QUESTION BANK ON
APPLICATION OF DERIVATIVE
MATHEMATICS

Select the correct alternative : (Only one is correct)
Q.11 Suppose x1 & x2 are the point of maximum and the point of minimum respectively of the function
f(x) = 2x3 9ax2 +12a2x+1 respectively, thenfortheequality 2
1
x =x2 to betrue thevalue of'a'must be
(A) 0 (B*) 2 (C) 1 (D) 1/4
[Hint: f (x) = 6 (x2  3 ax + 2 a2) = 6 (x  2 a) (x  a) = 0  x = 2 a or a
f (x) = 6 (2 x  3 a)
 
  



f a a
f a a
( )
( )
2

If a then x a
x a
If a then x a
x a
 

 

0
2
0 2
1
2
1
2
Now x1
2 = x2  a2 = 2 a  a = 2 other optionnot valid ]
Q.22 Point 'A' lies on the curve
2
x
e
y 
 and has the coordinate (x,
2
x
e
) where x > 0. Point B has the
coordinates (x, 0). If'O' isthe origin then the maximumarea ofthe triangleAOB is
(A)
e
2
1
(B)
e
4
1
(C)
e
1
(D*)
e
8
1
[Sol. A =
2
e
x
2
x

; A' = 




  
 2
2
x
2
x
e
·
x
2
e
2
1
=  
2
x
x
2
1
2
e
2


= 0  x =
2
1
gives A
Amax.
 Amax =
2
2
e 2
1

=
e
8
1
]
Q.33 The angle at which the curve y= KeKx intersects the y-axis is :
(A) tan1 k2 (B*) cot1 (k2) (C) sec1
1 4



 


k (D) none
[Hint:
dy
dx x



 0
= k2  tan  = k2  cot


2






 = k2



2






 = cot1 k2 = sin1 1
1 4
 k
 B ]
Q.46 {a1, a2, ....., a4, ......} is a progression where an =
n
n
2
3
200

. The largest termof this progression is :
(A) a6 (B*) a7 (C) a8 (D) none
[Hint: Let y =
x
x
2
3
200

;
dy
dx
= 2
3
3
)
(
)
(
200
x
x
400
x


Now if x > (400)1/3, y is decreasing and
if x < (400)1/3, y is increasing hence
y is greatest at x = (400)1/3.
But x  N hence practical maxima occurs at
x = 7 or x = 8 ; a7 =
49
543
; a8 =
64
712
]

Q.57 Theangle betweenthetangent linesto the graphofthe function f(x)=  
x
2
dt
)
5
t
2
( at thepoints where
the graph cuts the x-axis is
(A)
6

(B)
4

(C)
3

(D*)
2

Q.68 The minimumvalue ofthe polynomial x(x + 1) (x+ 2) (x + 3) is :
(A) 0 (B) 9/16 (C*)  1 (D) 3/2
[Hint: Note the graph of f(x). Least value coincideswithlocalminima
y = (x2 + 3x) (x2 + 3x + 2) = z (z + 2)
= (z + 1)2  1 = (x2 + 3x + 1)2  1
yleast =  1 ; this occurs where z = – 1 i.e. x2 + 3x + 1 = 0
or
dy
dx
= 2 (2x + 3) (x2 + 3x + 1) = 0
 x =
 
3 5
2
, 
3
2
or
 
3 5
2
Here x=
 
3 5
2
& x=
 
3 5
2
are the points oflocalminima and x = 
3
2
is the point oflocal
maxima . Localmaximumvalue =
9
16
]
Q.79 The minimumvalue of
 
tan
tan
x
x
 
6
is :
(A) 0 (B) 1/2 (C) 1 (D*) 3
[Hint: f(x) has a period equal to  & can take values (, )  3 is the local minimumvalue.
y =
 
 
2
6
2
6
sin cos
sin cos
x x
x x




=
 
 
sin sin
sin sin
2
6 6
2
6 6
x
x
 
 
 
 
= 1 +
 
1
2
6 6
sin sin
x  
 
y is minimum if 2x +

6
=

2
 x =

6
 ymin = 1 + 2 = 3 ]
Q.810 The difference betweenthe greatest and the least values ofthe function, f (x) = sin2x– xon 




 


2
,
2
(A*)  (B) 0 (C)
3
2
3 
 (D)
3
2
2
3 


[Hint: f 




 

2
is maximumand f 




 
2
is minimum  A
A ]

Q.911 The radius ofa right circularcylinder increases at therate of0.1 cm/min, and the height decreasesat the
rate of0.2cm/min. The rate ofchangeofthe volume ofthecylinder, incm3/min, whenthe radius is2 cm
and the height is 3 cmis
(A) – 2 (B) –
5
8
(C) –
5
3
(D*)
5
2
[Sol. Given, V = r2h
Differentiatingbothsides
dt
dV
= 






 h
dt
dr
r
2
dt
dh
r2
= r 






dt
dr
h
2
dt
dh
r
dt
dr
=
10
1
and
dt
dh
= –
10
2
dt
dV
= 






















10
1
h
2
10
2
r
r = )
h
r
(
5
r



Thus, when r = 2 and h = 3,
dt
dV
= )
3
2
(
5
)
2
(



=
5
2
]
Q.1014 If a variable tangent to the curve x2y= c3 makes intercepts a, b onx and y axis respectively, then the
value ofa2b is
(A) 27 c3 (B)
3
c
27
4
(C*)
3
c
4
27
(D)
3
c
9
4
[Hint : x2y = c3
x2
dx
dy
+ 2xy = 0 
dx
dy
=
x
y
2

equationoftangent at (x,y)
Y – y = )
x
X
(
x
y
2


Y = 0, gives , X =
2
x
3
= a
and X = 0 , gives , Y = 3y = b
Now a2b = y
3
.
4
x
9 2
= 
 3
2
c
4
27
y
x
4
27
(C) ]
Q.1118 Difference betweenthegreatest and the least values ofthe function
f (x) = x(ln x – 2) on [1, e2] is
(A) 2 (B*) e (C) e2 (D) 1
[Sol. y = x (ln x – 2)
y' = x 





x
1
+ (ln x – 2) = ln x – 1
dx
dy
= ln x – 1 = 0  x = e

now f (1) = – 2
f (e) = – e (least)
f (e2) = 0 (greatest)
 difference = 0 – (–e) = e Ans. ]
Q.1219 Let f (x) =


n
2
0
r
r
n
x
tan
x
tan
, n  N, where x  




 
2
,
0
(A*) f(x)is bounded and it takes bothofit's bounds and the range of f(x) contains exactlyoneintegral
point.
(B) f(x) is bounded andit takes bothofit's bounds and therangeof f(x) contains more thanoneintegral
point.
(C) f(x) is bounded but minimumand maximumdoesnot exists.
(D) f(x) is not bounded as the upper bound does not exist.
[Hint: Let tan x = t
 f (x) = n
2
4
n
t
....
t
....
t
1
t





=
1
t
1
t
.......
t
1
t
t
1
t
1
1
n
1
n
n
n
























 


1
n
2
1

[Equalityholds at x = /4 ]
also f (0) = 0  range of f (x) is 





1
n
2
1
,
0 ]
Q.1320 If f (x) = x3 + 7x – 1 then f (x) has a zero between x = 0 and x= 1. The theoremwhich best describes
this, is
(A) Squeeze playtheorem (B) Meanvalue theorem
(C)Maximum-Minimumvaluetheorem (D*) Intermediatevalue theorem
[Hint: f (0) = – 1 ; f (1) = 7. So f (0) and f (1) have opposite sign  (D)]
Q.1421 Consider the function f (x) =








0
x
for
0
0
x
for
x
sin
x
then the number of points in (0, 1) where the
derivative f(x) vanishes , is
(A) 0 (B) 1 (C) 2 (D*)infinite
[Hint: f (x) vanishes at points where sin
x

= 0 i.e.
x

= k , k = 1, 2, 3, 4, .....
hence x =
k
1
. Also f ' (x) = sin
x

–
x

cos
x

if x  0
Since the functionhas a derivativeat anyinterior point ofthe interval(0, 1), also continuous in[0,1] and
f(0) =f(1) henceRolle'stheoremis applicable to anyone ofthe interval 





1
,
2
1
, 





2
1
,
3
1
, ... 





 k
1
,
1
k
1
hence  some c in each of these intervalwhere f  (c) = 0  infinite points  (D) ]

Q.1523 The sumoflengths ofthehypotenuse and another sideofa right angledtriangle is given. Theareaofthe
triangle willbemaximumifthe angle betweenthemis :
(A)

6
(B)

4
(C*)

3
(D)
5
12

[Hint: A =
2
cx
2
c
x 2

f(x) = 4A2 = x2(c2 – 2cx) ]
Q.1625 Inwhichofthe following functions Rolle’stheoremis applicable?
(A) f(x) =






1
x
,
0
1
x
0
,
x
on [0, 1] (B) f(x) =










0
x
,
0
0
x
,
x
x
sin
on [–, 0]
(C) f(x) =
1
x
6
x
x2



on [–2,3] (D*) f(x) =













1
x
if
6
]
3
,
2
[
on
,
1
x
if
1
x
6
x
5
x
2
x 2
3
[Hint: (A) : discontinuous at x =1  not applicable
(B) : f(x) is not continuous at x =0 hence (B) is incorrect.
(C) : discontinuityat x = 1  not applicable
(D) : Notice that x3 – 2x2 – 5x + 6 = (x–1) (x2 –x –6). Hence , f(x) = x2 – x – 6 if 1
x  and f(1) = – 6
 f is continuous at x = 1. So f(x) = x2 –x– 6 throughout the interval [–2,3].
Also, note that f(–2) = f(3) = 0. Hence, Rolle’s theoremapplies. f(x) = 2x –1.
Setting f'(x)= 0 , we obtain x = 1/2 whichlies between –2 and 3. ]
Q.1727 Suppose that f(0) = – 3 and f' (x)  5 for allvalues ofx. Thenthe largest value whichf(2) canattain is
(A*) 7 (B) – 7 (C) 13 (D) 8
[Sol. Using LMVT in[0, 2]
0
2
)
0
(
f
)
2
(
f


