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L.K.Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm.. MOBILE: 9815527721,4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 1111
PAPER–A
IIT–JEE (2011)
(Integral Calculus solutions)
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MARKING SCHEME
1. For each question in Section I, you will be awarded 3 marks if you have darkened
only the bubble corresponding to the correct answer and zero mark if no bubble is
darkened .In all other cases, minus one (–1) mark will be awarded.
2. For each question in Section II, you will be awarded 3 marks if you darken only the
bubble corresponding to the correct answer and zero mark if no bubble is darkened.
Partial marks will be awarded for partially correct answers. No negative marks will be
awarded in this section.
3. For each question in Section III, you will be awarded 3 marks if you darken only the
bubble corresponding to the correct answer and zero mark if no bubble is darkened. In
all these cases, minus one (–1) mark will be awarded.
4. For each question in Section IV, you will be awarded 3 marks if you darken the
bubble corresponding to the correct answer and zero mark if no bubble is darkened. No
negative marks will be awarded for in this section. NAME OF THE CANDIDATE PHONE NUMBER
L.K. Gupta (Mathematics Classes)
Pioneer Education (The Best Way To Success)
S.C.O. 320, Sector 40– D, Chandigarh Ph: – 9815527721, 0172 – 4617721.
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 2222
Section I
This section contains 8 multiple choice questions. Each question has four choices A),
B), C) and D) out of which ONLY ONE is correct.
1. nlim
→∞
1/n 2/n (n 1)/n1 e e e......
n n n n
− + + + +
equals
(a) 0 (b) 1 (c) e–1 (d) none of these
Sol:
1/n 1/n 2 1/n n 1
n
1 e (e ) .......(e )lim
n
−
→∞
+ + +
= 1/n n
1/nn
1.[(e ) 1]lim
n(e 1)→∞
−
−
=1/nn
1(e 1)lim (e 1) 1 e 1
e 1
1 / n
→∞− = − × = −
−
2. 2 2
cc
sin t sin t
x yx 0y
1lim e dt e dt
x +→
−
∫ ∫ is equal to(where c is a constant)
(a) 2sin ye (b)
22 sin ysin ye (c) 0 (d) none of these
Sol:
2 2c c
sin t sin t
x 0y x y
1lim e dt e dt
x→+
−
∫ ∫ =
2 2x yc
sin sin t
x 0y c
1lim e dt e dt
x
+
→
+
∫ ∫ =
2x y
sin t
y
x 0
e dt
limx
+
→
∫
=
2sin (x y )
x 0
e 0lim
1
+
→
− =
2sin ye
3. Area enclosed by the curve y = f (x) defined parametrically as 2
2 2
1 t 2tx , y
1 t 1 t
−= =
+ + is
equal to
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 3333
(a) π sq. units (b) π / 2 sq.units (c)3π
sq.units4
(d) 3π
sq.units2
Sol:
Clearly t can be any real number
Let t = 2
2
1 tan θtan θ x
1 tan θ
−⇒ =
+
x cos 2 θ, and⇒ =
2
2 tan θy sin 2 θ
1 tan θ= =
+
2 2x y 1⇒ + =
Thus, required area = π sq.units.
4. { } { }15
1
The value of sgn( x )dx, where . denotes the fractional part function, is−
∫
(a) 8 (b) 16 (c) 24 (d) 0
Sol:
15 16
1 0
sgn({x})dx sgn({x 1})dx−
= −∫ ∫ (by property)
=16 1 1 1
0 0 0 0
sgn({x})dx 16 sgn({x})dx 16 sgn(x)dx 16 1.dx 16= = = =∫ ∫ ∫ ∫
5. The slope of the tangent to the curve y=
2x
1 2
x
cos t dt−
∫ at 4
1x
2=
(a) 4 8 3
π2 4
−
(b)
4 8 1π
3 4
−
(c)
4 8 1π
3 4
−
(d) none of these
Sol:
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 4444
2x
1 2
x
y cos t dt−= ∫∵
1 4 1 2dycos (x ).2x cos (x ).1
dx
− −∴ = −
⇒4
1 1
1 4x2
dy 1 2 1| cos cos
dx 2 2 2
− −
=
= −
= 3/4π π.2
3 4− =
4 8 1π
3 4
−
6. 12 9
5 3 3
(2x 5x )dx
(x x 1)
+
+ +∫ is equal to.
