiable(b) not differentiable at x = 0(c) differentiable in [–1, 1](d) none of these
37Differentiation
2015
1. If the function g xk x xmx x
( ),,
=+ ≤ ≤
+ < ≤
1 0 32 3 5
is
differentiable, then the value of k + m is
(a) 103 (b) 4
(c) 2 (d) 165 (JEE Main)
2014
2. If f and g are differentiable functions in [0, 1] satisfying f (0) = 2 = g(1), g(0) = 0 and f (1) = 6, then for some c ∈ ]0, 1[(a) 2 f ′(c) = 3g′(c) (b) f ′(c) = g′(c)(c) f ′(c) = 2g′(c) (d) 2f ′(c) = g′(c)
(JEE Main)
2013
3. If y = sec(tan–1 x), then dydx
at x = 1 is equal to
(a) 12
(b) 1
(c) 2 (d) 12
(JEE Main)
CompetitiveExams
2012
4. Consider the function, f(x) = |x – 2| + |x – 5|, x ∈ R Statement 1 : f ′(4) = 0 Statement 2 : f is continuous in [2, 5], differentiable
in (2, 5) and f (2) = f (5).
(a) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.
(b) Statement 1 is true, Statement 2 is false.(c) Statement 1 is false, Statement 2 is true.(d) Statement 1 is true, Statement 2 is true; Statement
2 is a correct explanation for Statement 1.
(JEE Main)
2011
5. d xdy
2
2equals to
(a) d y
dx
dydx
2
2
2
− (b) −
−d y
dx
dydx
2
2
3
(c) d y
dx
2
2
1
−
(d) −
− −d y
dx
dydx
2
2
1 3
(JEE Main)
38 Objective MHT-CET Mathematics
LeveL - 11. (a) : We have, f(0) = 0
lim ( ) lim sinx x
f x xx→ →
=
0 0
1
= 0So, the function f(x) is continuous at x = 0.
Now, Rf f h fh
h hhh h
′ = + − = −→ →
( ) lim ( ) ( ) lim sin( / )0 0 0 1 00 0
=→
lim sin( / ),h
h0
1 which does not exist.
Hence, f(x) is not differentiable at x = 0
2. (b) : We have, f(x) = x3 ⇒ f(3) = (3)3 = 27
lim ( ) lim ( )x h
f x h→ →+
= +3 0
33
= + + +
→lim ( )
hh h h
03 227 27 9
= 27
lim ( ) lim ( )x h
f x h→ →−
= −3 0
33
= − − + =
→lim ( )
hh h h
03 227 27 9 27
= 27\ = =
→ →+ −f f x f x
x x( ) lim ( ) lim ( )3
3 3
Hence, f(x) is continuous at x = 3.
Now, Rf ′(3) = lim ( ) ( )h
f h fh→
+ −0
3 3
= + −
→lim ( ) ( )
h
hh0
3 33 3
= + + + −
→lim ( )
h
h h hh0
3 227 27 9 27
= + + =
→lim ( )
hh h
02 27 9 27
Lff h f
hh
hh h′ =
− −−
= − −−→ →
( ) lim( ) ( )
lim ( ) ( )33 3 3 3
0 0
3 3
= − − + −
−→lim [ ]
h
h h hh0
3 227 27 9 27
= + − =
→lim ( )
hh h
02 27 9 27
Hints & explanations\ Rf ′ (3) = Lf ′(3) = 27So, the function f(x) is differentiable at x = 3
3. (a)
4. (a) : We have f (2) = |2 – 2| = 0.lim ( ) lim ( ) lim | |
x h hf x f h h
→ → →+= + = + −
2 0 02 2 2
= = =→ →
lim | | limh h
h h0 0
0
lim ( ) lim ( ) lim | |x h h
f x f h h→ → →−
= − = − −2 0 0
2 2 2
= − = =
→ →lim | | limh h
h h0 0
0
So, f(x) is continuous at x = 2.
Now, Rf ′(2) =+ −
→lim ( ) ( )
h
f h fh0
2 2
=
+ − −→
lim| |
h
hh0
2 2 0
= =
→lim ( )
h 01 1
Lff h f
hh′ =
− −−→
( ) lim( ) ( )
22 2
0
=
− − −−
= − = −→ →
lim| |
lim ( )h h
hh0 0
2 2 01 1
\ Rf ′(2) ≠ Lf ′(2)So, f(x) is not differentiable at x = 2.
