DIFFERENTIALEQUATION &AREAUNDER CURVE
There are 70 questions in this question bank.
Q.12/AUC Area common to the curve y = 9 2
x & x² + y² = 6 x is :
(A)
 3
4
(B)
 3
4
(C) 3  








3
4
(D*) 3  








3 3
4
[Hint: x2 + y2 = 9 ....(1) ; x2 + y2 – 6x = 0 ....(2)
On solving x=3/2 ; y2 = 9 –9/4 = 27/4
 y =
2
3
3
 A = dx
x
9
2
3
2
/
3
2
 
Q.22/DE Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation
corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is
K > 0, is
(A*)
dr
dt
 K = 0 (B)
dr
dt
 K = 0 (C)
dr
dt
 Kr (D) none
[Hint:
dt
dV
= – k4r2 ....(1)
but V =
3
r
3
4
 
dt
dV
= 4r2
dt
dV
....(2)
hence
dt
dr
= – K  (A) ]
Q.33/AUC If y = 2 sin x + sin 2x for 0  x  2, then the area enclosed by the curve and the x-axis is
(A) 9/2 (B*) 8 (C) 9 (D) 4
[Hint: A =  



0
dx
x
2
sin
x
sin
2
2 = 




0
0
dx
x
2
sin
2
dx
x
sin
4
= 8 + 0 = 8
Q.45/DEThegeneral solution of the differential equation, y + y (x)  (x). (x)= 0 where (x)isa known
functionis:
(A*) y = ce(x) + (x)  1 (B) y = ce+(x) + (x)  1
(C) y = ce(x) (x) + 1 (D) y = ce(x) + (x) + 1
where c is an arbitraryconstant .
[Sol.
dx
dy
+ y '(x) =  (x).'(x)
I.F. = )
x
(
dx
)
x
(
e
e 



hence y.e(x) =  e(x).(x).'(x) dx =  et.t dt where  (x) = t
= tet – et + C = (x).e(x) – e(x) + C
 y = ce–(x) + (x) – 1  A ]

Q.824/DE The solution of the differential equation, xy
dx
dy
= 2
2
x
1
y
1


(1 +x + x2) given that when x = 1, y= 0 is
(A) ln 1 2
 y = lnx + tan1 x 

2
(B*) ln
1 2
2
 y
x
= 2 tan1 x 

2
(C) ln
1 2
2







y
x
=

4
 2 tan1 x (D) none
[Hint:   2
y
1
dy
y
= dx
)
x
1
(
x
x
x
1 2
 


2
1
ln (1 + y2) = tan–1x + ln x + C
ln 2
2
x
y
1
= 2 tan–1x + C ]
Q.96/AUC The area of the figure bounded by the curves y = ex , y= e-x & the straight line x = 1 is
(A) e +
1
e
(B) e 
1
e
(C*) e +
1
e
 2 (D) none
Q.1035/DE Acurve is such that the area of theregion bounded bythe co-ordinate axes, the curve& the ordinate
of anypoint on it is equal to the cube of that ordinate .The curve represents
(A) apairofstraight lines (B) a circle
(C*) a parabola (D) an ellipse
[Sol. dx
)
x
(
f
x
0
 = y3
Differentiating
f (x) = 3y2.
dx
dy
y = 3y2
dx
dy
 y = 0 (rejected)
or 3y dy = dx
2
y
3 2
= x + c  parabola  C]
Q.117/AUC The area bounded by the curve y = x²  1 & the straight line x + y = 3 is :
(A)
9
2
(B) 4 (C)
2
17
7
(D*)
6
17
17
[Hint: 3 – x = x2 – 1  x2 + x – 4 = 0
x1 + x2 = –1
x1 x2 = – 4 ....(1)
A =  
 dx
1
x
)
x
3
(
2
1
x
x
2
 

 =  dx
x
x
4
2
1
x
x
2
 
 use(1)]

Q.1823/AUC The area common to y  x & x >  y and the curve x² + y² = 2 is :
(A)

4
(B)
3
2

(C)  (D*)

2
[Hint: dx
x
x
2
A
0
1
2
2







 