= f ' (c) where c  (0, 2)
2
3
)
2
(
f 
 5
f (2)  7  (A) ]
Q.1832 The tangent to the graphofthefunction y= f(x) at the point with abscissa x= a formswith the x-axis
an angle of /3 and at the point withabscissa x = b at an angle of/4, thenthe value ofthe integral,
a
b
 f (x) . f (x) dx is equal to
(A) 1 (B) 0 (C) 3
 (D*) –1
[ assume f (x) to be continuous ]
[Hint : f (a) = 3 and f (b) = 1 ; now put f (x) = t or I.B.P.
.
Now I =







f x
a
b
( )
2
=
   
  
f b f a
( ) ( )
2 2
2
=
1 3
2

=  1 ]

Q.1933 Let C be the curve y= x3 (where xtakes allrealvalues). The tangent atAmeets the curve again at B. If
the gradient at B is Ktimes the gradient atAthen K is equalto
(A*) 4 (B) 2 (C) – 2 (D)
4
1
[Sol.
dx
dy
= 3x2 = 3t2 at 'A'
 3t2 =
t
T
t
T 3
3


= T2 + Tt + t2
T2 + Tt – 2t2 = 0
(T – t)(T + 2t) = 0  T = t or T = – 2t (T = t is not possible)
now, mA =
t
t3
= t2 ; mB = T2
A
B
m
m
= 2
2
t
T
= 2
2
t
t
4
(using T = – 2t)
mB = 4 ]
Q.2034 The vertices ofa triangle are (0, 0), (x, cos x) and (sin3x, 0) where 0 < x <
2

. The maximum area for
sucha triangle insq. units, is
(A*)
32
3
3
(B)
32
3
(C)
32
4
(D)
32
3
6
[Hint: A(x) =
2
x
cos
x
sin3
; A
Amaximum occurs at x =
3

]
Q.2136 The subnormalat anypoint on the curve xyn = an + 1 is constant for :
(A) n = 0 (B) n = 1 (C*) n =  2 (D) no value ofn
Q.2238 Equation ofthe line through the point (1/2, 2) and tangent to the parabola y=
 x2
2
+ 2 and secant to
the curve y = 4 2
 x is :
(A*) 2x + 2y  5 = 0 (B) 2x + 2y  3 = 0 (C) y  2 = 0 (D) none
[Sol. Equation ofline through(1/2, 2) is y – 2 = m 






2
1
x .... (1)
Solve it with y = 2
2
x2


2
2
x
2
1
x
m
2
2











0
2
m
mx
2
x2



D = 0 gives

m2 – 4 





2
1







2
m
= 0  m2 + m = 0  m = 0 or m = – 1
with m = 0 , Tangent is y= 2 (rejected)
with m=1 , Tangent is y = – x +
2
5
]
Q.2339 The lines y = 
3
2
x and y = 
2
5
x intersect the
curve 3x2 + 4xy+ 5y2  4 = 0 at the points P and Q
respectively. The tangents drawn to the curve at P
and Q :
(A) intersect each other at angle of45º
(B) are parallelto each other
(C*) are perpendicular to each other
(D) none of these
[Hint:
dy
dx
= 










y
5
x
2
x
3
y
2

dy
dx x y



1 1
= 0 &
dy
dx x y



2 2
=  tangents are perpendicular]
Q.2442 The least value of'a' for which the equation,
4 1
1
sin sin
x x


= a has atleast one solution onthe interval(0, /2) is :
(A) 3 (B) 5 (C) 7 (D*) 9
[Hint: 2
2
)
x
sin
1
(
x
cos
x
sin
x
cos
4
dx
dy



 = 0 gives sinx =
2
3
note that f(x)  as x  0+ or x 

2
and between two maxima we have a minima.]
Q.2543 If f(x) = 4x3  x2  2x + 1 and g(x) = [  
Min f t t x x
x x
( ) : ;
;
0 0 1
3 1 2
   
  
then
g
1
4





 + g
3
4





 + g
5
4





 has the value equalto :
(A)
7
4
(B)
9
4
(C)
13
4
(D*)
5
2
[Hint: f  (x) = 12x2 – 2x – 2
= 2[6x2 – x – 1] = 2(3x + 1)(2x – 1)
Hence g(x) =





f x if x
f if x
x if x
( ) 0
1
2
1
2
1
2
1
3 1 2
 





  
  
]

Q.2645 Given: f (x) =
3
/
2
x
2
1
4 






 g (x) =







0
x
,
1
0
x
,
x
]
x
[
n
ta
h (x) = {x} k (x) = )
3
x
(
log2
5 
then in [0, 1] Lagranges Mean Value Theorem is NOT applicable to
(A*) f, g, h (B) h, k (C) f, g (D) g, h, k
[Hint: fis not differentiable at x=
2
1
g is not continuous in [0, 1] at x = 0
h is not continuous in [0, 1] at x = 1 & 0
k (x) =
5
n2
)
3
x
(
l
 = (x + 3)p where 2 < p < 3 ]
Q.2750 Two curves C1 : y = x2 – 3 and C2 : y = kx2 , R
k intersect each other at two different points. The
tangent drawn to C2 at one of the points of intersectionA (a,y1) , (a > 0) meets C1 again at B(1,y2)
 
2
1 y
y  . The value of‘a’ is
(A) 4 (B*) 3 (C) 2 (D) 1
[Sol. Point A(a, y1) lies on C1 and C2
hence y1 = a2 – 3 and y2 = ka2
 a2 – 3 = ka2 ....(1)
now y = kx2  kx
2
dx
dy


)
y
,
a
( 1
dx
dy



= 2ka =
a
1
y
y 1
2


( But y2 = 1 – 3 = – 2 )
=
a
1
)
3
a
(
2 2




 2ka = a
1
a
1
a
1 2




2ka = 1 + a .....(2)
Substituting k = 2
2
a
3
a 
from (1) in (2) we get a
1
a
)
3
a
(
a
2
2
2



 2a2 – 6 = a + a2
 a2 – a – 6 = 0  a = +3 , a = – 2 (rejected) ]
Q.2851 f (x) = dx
x
1
1
x
1
1
2
2
2
 










 thenf is
(A) increasing in (0, ) and decreasing in (– , 0)
(B) increasing in (– , 0) and decreasing in(0, 
(C*) increasing in (–  , 
(D) decreasing in (–  , 
[Hint: f  (x) = 2 –
2
2
x
1
1
x
1
1



= 2
2
2
x
1
x
1
1
)
x
1
(
2





= 0
x
1
x
1
x
2
1
2
2
2





R
x
 (C)]

Q.2952 The lower corner of a leafin a book is folded over so as to just reach the inner edge of the page. The
fraction ofwidthfolded over ifthe area ofthefolded part is minimumis :
(A) 5/8 (B*) 2/3 (C) 3/4 (D) 4/5
[Sol. 2A = xy sin ; 4A2 = x2 y2 sin2 ; f (x) =
1
x
2
x4

; f ' (x) = 0  x =
3
2
]
Q.3053 A rectangle with one side lying along the x-axis is to be inscribed inthe closed region ofthe xy plane
bounded bythe lines y = 0, y = 3x, and y = 30 – 2x. The largest area ofsuch a rectangle is
(A)
8
135
(B) 45 (C*)
2
135
(D) 90
[Sol. A = (x2 – x1)y
y = 3x1 and y = 30 – 2x2
A (y) = 







3
y
2
y
30
y
6A(y) = (90 – 3y – 2y)y = 90y – 5y2
6A' (y) = 90 – 10y = 0  y = 9 ; A''(y) = – 10 < 0
x1 = 3 ; x2 =
2
21
Amax= 





3
2
21
9 =
2
9
·
15
=
2
135
 (C) ]
Q.3154 Whichofthefollowing statement is truefor the function













0
x
x
4
3
x
1
x
0
x
1
x
x
)
x
(
f
3
3
(A) It ismonotonic increasing R
x 

(B) f (x) fails to exist for 3 distinct realvalues ofx
(C*) f (x) changes its signtwice as x varies from(– , )
(D) function attains its extreme values at x1 & x2 , such that x1, x2 > 0
[Sol.
COMMENTS :
function is inc. in (– , –2)  (0 , )
function is dec. in(–2,0)
x = –2  local maxima
x = 0  local minima
Derivable R
x
 – {0,1} – 











3
)
1
(
'
f
,
2
/
1
)
1
(
'
f
4
)
0
(
'
f
,
0
)
0
(
'
f
Continuous R
x
 . ]

Q.3256 Aclosedvesseltapers to a point bothat itstop E and itsbottomFand isfixed withEFverticalwhen the
depth ofthe liquid in it is x cm, the volume ofthe liquid in it is, x2 (15  x) cu. cm. The lengthEF is:
(A) 7.5 cm (B) 8 cm (C*) 10 cm (D) 12 cm
[Hint:
dv
dx
= 3x (10  x) = 0  x = 0 ; x = 10 ;
d v
dx x
2
2
10