5. (a) : lim ( ) lim ( ) lim ( )x h h
f x f h h→ → →+
= + = − −1 0 0
1 2 1
= − =
→lim ( )
hh
01 1
lim ( ) lim ( ) lim ( )x h h
f x f h h→ → →−
= − = − =1 0 0
1 1 1
f ( ) ( )1 2 1 1= − =
lim ( ) lim ( ) ( )x x
f x f x f→ →+ −
= = =1 1
1 1
So, f(x) is continuous at x = 1.
Now, Rf ′(1) =+ −
→lim
( ) ( )h
f h fh0
1 1
= − − −
=
− −
= −
→ →lim ( ) lim
h h
hh
hh0 0
2 1 1 1 1 1
Lff h f
hh′ =
− −−→
( ) lim( ) ( )
11 1
0
39Differentiation
= − −
−
=→
limh
hh0
1 1 1
\ Rf ′(1) ≠ Lf ′(1)So, f(x) is not differentiable at x = 1.
6. (a) : lim ( ) lim ( ) lim [( ) ]x h h
f x f h h→ → →+
= + = + −1 0 0
21 1 1
= + + − =
→lim ( )
hh h
021 2 1 0
lim lim ( ) lim [ ]x h h
f h h→ → →−
= − = − + =1 0 0
1 1 1 0
f ( ) ( )1 1 1 02= − =
lim ( ) lim ( ) ( )x x
f x f x f→ →+ −
= = =1 1
1 0
\ f(x) is continuous at x = 1.
Now, Rf ′(1) =+ −
→lim ( ) ( )
h
f h fh0
1 1
= + + − −
→lim ( )
h
h hh0
21 2 1 0
= + =
→lim ( )
hh
02 2
Lff h f
hh
hh h′ =
− −−
= − +−
= −→ →
( ) lim( ) ( )
lim ( )11 1 1 1 1
0 0
Q Rf ′(1) ≠ Lf ′(1)So, f(x) is not differentiable at x = 1.
7. (b) : f(2) = 1 + 2 = 3lim ( ) lim ( ) lim( )
x h hf x f h h
→ → →−= − = + − =
2 0 02 1 2 3
lim ( ) lim ( ) lim( )x h h
f x f h h→ → →+
= + = − − =2 0 0
2 5 2 3
lim ( ) lim ( ) ( )x x
f x f x f→ →− +
= = =2 2
2 3
\ f(x) is continuous at x = 2
Lf f h fh
hhh h
′ = − −−
=
+ − −−
→ →
( ) lim ( ) ( ) lim ( )2 2 2 1 2 30 0
= − −
−
=
−−
=→ →
lim limh h
hh
hh0 0
3 3 1
Rf f h fhh
′( = + −
→
2 2 20
) lim ( ) ( )
= − + −
→
lim ( )h
hh0
5 2 3
= − −
=
−
= −→ →
lim limh h
hh
hh0 0
3 3 1
Lf ′(2) ≠ Rf ′(2)\ f(x) is not differentiable at x = 2
8. (c) : f(0) = 0
lim ( ) lim cosx x
f x x x→ →=
=0 0
1 0
\ f(x) is continuous at x = 0
f f h fh
h hhh h
′(0 = + −
=
−
→ →) lim ( ) ( ) lim
cos
0 0
0 01 0
=
→
lim cos ,h h0
1 which does not exist.
... f ′(0) does not exist.\ f(x) is not differentiable at x = 0
9. (a) : f(2) = 2(2) – 3 = 4 – 3 = 1lim ( ) lim ( ) lim ( )
x h hf x f h f h
→ → →−= − = − − =
2 0 02 2 1 1
lim ( ) lim ( ) lim( )x h h
f x f h h→ → →+
= + = + − =2 0 0
2 4 2 3 1
\ f(x) is continuous at x = 2
Now, Rff h f
hh′ =
+ −
→
( ) lim( ) ( )
22 2
0
=
+ − −
→
lim( )
h
hh0
2 2 3 1
= + − =
→lim ( )
h
hh0
4 2 4 2
Lff h f
hh′ =
− −−
→
( ) lim( ) ( )
22 2
0
= − − −
−
→
limh
hh0
2 1 1
= −
−
=→
limh
hh0
1
Q Rf ′ (2) ≠ Lf ′ (2)\ f(x) is not differentiable at x = 2.
10. (b) : f(x) is continuous at x = 0, then
lim ( ) ( ) lim sinx x
pf x f x x p→ →
= ⇒
= ⇒ >
0 00 1 0 0
f(x) is differentiable at x = 0, if
lim ( ) ( )x
f x fx→
−−
0
00 exists
⇒
−
→lim
sin
x
px xx0
1 0exists
⇒
→
−lim sinx
px x01 1 exists ⇒ p – 1 > 0 or p > 1
If p ≤ 1, then lim sinx
px x→−
0
1 1 does not exist and
at x = 0, f(x) is not differentiable
40 Objective MHT-CET Mathematics
Hence for 0 < p ≤ 1, f(x) is continuous at x = 0 but not differentiable.