 + dx
x
x
2
A
1
0
2
 




 

 =
2

note that the area is equal to the sector AOB with central angle 900
 1/4 (the area of the circle) ]
Q.1952/DE Therealvalueofm forwhichthesubstitution, y= um willtransform thedifferential equation,
2x4y
dy
dx
+ y4 = 4x6 into a homogeneous equation is :
(A) m = 0 (B) m = 1 (C*) m = 3/2 (D) no value of m
[Hint: y = um 
dx
dy
= m um  1 . Hence 2 x4 . um . m um  1 .
dx
du
+ u4m = 4 x6 .
du
dx
=
4
2
6 4
4 2 1
x u
m x u
m
m


 4 m = 6  m =
3
2
and 2m – 1 = 2  m =
3
2
 (C) ]
Q.2024/AUC The area enclosed by x+ y = 1 & the axis of x is :
(A*) 1 (B)
1
2
(C) 2 (D) none
Q.2127/AUC The area bounded by x²+y²2x = 0 & y = sin
x
2
in the upper half of the circle is :
(A*)


2
4
 (B)


4
2
 (C) 


8
(D)


 2
2
Q.2254/DE The solution of the differential equation, x2
dy
dx
.cos
1
x
 ysin
1
x
= 1, where y1 as x  is
(A*) y = sin
1
x
– cos
1
x
(B) y =
x
x x
 1
1
sin
(C) y = cos
1
x
 sin
1
x
(D) y =
x
x x
 1
1
cos
[Hint:
dy
dx

y
x2 tan
1
x
=  sec
1
x
.
1
2
x
. IF = e x x
dx
 1
2
1
tan
= sec
1
x
 y . sec
1
x
=   





2
2
x
1
x
1
sec dx = tan
1
x
+ c
if y  1 then x  c =  1  y = sin
1
x
 cos
1
x
]

Q.2657/DE The latus rectum of the conicpassingthrough theorigin and having the propertythat normal at each
point (x, y) intersects the xaxis at ((x +1), 0) is :
(A) 1 (B*) 2 (C) 4 (D) none
[Sol. Given y
dx
dy
+ x = x + 1
c
x
2
y2


x = 0, y = 0  c = 0
 y2 = 2x  latus rectum= 2  B ]
Q.2735/AUC The line y = mx bisects the area enclosed by the curve y = 1 + 4x  x2 & the lines x = 0 ,
x =
3
2
& y = 0 . Then the value of m is :
(A*)
13
6
(B)
6
13
(C)
3
2
(D) 4
Q.2859/DE The solution of the differential equation, 2x2y
dy
dx
= tan (x2y2)  2xy2 given y(1) =

2
is
(A*) sinx2y2 = ex–1 (B) sin(x2y2) = x (C) cosx2y2 + x = 0 (D) sin(x2y2) = e.ex
[Sol. put x2y2 = z
Given )
y
x
(
tan
x
2
.
y
dx
dy
y
2
.
x 2
2
2
2


)
y
x
tan(
)
y
x
(
dx
d 2
2
2
2
 put x2 y2 = z
nowgivenexpression transformsto
dx
dz
= tan z
 dx =  dz
z
cot
x = ln (sin z) + c
when x = 1 , y =
2

 z =
2

 c = 1
 x = ln sin (x2y2) + 1
 ln sin(x2y2) = x – 1
sin (x2 y2) = ex–1]
Q.2936/AUC The area bounded by the curve y = f (x), the x-axis & the ordinates x =1 &
x = b is (b1)sin(3b+ 4). Then f(x) is :
(A) (x  1) cos (3x +4) (B) sin(3x+4)
(C*) sin (3x + 4) + 3 (x  1) . cos (3x +4) (D) none
[Sol: dx
)
x
(
f
b
1
 = (b – 1) sin (3b + 4)

Q.3344/AUC The area of the region of the plane bounded by max  
x y
,  1 & xy 
1
2
is :
(A)
1
2
+ ln 2 (B)
31
4
(C) 1 + 2 ln 2 (D*) 3 + ln 2
[Hint: shaded area in 1st quadrant
= dx
x
2
1
1
1
2
/
1
 