< 0  v is max at x = 10
 EF = 10 cm ]
Q.3357 Coffee is draining from a conical filter, height and diameter both 15 cms into a cylinderical coffee pot
diameter 15cm. The rate at whichcoffee drainsfromthe filter into thepot is 100 cucm/min.
Therateincms/minat whichthelevelinthepotisrisingat theinstant whenthecoffeeinthepot is10cm, is
(A)

16
9
(B)

9
25
(C)

3
5
(D*)

9
16
[Sol. For cylindricalpot
V = r2h
dt
dV
=  






dt
dr
r
2
·
h
dt
dh
r2
(r = constant,
dt
dr
= 0)
hence, 100 = r2
dt
dh
100 =  ·
4
225
·
dt
dh
(r =
2
15
cm)
dt
dh
=

225
400
=

9
16
cm/min ]
Q.3460 Let f(x) and g (x) be two differentiable functionin R and f(2) = 8, g (2) = 0, f(4) = 10 and g (4) = 8
then
(A) g ' (x) > 4 f ' (x)  x  (2, 4) (B) 3g ' (x) = 4 f ' (x) for at least one x  (2, 4)
(C) g (x) > f (x)  x  (2, 4) (D*) g ' (x) = 4 f ' (x) for at least one x  (2, 4)
[Hint: Consider h (x) = g (x) – 4 f (x), in [2, 4]
also h (2) = g (2) – 4 f (2) = – 32; h (4) = – 32
 h ' (x) = 0 for atleast one x  (2, 4) using Rolle's theorem ]
Q.3561 Let m and n be odd integers such that o < m < n. If f(x) = x
m
n
for x  R, then
(A) f(x) is differentiable everywhere (B) f  (0) exists
(C) fincreases on (0, ) and decreases on (–, 0) (D*) fincreases on R
[Hint : f  (x) =
m
n
x
m n
n







; for – ve exponents f ‘ (x) fails to exist m–n is even  (D) ]
Q.3662 Ahorse runs along a circle witha speed of20 km/hr .Alanternis at the centre ofthe circle .Afence is
along the tangent to the circle atthe point at whichthe horse starts .The speed withwhichthe shadow of
the horse move along the fence at the moment when it covers 1/8 ofthe circle inkm/hr is
(A) 20 (B*)40
(C) 30 (D) 60
[Hint: tan = x/r  x = r tan
 dx/dt = r sec2  (d/dt) = r sec2 = v sec2
where  = /8, dx/dt = v sec2(/4) = 2v = 40 km/hr ;  = 45º ]

Q.3765 Give the correct order ofinitials T or F for followingstatements. Use T ifstatement is true and F ifit is
false.
Statement-1: If f : R  R and c  R is such that f is increasing in (c – , c) and f is decreasing in
(c, c + )thenf has a localmaximumat c. Whereisa sufficientlysmallpositive quantity.
Statement-2 : Let f : (a, b)  R, c  (a, b). Then f cannot have both a local maximumand a point of
inflectionat x = c.
Statement-3 : The function f (x) = x2 | x | is twice differentiable at x= 0.
Statement-4 : Let f : [c – 1, c + 1]  [a, b] be bijective map such that f is differentiable at c then f–1
is also differentiable at f (c).
(A*) FFTF (B) TTFT (C) FTTF (D) TTTF
[Hint: S-1 : fmust be continuous  false S-2 : x = c is local max. & inflection point  false
S-4 : Consider f (x) = x3 at x = 0
f–1(x) = x1/3 , not differentiable at x = 0
Q.3866 Let f : [–1, 2]  R be differentiable such that 0  f ' (t)  1 for t  [–1, 0] and – 1  f ' (t)  0 for
t  [0, 2]. Then
(A*) – 2  f (2) – f (–1)  1 (B) 1  f (2) – f (–1)  2
(C) – 3  f (2) – f (–1)  0 (D) – 2  f (2) – f (–1)  0
[Sol. 0  f ' (t)  1 t  [–1, 0]
 0  

0
1
dt
)
t
(
'
f  1 ; 0  f (0) – f (–1)  1 ....(1)
now, –1  f ' (t)  0 t  [0, 2]
–2  
2
0
dt
)
t
(
'
f  0
–2  f (2) – f (0)  0 ....(2)
(1) + (2)
–2  f (2) – f (–1)  1
Alternatively: use LMVT once in [– 1, 0] and thenin [0, 2] ]
Q.3967 Acurve is represented bythe equations, x = sec2 t and y= cot t where t is a parameter. Ifthe tangent
at the point P on the curve where t = /4 meets the curve again at the point Q then PQ is equalto:
(A)
5 3
2
(B)
5 5
2
(C)
2 5
3
(D*)
3 5
2
[Hint: eliminating t gives y2 (x  1) = 1.
Equation of tangent at P(2, 1) is x + 2y= 4.
Solving with curve x = 5 & y =  1/2  Q (5, 1/2)  PQ =
3 5
2
]

Q.4069 For all a, b  R the function f (x) = 3x4  4x3 + 6x2 + ax + b has :
(A)no extremum (B*) exactlyone extremum
(C) exactlytwo extremum (D) threeextremum.
[ Hint :
dy
dx
= 12 x(x2  x + 1) + a &
d y
dx
2
2 = 12 (3x2  2x + 1) > 0

dy
dx
is an increasing function . But
dy
dx
is a polynomialof degree 3  exactlyone root ]
Q.4170 The set of values of p for whichthe equation ln x px = 0 possess three distinct roots is
(A*) 





e
1
,
0 (B) (0, 1) (C) (1,e) (D) (0,e)
[Hint: 1/x1 = p = y1/x1  y1 = 1 & x1 = e.
Hence slope of y = px i.e. p  (0, 1/e) for three roots ]
Q.4271 The sumof the terms ofan infinitelydecreasing geometric progressionis equalto the greatest value of
the function f(x) = x3 +3x– 9 onthe interval [–2, 3]. Ifthe difference betweenthe first and the second
termofthe progression is equalto f' (0) thenthe common ratio ofthe G.P. is
(A)
3
1
(B)
2
1
(C*)
3
2
(D)
4
3
Q.4374 The lateraledge ofa regularhexagonalpyramid is 1 cm. Ifthe volumeis maximum, thenits height must
be equalto :
(A)
1
3
(B)
2
3
(C*)
1
3
(D) 1
[Hint: x2 + h2 = 1 ; V =
1
3
6
3
4
. . x2 h =
3
2
h (1  h2)
V  (h) = 0  h =
1
3
 Vmax = 1/3 ]
Q.4475 The lateraledgeofa regular rectangular pyramidis 'a' cmlong. The lateraledge makesananglewith
the plane ofthe base. The value of for which the volume ofthe pyramid is greatest, is :
(A)

4
(B) sin1
2
3
(C*) cot1
2 (D)

3
[Hint: h = a sin  & x = a cos ; x2 + h2 = a2
V =
1
3
y2 h =
1
3
2 x2 h (note : 4x2 = 2y2  y2 = 2x2)
V () =
2
3
a2 cos2  . a sin =
2
3
a3 sin  cos2 
now V  () = 0  tan  =
1
2
; Vmax =
4 3
27
3
a
]
Q.4576 In a regular triangular prismthe distance fromthe centre ofone base to one ofthe vertices oftheother
base is l. The altitude ofthe prismfor which the volumeis greatest :
(A)

2
(B*)

3
(C)

3
(D)

4

[Hint: AG =
2
3
3
2
.
a
=
a
3
now l2 =
a2
3
+ h2 ; A =
=
3
4
a2
V (h) =
3
4
. 3 (l2  h2) h ; V  (h) = 0  h =

3
; Vmax =
3
2
]
Q.4678 Let f (x) =



1
x
if
)
2
x
(
1
x
if
x
3
5
3




thenthe number ofcriticalpointsonthe graph ofthe function is
(A) 1 (B) 2 (C*) 3 (D) 4
[Hint: A, B, C are the 3 criticalpoints of y = f (x) ]
Q.4779 The curve y  exy + x = 0 has a verticaltangent at :
(A) (1, 1) (B) (0, 1) (C*) (1, 0) (D) no point
[Hint:
dy
dx
 e xy y x
dy
dx






 + 1 = 0
dy
dx
(1  x exy) = y e xy  1
dy
dx
=
y e
xe
xy
xy


1
1
;
dy
dx
=  1  x e xy = 0
or 1  x (y + x) = 0
x =
1
x y

. Now verifyeach alternative ]
Q.4880 Number of roots ofthe equation x2 . e2 x = 1 is :
(A) 2 (B*) 4 (C) 6 (D) zero
[Hint: draw graphs of y = x2 . e2 x and y = 1
y =



0
x
if
e
·
x
0
x
if
e
·
x
x
2
2
x
2
2




]
Q.4981 The point(s) at each ofwhich the tangents to the curve y = x3  3x2  7x+ 6 cut off onthe positive
semiaxis OXa line segment halfthat onthe negativesemiaxis OY thenthe co-ordinatesthepoint(s) is/
are given by:
(A) (1, 9) (B*) (3, 15) (C) (1,  3) (D) none
[Hint: If OA = a ; OB = 2a  tan = 2
slope ofthe tangent is 2

dy
dx x y






1 1
= 2  2
1
x
3 – 6x1 – 7 = 2  2
1
x
3 – 2x1 – 3 = 0
 x1 = 3 or  1 (rejected)
 (3,  15)  B ]
[Also for the point P; x & y both positive or x positive & ynegative or x & yboth negative ]