(a) 2
5 3 2
x 2xc
(x x 1)
++
+ + (b)
10
5 2
xc
2(x 3x 1)+
+ +
(c)In 5 3 7 4| x x 1| (2x 5x ) c+ + + + + (d) none of the above
Sol:
Let I = 12 9
5 3 3
2x 5xdx
(x x 1)
+
+ +∫ = 3 6
3
2 5
2 5
x xdx
1 11
x x
+
+ +
∫
Put 2 5
1 11 t
x x+ + = ∴
3 6
2 5dx dt
x x
− − =
Then I = 3 2
dt 1c
t 2t− = +∫ =
2
2 5
1c
1 12 1
x x
+
+ +
= 10
5 3 2
xc
2(x x 1)+
+ +
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 5555
7. If { }21
f(x)cos xdx f(x) c2
= +∫ then f(x) is
(a) x + c (b) sin x + c (c) cos x + c (d) c
Sol:
'1f(x)cos x .2f(x)f (x)
2=
Then f '(x) cosx= ∴ f(x) sin x c= +
8. If
π/2
1
0
I cos(sin x)dx;= ∫π/2
2
0
I sin(cosx)dx= ∫ and
π/2
3
0
I cosxdx,= ∫ then.
(a) 1 2 3I I I> > (b) 2 3 1I I I> > (c) 3 1 2I I I> > (d) 1 3 2I I I> >
Sol:
x 0>∵ sin x x cos(sin x) cosx ....(1)∴ < ⇒ >
Also π
0 x2
< < 1 cosx 0∴ > > , sin(cosx)<cosx From Eqs (1)and (2) we get (2)
cos (sin x) > cos x >sin (cos x) Or
π/2 π/2 π/2
0 0 0
cos(sin x)dx cos x dx sin(cosx)dx> >∫ ∫ ∫
⇒ 1 3 2I I I> >
Section II.
This section contains 5 multiple choice questions. Each question has four choices A),
B), C) and D) out of which ONE OR MORE may be correct.
9. Which one of the following functions is/are homogeneous?
(a). 2 2
x yf (x,y)
x y
−=
+
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 6666
(b) 1 2
13 3x
f (x,y) x y tany
−−=
(c) 2 2 x/yf (x,y) x (ln x y ln y) ye= + − +
(d) 2 2
22x y x 2yf (x,y) x ln ln(x y) y tan
x 3x y
+ += − + +
−
Sol:
(a). 1
2 2 2
λ(x y)f (λx,λy) λ f (x,y)
λ (x y )
−−= =
+
⇒homogeneous of degree (−1).