 =
1
2
/
1
nx
2
1
x 


 l
=
2
1
n
2
1
2
1
l
 =
2
1
n
2
1
2
1
l

 2 times the shaded area = 1 – ln2
 Required area = 2 – (1–ln2) + 2 = 3 + ln2  D ]
Q.3474/DE If y= |
x
c
|
n
x
l (where c is an arbitraryconstant) is the general solution of the differential equation
dx
dy
=
x
y
+ 








y
x
then the function 








y
x
is :
(A) 2
2
y
x
(B) – 2
2
y
x
(C) 2
2
x
y
(D*) – 2
2
x
y
[Hint: ln c + ln |x| =
y
x
diff.w.r.t.x,
x
1
= 2
1
y
y
x
y 
dx
dy
x
y
x
y2


dx
dy
= 2
2
x
y
x
y
  








y
x
= – 2
2
x
y
 D ]
Q.3545/AUC The area of the region for which 0 < y < 3  2x  x2 & x > 0 is :
(A)  
3 2 2
1
3
 
 x x dx (B)  
3 2 2
0
3
 
 x x dx
(C*)  
3 2 2
0
1
 
 x x dx (D)  
3 2 2
1
3
 
 x x dx
Q.3677/DE Ifthefunctiony=e4x +2e–x isasolutionofthedifferential equation, K
y
dx
dy
13
dx
y
d
3
3


thenthe value
of K is :
(A) 4 (B) 6 (C) 9 (D*) 12

[Sol. y = c1cos(x + c2) – (c3
4
c
x
e 

) + (c5sin x)
 y = c1 (cos x cos c2 – sin x sin c2) – (c3
4
c
e e–x)+ (c5 sin x)
 y = (c1 cos c2 ) cosx – (c1 sin c2 – c5) sinx – (c3
4
c
e ) e–x
 y = l cosx + m sinx – n e–x ....(i) where l, m, n are arbitraryconstant

dx
dy
= – l sinx + m cosx + n e–x ....(ii)
 2
2
dx
y
d
= – l cosx – m sinx – n e–x ....(iii)
 3
3
dx
y
d
= l sinx – m cosx + n e–x ....(iv)
(i)+(iii)gives 2
2
dx
y
d
+ y = – 2n e–x ....(v)
(ii) + (iv) gives
dx
dy
dx
y
d
3
3
 = 2n e–x ....(vi)
From (v) and (vi) we get 










 y
dx
y
d
dx
dy
dx
y
d
2
2
3
3
or 0
y
dx
dy
dx
y
d
dx
y
d
2
2
3
3



 is therequired differential equation ]
Q.4263/AUC Let f(x) = x  x2 & g(x) = ax . If the region above the graph of g and below the graph of f has
an area equal to 9/2 then 'a' is equal to :
(A) 2 (B*) 4 (C*)  2 (D) 3
Q.4380/DE The curve, with the propertythat the projection of the ordinate on the normal is constant 'a', is
(A*) c
y
a
y
n
a
x 2
2






 

 l (B) c
y
a
x 2
2



(C) (y – a)2 = cx (D) ay = tan–1 (x + c)
[Sol. Ordinate = PM. Let P  (x, y)
Projection of ordinateon normal = PN
 PN = PM cos = a (given)
 a
tan
1
y
2



 y = 2
1
)
y
(
1
a 

a
a
y
dx
dy
2
2

   


dx
a
y
dy
a
2
2  c
x
|
a
y
y
|
n
a 2
2




l ]
Q.4464/AUC The area bounded by the curves y =  x and x =  y where x, y  0
(A) cannot be determined
(B*) is 1/3
(C) is 2/3
(D) is same as that of the figure bounded by the curves y = x ; x  0 and x = y ; y  0

Q.4868/AUC Consider the graph of continuous function y = f(x) for x  [a  b, a + b] a, b  R+ and b > a. If
the origin is shifted to (a + b, 0) such that new axes are parallel to the old axes, then the area bounded
bythe given curve, the X-axis and the new ordinates X = a, X = b can be written as :
(A)
b a
a b