Q.5082 A curve withequation ofthe form y= ax4 + bx3 + cx + d has zero gradient at the point (0, 1) and also
touchesthe xaxisat thepoint (1, 0) thenthe values of x forwhichthe curve hasa negative gradient
are :
(A) x >  1 (B) x < 1 (C*) x <  1 (D)  1  x  1
Q.5184 Number ofsolution(s) satisfying the equation, 3x2  2x3 = log2 (x2 + 1)  log2 x is :
(A*) 1 (B) 2 (C) 3 (D) none
[Hint: Compare the greatest value ofthe functionappearing onthe left handside ofthe equationwiththeleast
value appearing on the right hand side .  x = 1
Note that L.H.S. has a maximumvalue 1 at x = 1 but R.H.S. side has a minimum value at x = 1
 x = 1 can be the onlysolution ]
Q.5286 Consider the function
f (x) = x cos x– sin x, thenidentifythe statement whichis correct .
(A) f is neither odd nor even (B*) f is monotonic decreasing at x = 0
(C) f has a maxima at x =  (D) f has a minima at x = – 
[Hint: f ' (x) = – x sin x = 0 when x = 0 or 
0
)
)(
)(
(
)
0
(
'
f
0
)
)(
)(
(
)
0
(
f '















no sign change
This also implies that fis decreasing at x = 0  (B) is correct
f "(x) = – (x cos x + sin x)
f " () = – (–) > 0 minima at x = 
f " (– ) = – () < 0 maxima at x =  ]
Q.5391 Consider the two graphs y= 2x and x2 – xy + 2y2 = 28. The absolute value ofthe tangent ofthe angle
betweenthe two curves at the points where theymeet, is
(A) 0 (B)
2
1
(C*) 2 (D) 1
[Hint: y = 2x ; x2 – xy + 2y2 = 28
solving the point ofintersection are (2, 4) and (–2, –4)
For 1st curve,
dx
dy
= 2 = m1 ....(1)
for 2nd curve,
dx
dy
=
x
y
4
x
2
y


= 0 = m2 ....(2)
 tan = 2Ans. ]
Q.5492 The x-intercept ofthe tangent at anyarbitrarypoint ofthe curve 2
2
y
b
x
a
 = 1 isproportionalto:
(A) square ofthe abscissa ofthe point oftangency
(B) square root ofthe abscissa ofthe point oftangency
(C*) cube ofthe abscissa ofthe point oftangency
(D) cube root ofthe abscissa ofthe point oftangency.
[Hint: 2
2
y
b
x
a
 = 1  ay2 + bx2 = x2y2 .....(1)

– 3
x
a
2
– 3
y
b
2
dx
dy
= 0 
dx
dy
= – 3
3
bx
ay
equationoftangent
Y – y = – 3
3
bx
ay
(X – x)
for x-intercept, put Y= 0
 X = 2
3
ay
bx
+ x
X = x 




 
2
2
2
ay
ay
bx
= x 





2
2
2
ay
y
x
=
a
x3
 x-intercept is proportionalto cube ofabscissa ]
Q.5593 For the cubic, f (x) = 2x3 + 9x2 + 12x + 1 whichone of the following statement, does not hold good?
(A) f(x)is nonmonotonic
(B) increasing in (– , – 2)  (–1, ) and decreasing is (–2, –1)
(C*) f: R  R is bijective
(D) Inflectionpoint occurs at x= – 3/2
[Sol. f (x) = 2x3 + 9x2 + 12x + 1
f '(x) = 6[x2 + 3x + 2] = (x + 2)(x + 1)
Hence x = – 2 is maxima and x = – 1 is minima
f ''(x) = 2x + 3 = 0
 x = – 3/2 is theinflection point
obviouslyfis monotonic
 f is many one
hence it is not bijective. ]
Q.5695 The function 'f' is defined by f(x) = xp (1  x)q for all x  R, where p,q are positive integers, has a
maximumvalue, for x equalto :
(A) pq
p q

(B) 1 (C) 0 (D*)
p
p q

[Hint: change ofsignoffirst order derivative ]
Q.5798 Let hbe a twice continuouslydifferentiablepositive function onanopenintervalJ. Let
g(x) = ln 
)
x
(
h for each x  J
Suppose  2
)
x
(
'
h > h''(x) h(x) for each x  J. Then
(A) g is increasing on J (B) g is decreasing on J
(C) g is concave up on J (D*) g is concave down on J
[Sol. Given g(x) = ln 
)
x
(
h
g ' (x) =
)
x
(
h
)
x
(
'
h
g''(x) =
)
x
(
h
))
x
(
'
h
(
)
x
(
'
'
h
)
x
(
h
2
2

< 0 (given)
 g''(x) <0  g (x) is concave down ]

Q.5899 Let f (x) =




2
1
x
if
0
2
1
x
if
1
x
2
)
1
x
6
)(
1
x
(





then at x =
2
1
(A) f has a localmaxima (B) f has a localminima
(C*) f has an inflection point (D) f has a removable discontinuity
[Hint: f ' (x) = 2
2
)
1
x
2
(
5
x
12
x
12



> 0  x  R   ;
hence f is increasing  x R]
Q.59104 Let f (x) and g (x) be two continuous functions defined from R  R, such that f (x1) > f (x2) and
g (x1) < g (x2),  x1 > x2 , then solution set of  
)
2
(
g
f 2


 >  
)
4
3
(
g
f 
 is
(A) R (B)  (C*) (1, 4) (D) R – [1, 4]
[Hint: obviouslyfis increasing and g is decreasing in(x1, x2)
hence  
)
2
(
g
f 2


 >  
)
4
3
(
g
f 
 as f is increasing
 g(a2 – 2) > g(3 – 4)
 2 – 2 < 3 – 4 as g is decreasing
2 – 5 + 4 < 0
( – 1)( – 4) < 0  a  (1, 4) ]
Q.60108 If f(x) =
x
x2
 (t  1) dt , 1  x  2, then global maximumvalue of f(x) is :
(A) 1 (B) 2 (C*) 4 (D) 5
[ Hint : f ' (x) = 2x3  3x + 1 this is always positive in (1, 2)
 increasing in[1, 2]
 f(2) willbe the greatest value ]
Q.61110 Aright triangleis drawninasemicircleofradius
2
1
withoneofitslegs along thediameter.Themaximum
area ofthe triangle is
(A)
4
1
(B*)
32
3
3
(C)
16
3
3
(D)
8
1
[Hint: (BC)(CE) = x (1 – x) (property of circle) [AC = x ; CD = 1 – x]
but BC = CE
 BC = )
x
1
(
x 
A=
2
x
x
x 2

. Now maximiseA.
Amax occurs at x =
4
3
]

Q.62111 At anytwo points of the curve represented parametrically by x = a (2 cos t  cos 2t) ;
y = a (2 sin t  sin 2t) the tangents are parallel to the axis of x corresponding to the values of the
parameter t differing fromeachother by:
(A*) 2/3 (B) 3/4 (C) /2 (D) /3
[Hint:
dy
dx
=
cos cos
sin sin
2
2
t t
t t


= 0  cos 2t = cos t  cos 2t = cos (2 t)  t = 2/3]
Q.63112 Let xbethe lengthofone oftheequalsides ofanisosceles triangle, and let  bethe anglebetweenthem.
Ifx is increasing at the rate (1/12) m/hr, and  is increasing at the rate of/180 radians/hr then the rate
in m2/hr at which the area of the triangle is increasing when x = 12 m and  = /4
(A) 21/2 




 

5
2
1 (B)
2
73
· 21/2 (C)
5
2
3 2
1

 (D*) 21/2 




 

5
2
1
[Sol. A =
2
1
x2 sin  2A = x2 sin
dt
dA
2 = x2cos
dt
d
+ sin 2x
dt
dx
dt
dA
2 = (144) 





2
1
180

+
2
1
· 2 · 12 ·
12
1
=
2
15
12
+
2
2
dt
dA
=
2
5
2
+
2
1
=
5
2 
+
2
2
= 







2
1
5
2 ]
Q.64114 Ifthe function f (x) =
4
x
x
x
3
t 2



, where 't' is a parameter has a minimum and a maximumthen the
range ofvalues of't' is
(A) (0, 4) (B) (0, ) (C*) (– , 4) (D) (4, )
[Sol. f (x) =
4
x
x
x
3
t 2



; f ' (x) = 2
2
)
4
x
(
)
x
x
3
t
(
)
x
2
3
)(
4
x
(






for maximumor minimum, f'(x) = 0
– 2x2 + 11x – 12 – t – 3x + x2 = 0
– x2 + 8x – (12 + t) = 0
for one M and m,
D > 0
64 – 4(12 + t) > 0
16 – 12 – t > 0  4 > t or t < 4 ]
Q.65118 The least area ofa circle circumscribing anyright triangle ofarea S is :
(A*)  S (B) 2  S (C) 2  S (D) 4  S
[Sol. S =
2
xy
= constant
Area of the circles (A) = r2 =
4
)
y
x
( 2
2


; (x2 + y2 = 4r2)
A (x) =
















2
2
x
S
2
x
4

A(x) = 3
2
x
s
8
x
2  = 0  x4 = 4S2  x2 = 2S
S2 =
4
y
x 2
2
=
4
Sy
2 2
 y2 = 2S
 least area of circle = r2 = S
)
y
x
(
4
2
2




Ans. ]
Q.66121 Apoint is moving along the curve y3 = 27x. The intervalinwhich the abscissa changes at slower rate
thanordinate, is
(A) (–3 , 3) (B) (–  ,  ) (C*) (–1, 1) (D) (– , –3) (3, )
[Hint: y3 = 27 x 
dt
dx
27
dt
dy
y
3 2