11. (b) :
f x
xx
xx
x
( )
,,,,
,
=
− − ≤ < −1− − ≤ <
≤ <≤ <≤
2 21 1 00 0 11 1 2
2 2
ifififif
if
f(1) = 1lim ( ) lim ( ) lim( )
x h hf x f h
→ → →−= − = =
1 0 01 0 0
lim ( ) lim ( ) lim( )x h h
f x f h→ → →+
= + = =1 0 0
1 1 1
lim ( ) lim ( ) ( )x x
f x f x f→ →− +
≠ =1 1
1
\ f(x) is not continuous at x = 1
Lf f h fhh
′(1) = − −−
→
lim ( ) ( )0
1 1
= −
−
→
limh h0
0 1=
= ∞→
limh h0
1
Rf f h fh hh h
′ = + −
=
−
=→ →
( ) lim ( ) ( ) lim1 1 1 1 1 00 0
... Rf ′(1) ≠ Lf ′(1)\ f(x) is not differentiable at x = 1
12. (c) : f(0) = 0
lim ( ) lim( )
x hf x h
→ →−= =
0 00
lim ( ) lim( )x h
f x→ →+
= =0 0
0 0
lim ( ) lim ( ) ( )x x
f x f x f→ →− +
= =0 0
0
\ f(x) is continuous at x = 0
Lf f h fhh
′( = − −−
→
0 0 00
) lim ( ) ( )
= − −
−
→
limh
hh0
0 0 = −−
=→
limh
hh0
1
Rf f h fh hh h
′ = + −
=
−
=→ →
( ) lim ( ) ( ) lim0 0 0 0 0 00 0
... Rf ′(0) ≠ Lf ′(0)\ f(x) is not differentiable at x = 0
13. (a) : Every differentiable function is continuous but every continuous function is not necessarily differentiable.
14. (b) :
f xx x
x x( )
( ),,
=− − <
− ≥1
1 11
forfor
f(1) = 0lim ( ) lim ( ) lim[ ( )]
x h hf x f h h
→ → →−= − = − − − =
1 0 01 1 1 0
lim ( ) lim ( ) lim( )x h h
f x f h h→ → →+
= + = + − =1 0 0
1 1 1 0
lim ( ) lim ( ) ( )x x
f x f x f→ →+ −
= =1 1
1
\ f(x) is continuous at x = 1
Lf f h fhh
′ = − −−
→
( ) lim ( ) ( )1 1 10
= − − − −
−
→
lim ( )h
hh0
1 1 0 = −
= −→
limh
hh0
1
Rf f h fhh
′(1 = + −
→
) lim ( ) ( )0
1 1
= + − −
=
=→ →
lim ( ) limh h
hh
hh0 0
1 1 0 1... Rf ′(1) ≠ Lf ′(1)\ f(x) is not differentiable at x = 1
15. (c) : f(1) = 1lim ( ) lim ( ) lim( )
x h hf x f h h
→ → →+= + = + − =
1 0 01 2 2 1 1
lim ( ) lim ( ) lim ( )x h h
f x f h f h→ → →−
= − = − =1 0 0
1 1 1
lim ( ) lim ( ) ( )x x
f x f x f→ →− +
= =1 1
1
\ f(x) is continuous at x = 1
Lf f h fhh
′ = − −−
→
( ) lim ( ) ( )1 1 10
= − −
−
=
−−
=→ →
lim limh h
hh
hh0 0
1 1 1
Rf f h fh
hhh h
′( = + −
=
+ − −
→ →
1 1 1 2 1 1 10 0
) lim ( ) ( ) lim ( )
= + −
=
=→ →
lim limh h
hh
hh0 0
2 2 2 2 2... Rf ′(1) ≠ Lf ′(1)\ f(x) is not differentiable at x = 1
16. (a) : Given
f x x a x a x a
x a( ) ( )cos ,
,= − −
≠
=
1
0
lim ( ) ( )cosx a h
f x a h a a h a→ →= + − + −
lim
0
1
=
=→
lim cosh
h h0
1 0
lim cos
hh h h
→= ≤ ≠ 0
0
0 1 1and for all
Hence limx a
f x f a→
=( ) ( ) ⇒ f(x) is continuous at x = a.
Also, f a f a h f ahh
′( = + −→
) ( ) ( )lim0
41Differentiation
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