2
2
f(x) dx (B)



a
b
f(x)dx (C*)
a
b
 f(a + b  x)dx (D)
1
2 a b
a b


 f(x)dx
[Hint: Required area =Area of the regionABB'PA'
= 


a
b
dX
)
X
(
f = 
b
a
dx
)
x
(
f
=  

b
a
dx
)
x
b
a
(
f  (C) ]
x – (a + b) = X
if X = – a then x = b
X = – b then x = a ]
Q.4971/AUC
The curvilineartrapezoid is bounded bythe curvey= x2 +1 and thestraight lines x=1 andx=2.The
co-ordinates ofthepoint (on thegivencurve)with abscissax [1,2] wheretangent drawn cutofffrom
the curvilineartrapezoidan ordinarytrapezium ofthegreatest area,is
(A) (1,2) (B) (2,5) (C*) 





4
13
,
2
3
(D) none
[Hint:
1
1y
x
dx
dy



= 2x1
T : y – y1 = 2x1 (x–x1) ....(1)
For co-ordinates of the point P put x = 1 in (1)
y = y1 + 2x1 (1–x1)
= 1 + 2
1
x +2x1 + 2 2
1
x = 1 + 2x1 – 2
1
x
Hence P(1, 1 + 2x1 – 2
1
x )
For Q put x= 2 in (1)
y = y1 + 2x1 (2 – x1) = 1 + 2
1
x + 4x1 – 2 2
1
x = 1 + 4x1 – 2
1
x
 A = (1 + 2x1 – 2
1
x ) + (1+ 4x1 – 2
1
x ) = 1 + 3x1 – 2
1
x
1
dx
dA
= 3 – 2x1  x1 = 3/2 ]
Q.5089/DE If 
x
a
dt
)
t
(
y
t = x2 + y (x) then yas a function of x is
(A*) y = 2 – (2 + a2) 2
a
x 2
2
e

(B) y = 1 – (2 + a2) 2
a
x 2
2
e

(C) y = 2 – (1 + a2) 2
a
x 2
2
e

(D) none

[Sol. (a, 0) lies on the given curve
 0 = sin2a – 3 sina  sina = 0 or cosa = 3 /2
 a =
6

(as a >0 and thefirst point of intersection with positive X-axis)
and A = 


6
/
0
dx
)
x
sin
3
x
2
(sin =
6
/
0
x
cos
3
2
x
2
cos









= 3
4
7
3
2
1
2
3
4
1



















 4A + 8 cosa = 7 ]
Q.5390/DE General solution of the differential equation 2y' ln x +
x
y
= y–1cos x is
(A*) y2 ln x = c + sin x (B) y2 ln x = c – sin x
(C) y2 ln x = c + cos x (D) y2 ln x = c – cos x
[Hint: put y2 = t ]
Q.5473/AUC
Area of the region enclosed between the curves x = y2 – 1 and x = |y| 2
y
1 is
(A) 1 (B) 4/3 (C) 2/3 (D*) 2
[Hint: dy
)
1
y
(
y
1
y
2
A
1
0
2
2
 




 



= 2 ]
Q.5591/DE Solution of the equation 
1
0
dt
)
t
x
(
y = n y(x) is
(A) y = c x1/n (B*) y = c n
n
1
x

(C) y = c n
1
1
x

(D) y = c x– 1/n
[Hint: start x t = y ]
Q.5677/AUC
Let y= g (x) be the inverse of a bijective mapping f : R  R f (x) = 3x3 + 2x. The area bounded by
the graph of g (x), the x-axis and the ordinate at x = 5 is :
(A)
4
5
(B)
4
7
(C)
4
9
(D*)
4
13
[Hint: noteforinversefunctionyaxis willbethe x axis andx axis will bethe yaxis
requiredarea =Area of rectangle – 
1
0
dx
)
x
(
f
= 5 –  
1
0
3
dx
)
x
2
x
3
(
= 5 – (
4
3
+ 1)
= 3
4
1
=
4
13
Ans ]