But 1
dt
dy
dt
dx
  1
9
y2
  – 3 < y < 3 for )
3
,
3
(
y 
 , )
1
,
1
(
x 
  (C) ]
Q.67123 Let f(x) and g (x) are two function which are defined and differentiable for allx x0. If f(x0) = g (x0)
and f ' (x) > g ' (x) for all x > x0 then
(A) f (x) < g (x) for some x > x0 (B) f (x) = g (x) for some x > x0
(C) f(x) > g (x) only for some x> x0 (D*) f (x) > g (x) for allx > x0
[Sol. Consider  (x) = f (x) – g (x)   '(x) = f ' (x) – g ' (x)
 (x) is also continuous and derivable in [x0, x]
using LMVT for Q(x) in [x0, x]
'(x) =
0
0
x
x
)
x
(
)
x
(




.
since ' (x) = f ' (x) – g ' (x) are f ' (x) – g ' (x) > 0
  '(x) > 0
hence  (x) –  (x0) > 0
 (x) >  (x0)
f (x) – g (x) > 0
are f (x0) = f (x0) – f (x0) = 0  (D) ]
Q.68126 P and Q are two points ona circle ofcentre C and radius , theangle PCQ being 2 thenthe radius of
the circle inscribedinthe triangle CPQis maximumwhen
(A)
2
2
1
3
sin


 (B*)
2
1
5
sin


 (C)
2
1
5
sin


 (D)
4
1
5
sin



[Sol. r =
s

where  =Area of triangle CPQ and s = semiperimeter of CPQ.
r =
s
2
2
sin
2


=






sin
2
2
2
sin
2
=




sin
1
2
sin
.
2

Consider f () =



sin
1
2
sin
f () = 2
)
sin
1
(
cos
.
2
sin
2
cos
2
)
sin
1
(








= 0
2(1 + sin)(1 – 2sin2) – 2sin (1 – sin2) = 0
2(1 – 2sin2) = 2 sin (1 – sin)
1 – 2sin2 = sin – sin2
sin2 + sin – 1 = 0
sin =
2
4
1
1 


 sin =
2
1
5 
]
Q.69132 The line whichis parallelto x-axis and crosses the curve y= x at anangle of

4
is
(A) y = 
1
2
(B) x =
1
2
(C) y =
1
4
(D*) y =
1
2
[Sol. y = x and y = c; solving x = c2
y = x ;
x
2
1
dx
dy

c
2
1
dx
dy
p




= 1  c =
2
1
Equation ofthe line is y=
2
1
]
Q.70134 The function S(x) =  






 
x
0
2
dt
2
t
sin has two criticalpoints in the interval [1, 2.4]. One of the critical
points is a localminimumand theother is a localmaximum. The localminimumoccurs at x=
(A) 1 (B) 2 (C*) 2 (D)
2

[Hint: S(x) =  






 
x
0
2
dt
2
t
sin ; S ' (x) = 






 
2
x
sin
2
= 0
2
x2

= n  x2 = 2n (1  x2  5.76 as is given)
hence n = 1 or 2
x = 2 or x = 2 ; S''(x) = cos 






 
2
x2
. x
S''( 2 ) < 0 and S''(2) > 0  minima at x = 2 ]
Q.71138 For asteamer the consumption ofpetrol(per hour) varies as the cube ofits speed (inkm). Ifthe speed
of the current is steady at C km/hr then the most economicalspeed of the steamer going against the
current willbe
(A) 1.25 C (B*) 1.5 C (C) 1.75C (D) 2 C

[Hint: Time ofjourney =
C
V
d

whenV is the speed instillwater.
.
 petrolburnt per hour = kV3
 fuels cost =
C
V
d
kV3

= kd
C
V
V3

= f (v)
now proceed.]
Q.72139 Let f and g be increasing and decreasing functions, respectively from [0 , ) to [0 , ). Let
h (x) = f [g(x)] . If h(0) = 0, then h(x)  h(1) is :
(A*) always zero (B) strictly increasing
(C) always negative (D) always positive
Q.73140 The set of value(s) of 'a' for which the function f (x) =
a x3
3
+ (a + 2) x2 + (a  1) x + 2 possess a
negative pointofinflection.
(A*) (,  2)  (0, ) (B) { 4/5}
(C) ( 2, 0) (D) emptyset
[Hint: For negative point ofinflection 
b
a
2
< 0 ;
alternatively
d y
dx
2
2
= 0, get 'x' and put x < 0
f (x) = a x2 + 2 (a + 2) x + (a  1)
f (x) = 2 ax + 2 (a + 2) = 0  x = 
a
a
 2
< 0 =
a
a
 2
> 0 ]
Q.74141 Afunction y = f(x) is given by x =
1
1 2
 t
& y =
1
1 2
t t
( )

for all t > 0 then f is :
(A) increasing in (0, 3/2) & decreasing in (3/2, )
(B*) increasing in (0, 1)
(C) increasing in (0, )
(D) decreasing in (0, 1)
[Sol. 2
)
t
1
(
t
2
dt
dx



2
2
2
2
)
t
1
(
t
t
3
1
dt
dy




0
t
2
t
3
1
dx
dy
3
2


 for t > 0  yis increasing for every )
1
,
0
(
x ]
Q.75144 The set ofallvalues of 'a' for which the function,
f(x) = (a2  3a + 2) cos sin
2 2
4 4
x x






 + (a  1) x + sin 1 does not possess critical points is:
(A) [1, ) (B*) (0, 1)  (1, 4) (C) ( 2, 4) (D) (1, 3)  (3, 5)

Q.76150 Read the followingmathematicalstatements carefully:
I. Adifferentiable function 'f' with maximum at x = c  f ''(c) < 0.
II. Antiderivative ofa periodic functionis also a periodic function.
III. If f has a period T then for any a  R. 
T
0
dx
)
x
(
f =  
T
0
dx
)
a
x
(
f
IV. If f (x) has a maxima at x = c, then 'f ' is increasing in (c – h, c) and decreasing in (c, c + h)
as h  0 for h > 0.
Now indicate the correct alternative.
(A*) exactlyone statement is correct. (B) exactlytwo statements are correct.
(C) exactlythree statements are correct. (D)Allthefour statements are correct.
[Sol. I. consider the function f (x) =  x4 , f (x) =  4 x3 & f (x) =  12 x2.
Here f(x) has a maxima at x = 0 but f (0) = 0  False.
II. f (x) = cos x + 1 is periodic with period 2 but   dx
)
1
x
(cos = sin x + x is not periodic.
III.  
T
0
dx
)
a
x
(
f , let x + a = y ; 
T
a
a
dy
)
y
(
f = 
0
a
dy
)
y
(
f + 
T
0
dy
)
y
(
f + 
T
a
T
dy
)
y
(
f
consider 
T
a
T
dy
)
y
(
f ; y = T + v
 
a
0
dv
)
T
v
(
f = 
a
0
dv
)
v
(
f . Hence 
T
0
dx
)
x
(
f =  
T
0
dx
)
a
x
(
f  True.
rue.
IV. The statement can be true onlyif'f' is continuous at x = c . Consider the function
f (x) =
x if x
if x
x if x


 





0
1 0
0
This has a maxima at x = 0 however this
does not satisfyconditions stated in the problem  False ]
Q.77152 Ifthe point ofminima ofthe function, f(x) = 1 + a2x– x3 satisfythe inequality
x x
x x
2
2
2
5 6
 
 
< 0, then 'a' must liein the interval:
(A)  
3 3 3 3
, (B)  
 
2 3 3 3
,
(C)  
2 3 3 3
, (D*)    
 
3 3 2 3 2 3 3 3
, ,

[Sol. f  (x) = 0  x =
a
3
or –
a
3
f  (x) = –6x
Case I if a > 0  x = –
a
3
is minima
Case II if a < 0  x =
a
3
is minima
put x =
a
3
and thenx = –
a
3
in the giveninequalityto get the result]

Q.78153 The radius ofa right circular cylinder increases at a constant rate. Itsaltitude is a linear function ofthe
radius andincreases three times as fast as radius. Whenthe radius is 1cmthe altitude is 6 cm. Whenthe
radius is 6cm, the volume is increasing at the rate of1Cu cm/sec. Whenthe radius is 36cm, the volume
is increasing at a rate ofn cu. cm/sec. The value of'n' is equalto:
(A) 12 (B) 22 (C) 30 (D*) 33
[Sol. c
dt
dr
 and h = ar + b ; also
dt
dh
= 3
dt
dr
(given)

dt
dr
3
dt
dr
a   a = 3
hence h = 3r + b
when r = 1 ; h = 6  6 = 3 + b  b = 3
 h = 3 (r + 1)
V =  r2 h = 3 r2(r +1) = 3 (r3 + r2)
dt
dV
= 3 (3r2 + 2r)
dt
dr
where r = 6 ;
dt
dV
= 1 cc/sec
 1 = 3 (108 + 12)
dt
dr
 360 
dt
dr
= 1
again when r = 36 ,
dt
dV
= n
n = 3 ((3.36)2 + 2.36 )
dt
dr
n = 3 . 36 (110) . 
360
1
n = 33  (D) ]
Q.79154 Two sides ofa triangle are to have lengths 'a' cm& 'b'cm. Ifthe triangle is to have the maximumarea,
then the length ofthe medianfromthe vertex containing the sides 'a' and 'b' is
(A*)
1
2
2 2
a b
 (B)
2
3
a b