Q.609/DE Equationofacurvepassingthrough theorigin iftheslopeofthe tangent drawnat anyofitspoint(x, y)
is cos(x + y) + sin(x + y), is
(A) y = 2 tan–1(ex – 1) + x (B*) y = 2 tan–1(ex – 1) – x
(C) y = 2 tan–1(ex) – x (D) y = 2 tan–1(ex) + x
[13th test (24-03-2005)]
[Sol.
dx
dy
= cos(x + y) + sin(x + y) ; put x + y = u ; 1 +
dx
du
dx
dy

dx
du
– 1 = cos u + sin u 
dx
du
= (1 + cos u) + sin u = 2cos2
2
u
+ 2sin
2
u
cos
2
u
= 2cos2
2
u
(1 + tan
2
u
)
du
2
u
tan
1
2
2
u
sec2








= dx ;  tan
2
u
= t ; 
 

dx
t
1
dt
x = ln (1 + t) + C t = 0 ; C = 0
1 + t = ex  t = ex – 1 ; 




 
2
y
x
tan = ex – 1 
2
y
x 
= tan–1(ex – 1)
y = 2 tan–1(ex – 1) – x ]
Q.6165/auc The area bounded bythe curves y= x (1  lnx) ; x = e1 and positive X-axis between x = e1 and
x = e is :
(A)
e e
2 2
4
5








(B*)
e e
2 2
5
4








(C)
4
5
2 2
e e








(D)
5
4
2 2
e e








[Hint: y = x (1 – lnx) = 0  x = e (as x > 0)
dx
dy
= – lnx  in (0,1) and  in (1, )
also 0
x
Lim

x (1 – lnx) = 0
A =  
e
e
/
1
dx
)
nx
1
(
x l ]
Q.6279/auc Which one of the following DOES NOT represent the area enclosed bythe curves
y = sec–1x , y = cosec–1x and the line x – 1 = 0 ?
(A*) dx
)
1
x
ec
(cos
2
/
0


 (B) 2 dx
)
1
x
(sec
4
/
0



(C) dx
)
1
x
ec
(cos
2
2
/
4
/



 (D) dx
)
x
sec
x
ec
(cos
2
1
1
1





Q.6586/auc Area bounded by the curve y = min {sin2x, cos2x} and x-axis between the ordinates x = 0 and
x =
4
5
is
(A)
2
5
square units (B)
4
)
2
(
5 

square units
(C*)
8
)
2
(
5 

square units (D) 







2
1
8
square units
[Hint: A = 









4
5
4
3
2
4
3
4
2
4
0
2
dx
x
sin
dx
x
cos
dx
x
sin ]
Q.6615/DE Waterisdrainedfrom averticalcylindrical tank byopeningavalveat thebase ofthetank.It isknown
that therateat whichthewaterlevel drops isproportional to the squareroot of waterdepth y, where the
constant of proportionalityk > 0 depends on the acceleration due to gravity and the geometryof the
hole. If t is measured in minutes andk =
15
1
then the time to drain thetank if the wateris 4 meter deep
to start with is
(A)30min (B)45min (C*) 60 min (D)80 min
[Hint:
dt
dy
= – k y

0
4
y
dy
= – 
t
0
dt
k
0
4
y
2 = – kt = –
15
t
0 – 4 = –
15
t
 t = 60 minutes  (C) ]
Q.6787/auc If the area bounded between x-axis and the graph of y = 6x – 3x2 between the ordinates x = 1 and
x = a is 19 square units then 'a' can take the value
(A) 4 or – 2
(B) two values are in (2, 3) and one in (–1, 0)
(C*) two values one in (3,4) and one in (–2,–1)
(D) none of these
[Hint: I =   dx
)
x
3
x
6
( 2
=
3
x
3
2
x
6 3
2
 = 3x2 – x3 = x2(3 – x)
A1 = I(2) – I(1) = 4 – 2 = 2 units
A2 = I(2) – I(3) = 4 – 0 = 4 units
A3 = I(3) – I(4) = 0 – (–16) = 16 units
 onevalue of awill lie in (3, 4) usingsymmetric otherwill lie in (–2, –1) ]