(C)
a b
2 2
2

(D)
a b
 2
3
[Hint: A= (1/2)ab sin Ais maximum if = 900 ]
Q.80155 Let a > 0 and fbe continuous in [– a, a]. Suppose that f '(x) exists and f' (x)  1 for allx  (– a, a). If
f (a) = a and f (– a) = – a then f (0)
(A*) equals 0 (B) equals
2
1
(C) equals 1 (D)isnotpossibleto determine
[Hint: Use LMVT once in [–a, 0] and then in [0, a] and use the fact f '(x)  1 ]

Q.81157 The lines tangent to the curves y3 – x2y+ 5y– 2x = 0 and x4 – x3y2 + 5x+ 2y= 0 at the origin intersect
at anangle equalto
(A)

6
(B)

4
(C)

3
(D*)

2
Q.82161 The cost function atAmericanGadget is
C(x) = x3 – 6x2 + 15x (x inthousands ofunits and x > 0)
The productionlevelat whichaverage cost is minimumis
(A) 2 (B*) 3 (C) 5 (D) none
[Hint: Average cost
C x
x
( )
= x2 – 6x + 15
 f (x) = x2 – 6x + 15
f ' (x) = 2x – 6 = 0  x = 3 ]
Q.83163 Arectangle has one side on the positive y-axis and one side on the positive x - axis. The upper right
hand vertexon the curve y=
nx
x2 . The maximumarea ofthe rectangle is
(A*) e–1 (B) e– ½ (C) 1 (D) e½
[Hint: The graph ofy =
nx
x2 is as shown
A = xy =
nx
x
dA
dx
nx
x


1
2

= 0  x = e ; A
Amax = e ·
1
2
e
=
1
e
]
Q.84165 Aparticle moves along the curve y = x3/2 in the first quadrant in sucha waythat its distance from the
origin increases at the rate of11 units per second. The value of
dx
dt
when x = 3 is
(A*) 4 (B)
9
2
(C)
3 3
2
(D) none
[Hint: r2 = x2 + y2 ; r
d r
d t
= x
dx
dt
+ y
dy
dt
where x = 3; y = 3 3 and r = 6; hence 6 · 11 = 3
dx
dt
+ 3 3
dy
dt
but
dy
dt
=
3
2
x
dx
dt
=
3 3
2
dx
dt
]
Q.85166 Number ofsolution ofthe equation 3tanx + x3 = 2 in 




 
4
,
0 is
(A) 0 (B*) 1 (C) 2 (D) 3

[Hint: f(x) = 3tanx + x3 thenf(x) = 3sec2x+ 3x2 > 0 hence f(x) . Thusf(x) assume the value2 exactlyonce.
Also f(0) = 0 and 




 
4
f > 2 so by intermediate value theoremf(c) = 2 for some c in 




 
4
,
0  (B) ]
Q.86173 Let f (x) = ax2 – b | x |, where a and b are constants. Then at x = 0, f (x) has
(A*) a maxima whenever a > 0, b > 0 (B) a maxima whenever a > 0, b < 0
(C) minima whenever a > 0, b > 0 (D) neithera maximanor minimawhenever a > 0, b< 0
[Hint: for a > 0, b > 0 for a > 0, b < 0
]
Q.87174 Let f (x) =  






x
1
dt
t
t
n
)
t
(
n
t
l
l (x > 1) then
(A) f(x) has one point ofmaxima and no point ofminima.
(B) f' (x) has two distinct roots
(C) f(x) has one point ofminima and no point ofmaxima
(D*) f(x) is monotonic
[Hint: f ' (x) = x · ln x –
x
x
n
l
=
x
x
n
l
(x – 1)(x + 1)
f ' (1–) > 0 and f ' (1+) > 0
 f ' (x) > 0,  x > 1
 f(x)is monotonic ]
Q.88175 Consider f (x) = | 1 – x | 1  x  2 and
g (x) = f (x) + b sin
2

x, 1 < x < 2
then whichofthe following is correct?
(A) Rolles theorem is applicable to both f, g and b =
2
3
(B) LMVT is not applicable to f and Rolles theoremifapplicable to g with b =
2
1
(C*) LMVT is applicable to f and Rolles theoremis applicable to g with b = 1
(D) Rolles theoremis not applicable to both f, g for anyreal b.
[Sol: f (x) = x – 1, 1  x  2
g (x) = x – 1 + b sin
2

x, 1  x  2
f (1) = 0 ; f (2) = 1  Rolle's theorem is not applicable to ' f ' but LMVT is applicable to f.
( x – 1 is continuous and differentiable in [1, 2] and (1, 2) respectively)
Now g (1) = b ; g (2) = 1 and
Function x – 1, sin
2

x are bothcontinuous in [1, 2] and (1, 2)
 For Rolle's theoremto be applicable to g.
We must have b = 1 ]

Q.89176 Consider f (x) =  






x
0
dt
t
1
t and g (x) = f  (x) for x 





 3
,
2
1
If P is a point onthe curve y= g(x) such that the tangent to this curve at P is parallelto a chord joining
the points 













2
1
g
,
2
1
and (3, g(3) ) ofthe curve, then the coordinates ofthe point P
(A) can't be found out (B) 





28
65
,
4
7
(C) (1, 2) (D*) 







6
5
,
2
3
[Sol: f (x) =  






x
0
dt
t
1
t  f ' (x) = x +
x
1
 g (x) = x +
x
1
for x  





3
,
2
1
g 





2
1
= 2 +
2
1
=
2
5
, g (3) = 3 +
3
1
=
3
10
]
Let P   
)
c
(
g
,
c , c  





3
,
2
1
ByLMVT, g' (c) =
2
1
3
2
1
g
)
3
(
g









2
1
3
2
5
3
10
c
1
1 2




 c2 =
2
3
 c =
2
3
 g (c) = 6
5
2
3
1
2
3


 P 








6
5
,
2
3
]
Q.90177 The co-ordinates of the point on the curve 9y2 = x3 where the normal to the curve makes equal
intercepts withthe axes is
(A) 





3
1
,
1 (B)  
3
,
3 (C*) 





3
8
,
4 (D) 







5
6
5
2
,
5
6
[Hint: normalmakes equalintercepts
 1
dx
dy
1
1 y
,
x




]
Q.91178 The angle made bythe tangent ofthe curve x= a (t + sint cost) ; y= a (1 + sint)2 withthe x-axisat any
point onit is
(A*)  
t
2
4
1

 (B)
t
cos
t
sin
1
(C)  


t
2
4
1
(D)
t
2
cos
t
sin
1

[Sol.
dt
dx
= a +
2
a
2cos2t = a [1 + cos2t) = 2a cos2t
dt
dy
= 2a (1 + sint ) · cos t
dx
dy
=
t
cos
a
2
t
cos
·
)
t
sin
1
(
a
2
2

=
t
cos
)
t
sin
1
( 
tan =
   
 
   
2
t
sin
2
t
cos
2
t
sin
2
t
cos
2
2
2


=
2
t
tan
1
2
t
tan
1


= tan 







2
t
4
  =
4
t
2


]
Q.92179 If f (x) = 1 + x +  
 
x
1
2
dt
nt
2
t
n l
l , then f (x) increases in
(A*) (0, ) (B) (0, e–2)  (1, ) (C) no value (D) (1, )
Q.93180 The function f(x) =
)
x
e
(
n
)
x
(
n



l
l
is :
(A) increasingon [0, )
(B*) decreasing on [0, )
(C) increasing on [0, /e) &decreasing on [/e, )
(D) decreasing on [0, /e) &increasing on [/e, )
Directions for Q.94 to Q.96
Suppose you do notknow the functionf(x), howeversomeinformationabout f (x)is listedbelow. Read
thefollowing carefullybefore attempting the questions
(i) f (x) is continuous and defined for allrealnumbers
(ii) f '(–5) = 0 ; f '(2) is not defined and f '(4) = 0
(iii) (–5, 12) is a point whichlies on the graphof f(x)
(iv) f ''(2) is undefined, but f ''(x) is negative everywhere else.
(v) the signs of f '(x) is given below
Q.94181 On the possible graph of y = f (x) we have
(A) x = – 5 is a point of relative minima.
(B) x = 2 is a point ofrelative maxima.
(C) x = 4 is a point ofrelative minima.
(D*) graph ofy= f (x) must have a geometricalsharp corner.
Q.95 Fromthe possible graph of y = f (x), we cansay that
(A) There isexactlyone point ofinflectionon the curve.
(B) f (x) increases on – 5 < x < 2 and x > 4 and decreases on –  < x < – 5 and 2 < x < 4.
(C*) The curve is always concave down.
(D) Curve always concave up.
[Sol. At x = – 5 f'(x) changes from+ ve to – ve and x = 4, f' (x) change sign for + ve to – ve hence maxima
at x= – 5 and4. fis continuousand f' (x) is not defined hence x= 2 must be geometricalsharpcorner]

Q.96 Possible graph of y= f (x) is
(A) (B)
(C*) (D)
Directions for Q.97 to Q.100
Consider the function f (x) =
x
x
1
1 





 then
Q.97182 Domain of f (x) is
(A) (–1, 0)  (0, ) (B) R – { 0 }
(C*) (–, –1)  (0, ) (D) (0, )
Q.98 Whichoneofthe following limitstends to unity?
(A) )
x
(
Lim
x
f


(B*) )
x
(
Lim
0
x
f


(C) )
x
(
Lim
1
x
f



(D) )
x
(
Lim
x
f


Q.99 Thefunctionf(x)
(A) has a maxima but no minima (B) has aminima but no maxima
(C) has exactlyone maxima and one minima (D*) has neither a maxima nor a minima
Q.100 Range ofthefunctionf(x) is
(A) (0, ) (B) (– , e) (C) (1, ) (D*) (1, e)  (e, )
Q.101187 Acube ofice melts without changing shapeat the uniformrate of 4 cm3/min. Therate ofchangeofthe
surface area ofthe cube, in cm2/min, when the volume ofthe cube is 125 cm3, is
(A) – 4 (B*) – 16/5 (C) – 16/6 (D) – 8/15
[Sol.
dt
dV
= 4cm3/min;
dt
dS
= ? where V = 125 cm2
V = x3 ; S = 6x2;
dt
dV
= 3x2
dt
dx
– 4 = 3x2
dt
dx
....(1) ;
dt
dS
= 12x
dt
dx
– 4 = 3x2
dt
dS
·
x
12
1
dt
dS
= –
x
16
; where V = 125 = x3  x = 5
dt
dS
= –
x
16
cm2/min 
dt
dS
= –
5
16
cm2/min Ans. ]

Q.102189 Let f (1) = – 2 and f ' (x)  4.2 for 1  x  6. The smallest possible value of f (6), is
(A) 9 (B) 12 (C) 15 (D*) 19
[Sol. Using LMVT  some c  (1, 6) s.t.
f ' (c) =
5
)
1
(
f
)
6
(
f 
=
5
2
)
6
(
f 
 4.2
f (6) + 2  21
f (6)  19 Ans. ]
Q.103190 Which ofthe following six statements are true about the cubic polynomial
P(x) = 2x3 + x2 + 3x – 2?
(i) It has exactlyone positive realroot.
(ii) It haseither one or threenegative roots.
(iii) It has a root between 0 and 1.
(iv) It must haveexactlytwo realroots.
(v) It has a negative root between – 2 and –1.
(vi) It has no complex roots.
(A) only(i), (iii) and (vi) (B) only(ii), (iii) and (iv)
(C*) only(i) and (iii) (D) only(iii), (iv) and (v)
[Sol. First we notice that the function is increasing bychecking the derivative P'(x) = 6x2 + 2x + 3 > 0.
i.e. P (x) willhave exactlyonerealroot (and two complex roots)
Since P(0) = – 2 and P(1) = 4 the root is between 0 and 1 by the Intermediate Value theorem.
Thus onlystatement (i) and (iii) are true, the correct answer is (D) ]
Q.104191 Given that f (x) is continuouslydifferentiable on a  x  bwhere a < b, f (a) < 0 and f (b) > 0, which
ofthe followingare always true?
(i) f (x) is bounded on a  x  b.
(ii) The equation f (x) = 0 has at least one solution in a < x < b.
(iii) The maximumand minimumvalues of f (x) on a  x  boccur at points where f '(c) = 0.
(iv) There is at least one point c with a < c < b where f ' (c) > 0.
(v) There is at least one point d with a < d < b where f ' (c) < 0.
(A) only(ii)and (iv) are true (B) allbut (iii) are true
(C) allbut (v) are true (D*) only(i), (ii) and (iv) are true
[Sol. (i) This statement is true, everycontinuousfunction is bounded ona closed interval
(ii) True again, byIntermediateValueTheorem
(iii) Not true, because maximum and / or minimum values could also occur at a or b, without the
derivatives being 0.
(iv) True. Bythe MeanValue Theorem there exist a point between a and b where the derivative is
exactly
a
b
)
a
(
f
)
b
(
f


, aclearlypositive value.
(v) Not alwaystrue, for examplethefunctionmight be strictlyincreasingguarenteeingthederivative
to be always positive.
Thus the true statements are (i), (ii) and (iv) and the correct answer is (D) ]
Q.105193 Consider the function f (x) = 8x2 – 7x + 5 on the interval [–6, 6]. The value of c that satisfies the
conclusionofthemeanvalue theorem, is
(A) – 7/8 (B) – 4 (C) 7/8 (D*) 0
[Sol. f ' (c) = 16c – 7 =
12
)
6
(
f
)
6
(
f 

=
12
)
5
6
.
7
36
.
8
(
)
5
6
·
7
36
·
8
( 




= –
12
6
·
7
·
2
= – 7
16c = 0  c = 0 Ans. ]
Q.106196 Consider the curve represented parametricallybythe equation

x = t3 – 4t2 – 3t and
y = 2t2 + 3t – 5 where t  R.
IfHdenotes the number ofpoint onthe curvewhere the tangent ishorizontaland V thenumber ofpoint
where the tangent is verticalthen
(A) H = 2 and V = 1 (B*) H = 1 and V = 2
(C) H = 2 and V = 2 (D) H = 1 and V = 1
[Hint:
dt
dx
= 3t2 – 8t – 3;
dt
dy
= 4t + 3

dx
dy
=
3
t
8
t
3
3
t
4
2



Hence H means 4t + 3 = 0  t = – 3/4  H = 1
V means 3t2 – 8t – 3 = 0  t = 3 & t = – 1/3  V = 2 ]
Q.107197 At thepoint P(a, an) onthe graphof y= xn (n N)inthe first quadrant a normalis drawn.The normal
intersects the y-axis at the point (0, b). If
2
1
b
Lim
0
a


, then n equals
(A) 1 (B) 3 (C*) 2 (D) 4
[Sol. y = xn
dx
dy
= n xn – 1 = nan – 1
slope of normal= – 1
n
na
1

equation of normal y – an = – 1
n
na
1
 (x – a)
put x = 0 to get y-intercept
y = an + 2
n
na
1
 ; Hence b = an + 2
n
na
1

b
Lim
0
a
=





2
n
if
2
n
if
2
1
2
n
if
0




 (A) ]
Q.108198 Suppose that f is a polynomial ofdegree 3 and that f''(x)  0 at anyofthe stationarypoint. Then
(A) f has exactlyone stationarypoint. (B) f must have no stationarypoint.
(C) f must have exactly2 stationarypoints. (D*) f has either 0 or 2 stationarypoints.
[Sol. The derivative ofa degree 3 polynomialis a quadratic. This must have either 0, 1 or 2 roots. Ifthis has
preciselyone root, then this must be repeated. Hence we have f' (x) = m(x – )2, where  is repeated
root and m R. So our originalfunctionfhas a criticalpoint at x= .
Also f ''(x) = 2m(x–), in whichcase f ''() = 0. But we are told that the 2nd derivative is nonzero at
criticalpoint. Hence there must beeither 0 or 2 criticalpoint. ]

Q.109201 Let f (x) = 


0
x
for
8
x
0
x
for
x
2
2




. Then x intercept ofthe line that is tangent to the graphof f (x) is
(A) zero (B*) – 1 (C) –2 (D) – 4
[Sol. Let y= mx + c be a tangent to f (x)
y = x2 + 8 for x  0
mx + c = x2 + 8
x2 – mx + 8 – c = 0 (for the line to be tangent D = 0)
 m2 = 4(8 – c) ....(1)
again y = – x2 for x < 0
mx + c = – x2
x2 + mx + c = 0
D = 0  m2 = 4c ....(2)
from(1) and (2)
c = 4, m = 4
 y = 4x + 4
put y = 0  x = – 1 Ans. ]
Q.110202 Suppose that f is differentiable for allxand that f '(x)  2 for allx. If f (1) = 2 and f(4) = 8 then f (2)
has the value equalto
(A) 3 (B*) 4 (C) 6 (D) 8
[Sol. Using LMVT for f in [1, 2]
 c  (1, 2)
1
2
)
1
(
f
)
2
(
f


= f ' (c)  2
f (2) – f (1)  2  f (2)  4 ....(1)
again using LMVT in[2, 4]
 d  (2, 4)
2
4
)
2
(
f
)
4
(
f


= f ' (d)  2
 f (4) – f (2)  4
8 – f (2)  4
4  f (2)  f (2)  4 ....(2)
from(1) and (2) f (2) = 4 Ans. ]
Q.111204 There are 50 apple trees in an orchard. Each tree produces 800 apples. For each additional tree
planted intheorchard, the output peradditionaltree drops by10apples. Number oftreesthat should be
added to the existing orchard formaximising the output ofthe trees, is
(A) 5 (B) 10 (C*) 15 (D) 20
[Sol. Let x tree be added then
P(x) = (x + 50) (800 – 10x)
now P'(x) = 0  x = 15 Ans. ]
Q.112205 The ordinate ofallpoints onthe curve y=
x
cos
3
x
sin
2
1
2
2

where the tangent is horizontal, is
(A) always equalto 1/2
(B) always equalto 1/3
(C) 1/2 or 1/3 according as n is an even or an odd integer.
(D*) 1/2 or 1/3 according as n is an odd or an even integer.

[Sol. y' = + 2
2
)
x
cos
2
(
x
cos
x
sin
2
·
1

= 0
sin 2x = 0
x =
2
n
, n  I
if n is odd then y =
x
cos
2
1
2

=
2
1
if n is even then y =
x
cos
2
1
2

=
3
1
]
Select the correct alternatives : (More than one are correct)
Q.113503 The equationofthe tangent to the curve
x
a
y
b
n n





 





 = 2 (nN) at the point withabscissa equalto
'a' can be :
(A*)
x
a
y
b





 





 = 2 (B*)
x
a
y
b





 





 = 2 (C)
x
a
y
b





 





 = 0 (D)
x
a
y
b





 





 = 0
Q.114504 The function y =
2 1
2
x
x


(x  2) :
(A*) is its own inverse (B*) decreases for all values ofx
(C) has a graph entirelyabove x-axis (D) is bound for all x.
Q.115505 If
x
a
y
b
 = 1 is a tangent to the curve x = Kt, y =
K
t
, K > 0 then :
(A*) a > 0, b > 0 (B) a > 0, b < 0 (C) a < 0, b > 0 (D*) a < 0, b < 0
Q.116506 The extremumvalues of the function f(x) =
1
4
1
4
sin cos
x x
 
 , where x  R is :
(A*)
4
8 2

(B)
2 2
8 2

(C*)
2 2
4 2 1

(D)
4 2
8 2

[Hint: f (x) = 0  (sin x + cos x) (non zero quantity) = 0
 tan x =  1  x = 3/4 or 7/4 .
Global Min = x = 2 n+ (3/4) ; global max = x = 2 n+ (7/4)
M =
4
8 2

; m =
4
8 2

]
Q.117507 The function f(x) = 1 4
0

 t
x
dt is suchthat :
(A*) it is defined on the interval[1, 1] (B*) it is anincreasing function
(C*) it is an odd function (D*) the point (0, 0) is the point ofinflection
[Hint: f  (x) = 1 4
 x > 0 in (–1, 1)  f is 
Now f (x) + f (– x) = dt
t
1
dt
t
1
x
0
4
x
0
4





  










 
 dy
y
1
dt
t
1
y
0
4
x
0
4
( t = – y)
= 0  f (x) is odd

again f  (x) =


4
2 1
3
4
x
x
whichvanished at x = 0 and changes sign (0, 0) isinflectionsincefis
well defined in [–1, 1]  A, B, C, D ]
Q.118509 Let g(x) = 2f (x/2) + f(1  x) and f (x) < 0 in 0  x  1 then g(x) :
(A) decreases in [0, 2/3) (B*) decreases in (2/3, 1]
(C*) increases in [0, 2/3) (D) increases in (2/3, 1]
[Hint: f (x) < 0  f (x) is decreasing . Now g (x) = f (x/2)  f (1  x) .
For g(x) to be increasing g (x) > 0  f (x/2) > f (1  x)
 (x/2) < (1  x)  x < 2/3 ]
Q.119510 The abscissa ofthe point onthe curve xy = a+ x, the tangent at whichcuts offequalintersects from
the co-ordinate axes is : ( a > 0)
(A*)
a
2
(B*) 
a
2
(C) a 2 (D)  a 2
[Hint: xy = a2 + x2 + 2ax  y =
a
x
2
+ x + 2a 
dy
dx
= 
a
x
2
2 + 1 =  1
 2x2 = a2  x = ±
a
2
]
Q.120512 The function
sin ( )
sin ( )
x a
x b


has no maxima or minima if:
(A*) b  a = n  , n  I (B*) b  a = (2n + 1)  , n  I
(C*) b  a = 2n  , n  I (D) none of these .
[Sol. f (x) =
)
b
x
sin(
)
a
x
sin(


f (x) =
)
b
x
(
sin
)
b
x
cos(
)
a
x
sin(
)
a
x
cos(
)
b
x
sin(
2







=
)
b
x
(
sin
)
a
b
sin(
2


If sin(b – a) = 0 then f (x) = 0  f(x) willbe constant
i.e. b – a = n or n or b – a = (2n + 1) or b – a = 2n
then f(x) has no minima ]
Q.121513 The co-ordinates ofthe point Pon the graph ofthe functiony= e–|x| wherethe portion ofthe tangent
intercepted between the co-ordinate axes has the greatest area, is
(A*) 1
1
,
e





 (B*) 






1
1
,
e
(C) (e, e–e) (D) none
[Hint: Find the co-ordinates ofthe point P (x > 0) and use the fact that f(x) = e–|x| is even]
Q.122514 Let f(x) = (x2  1)n (x2 + x + 1) then f(x) has local extremum at x = 1 when :
(A*) n = 2 (B) n = 3 (C*) 4 (D*) n = 6
[Hint: note that n must be an even integer
f ' (x) = (x2  1)n (2x + 1) + (x2 + x + 1) n(x2 – 1)n – 1 · 2x]
Q.123518 For the function f(x) = x4 (12lnx  7)
(A*) the point (1, 7) is the point ofinflection (B*) x = e1/3 is the point of minima
(C*) the graph is concave downwards in (0, 1) (D*) the graph is concave upwards in (1, )
[Hint:
dy
dx
= 16 x3 (3 lnx  1) &
d y
dx
2
2 = x2 (9 lnx) ]

Q.124519 The parabola y = x2 + px + q cuts the straight line y = 2x  3 at a point with abscissa 1. If the
distance between the vertex ofthe parabola and the xaxis is least then :
(A) p = 0 & q =  2
(B*) p =  2 & q = 0
(C) least distance between the parabola and xaxis is 2
(D*) least distance between the parabola and xaxis is 1
[Sol. When x = 1 ; y = – 1 ( from the line)
This must lie on the parabola y = x2 + px + q  – 1 = 1 + p + q  p + q = – 2
 Now distance of the vertex ofthe parabola fromthe x-axis is
d = q
2
p
4
p
2
p
f
2
2









 =
4
p
q
2

Substituting q = –2 – p here
f (p) = – 2 – p
4
p2

Hence f ' (p) = – 1 –
2
p
= 0  p = – 2
hence q = 0
note that least distance ofthe vertex fromx-axis is 1 ]
Q.125521 The co-ordinates of the point(s) on the graph of the function, f(x) =
x x
3 2
3
5
2
 + 7x - 4 where the
tangent drawncut offintercepts fromthe co-ordinate axes whichare equalinmagnitudebut oppositein
sign, is
(A*) (2, 8/3) (B*) (3, 7/2) (C) (1, 5/6) (D) none
[Sol. Since intercepts are equalin magnitude but opposite insign 
P
dx
dy
= 1
now
dx
dy
= x2 – 5x + 7 = 1  x2 – 5x + 6 = 0 x = 2 or 3 ]
Q.126524 On whichofthe following intervals, the function x100 + sin x  1 is strictlyincreasing.
(A) (1, 1) (B*) (0, 1) (C*) (/2, ) (D*) (0, /2)
[Sol. f  (x) = 100 x99 + cosx
for )
1
,
0
(
x and 




 
2
,
0 , cosx and x are both +ve 
for 







 ,
2
x , x > 1 hence 100 x99 obviously > cosx  ]
Q.127525 Let f(x) = 8x3 – 6x2 – 2x + 1, then
(A) f(x) = 0 has no root in (0,1) (B*) f(x) = 0 has at least one root in (0,1)
(C*) f(c) vanishes for some )
1
,
0
(
c (D) none
[Hint: Consider g(x) whichis the integraloff(x) and applyRolle’s theoremin it

1
0
dx
)
x
(
f = 0  f (x) = 0 has at least one root ]

Q.128526 Equation of a tangent to the curve ycot x = y3 tanx at the point where the abscissa is

4
is :
(A*) 4x + 2y =  + 2 (B*) 4x  2y =  + 2 (C) x = 0 (D*) y = 0
[Hint: y cot2x – y3 = 0  y (y2 – cot2x) = 0
hence the curve is y = 0 or y2 = cot2x ]
Q.129528 Let h(x) = f(x)  {f(x)}2 + {f(x)}3 for every real number 'x' , then :
(A*) 'h' is increasing whenever 'f' is increasing
(B) 'h' is increasing whenever 'f' is decreasing
(C*) 'h' is decreasing whenever 'f' is decreasing
(D) nothing can be said ingeneral.
[Hint: Here h(x) = f(x)  [{f(x)}2 + {f(x)}3 ]
 h (x) = f (x)  2 f(x) f (x) + 3 {f(x)}2 f (x)
= f (x) [1  2 f(x) + 3 {f (x)}2]
= f (x) (3 y2  2 y + 1) where y = f(x)
The discriminant of 3 y2  2 y + 1 = 4  12 =  8 < 0
and so its sign is the same as both co-efficient of y2 i.e. > 0
 h (x) = f (x) (a positive quantity)
 sign of h (x) is the same as that of f (x) either h (x) > 0 & f (x) > 0
or h (x) < 0 & f (x) < 0 . Hence h(x) & f(x) increase and decrease together. ]
Q.130529 Ifthe side ofa triangle varyslightlyinsuch a waythat its circumradius remains constant, then,
da
A
db
B
dc
C
cos cos cos
  is equal to :
(A) 6R (B) 2R (C*) 0 (D*) 2R(dA + dB + dC)
[Hint: Given
a
A
b
B
c
C
sin sin sin
  = 2R (say)
 da = 2R cos A d A, d b = 2R cos B d B , d c = 2R cos C d C
 da
A
db
B
dc
C
cos cos cos
  = 2R (dA + dB + dC) ....(1)
Also A + B + C =  So , dA + dB + dC = 0 ....(2)
From equations (1) & (2) we get da
A
d b
B
dc
C
cos cos cos
  = 0 ]
Q.131532 In whichofthe following graphs x = c is the point ofinflection .
(A*) (B*) (C) (D*)
[Sol. At point ofinflection concavityofthe curve changes disregards to anyother factor  ABD ]

Q.132534 An extremum value of y =
0
x
 (t  1) (t  2) dt is :
(A*) 5/6 (B*) 2/3 (C) 1 (D) 2
[Hint:
dy
dx
= (x  1) (x  2) = 0 when x = 1 & x = 2 .
d y
dx
2
2 = 2 x  3 ;
d y
dx x
2
2
1




< 0  y is maximum at x = 1 and
d y
dx x
2
2
2




< 0  y is minimum at x = 2
Hence extreme values are ymax =
0
1
 (t2  3 t + 2) d t & ymin =
0
2
 (t2  3 t + 2) d t ]