(b). 1/3 2/3 1 xf (λx,λy) (λx) (λ y) tan
y
− −=
1/3 1/3 2/3 1 xλ x y tan
y
− − −=
1
3λ f (x,y)−
=
⇒homogeneous
(c). ( )2 2 2 x/yf (λx,λy) λx ln λ (x y ) ln λy λ ye= + − +
2 2
x/yλ (x y )λx ln λye
λy
+ = +
( )2 2 x/yλ x ln x y ln y ye = + − +
λ f (x,y)=
⇒homogeneous
(d). 2 2 2 2
2 22λ x λ y x 2yf (λx,λy) λx ln λ x tan
λ x λ(x y) 3x y
+ += +
+ −
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 7777
2 22 22x y x 2y
λx ln λ x tanx(x y) 3x y
+ += +
+ −
⇒non homogeneous
10. If f (x), g(x) be twice differential functions on [0, 2] satisfying f ‘’ (x) = g’’ (x), f ‘(1)
=2g’(1) = 4 and f (2) = 3g(2) = 9, then
(a). f(4) −g(4) = 10 (b). f (x) g(x) 2 2 x 0− < ⇒ − < <
(c). f (2) g(2) x 1= ⇒ = − (d). f (x) g(x) 2x− = has real root
Sol: (a), (b), (c)
We have f’’(x) = g’’(x). On integration, We get f’(x) = g ‘(x) + C …. (i)
Putting x = 1, we get
f’ (1) = g’ (1) + C 4 2 C C 2⇒ = + ⇒ =
f '(x) g'(x) 2∴ = +
Integrating w.r.t. x, we get f(x) = g(x) + 2x + c1 ….. (ii)
Putting x = 2, we get
f (2) = g (2) + 4 + c1 1 19 3 4 c c 2⇒ = + + ⇒ =
f(x) g(x) 2x 2. Putting x 4, we get f(4) g(4) 10∴ = + + = − =
f (x) g(x) 2 2x 2 2 x 1 1 2 x 0− < ⇒ + < ⇒ + < ⇒− < <
Also f(2) g(2) x 1= ⇒ = −
f (x) g(x) 2x− = has no solution.
11. Identify the statement (s) which is/are true.
(a). y/x yf (x,y) e tan
x= + is homogeneous of degree zero.
(b). 2
1y y yx ln dx sin dy 0
x x x
−+ = is homogeneous differential equation.
(c). 2f (x,y) x sin x cos y= + is not homogeneous.
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 8888
(d). 2 2 2 3(x y )dx (xy y )dy 0+ − − = is a homogeneous differential equation.
Sol: (a), (b), (c)
12. The solution of 2 2
2 2
xdx ydy 1 x y
xdy ydx x y
+ − −=
− +is
(a). { }2 2 1x y sin tan (y / x) C−+ = + (b). { }2 2 1x y cos (tan y / x) C−+ = +
(c). 2 2 1x y (tan (sin y / x) C)−+ = + (d). ( )1 2 2y x tan c sin x y−= + +
Sol: (a), (b)
The D. E can be re-written as
( ) 2 22 2
x dx y dy x dy y dx
x y1 x y
+ −=
+− +
( )1 2 2
2 2
xdy ydxSince d tan ( y / x) ,and d x y
x y
− −= +
+
2(xdx y dy),= +
( )
( )
2 2
2 22 2 2 2
1d x y
xdy ydx2we havex yx y 1 x y
+−
∴ =++ − +
{ }1d tan (y / x)−=
2 2 2
2
t dtPut x y t in t he L.H.S and get
t 1 t+ =
−
{ }1d tan (y / x)−=
Integrating both sides, we get
1 1sin t tan (y / x) c− −= +
( )1 2 2 1i.e.,sin x y tan (y / x) c− −+ = +
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 9999
13. Let f(x) = x t
2
1
3
1 t+∫ dt, where x > 0, then
(a). for0 α β,f(α) f(β)< < < (b). for0 α β,f(α) f(β)< < >
(c). 1f(x) π / 4 tan x, x 1−+ < ∀ ≥ (d). 1f(x) π / 4 tan x, x 1−+ > ∀ ≥
Sol.
x x
2 2 2
3 3 1f '(x) 0 x 0 f '(x) , x 1
1 x 1 x 1 x= > ∀ > ⇒ = > ∀ ≥
+ + +
x x
2
1 1
1f '(x) dx dx
1 x⇒ >
+∫ ∫
1 1 1f(x) tan x tan 1 f(x) π / 4 tan x− − −⇒ > − ⇒ + >
Section III [Linked comprehension type]
This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice
questions and based upon the second paragraph 2 multiple choice questions have to be
answered. Each of these questions has four choices A), B), C) and D) out of which ONLY
ONE is correct.
Paragraph for question 14 to 16
Let the definite integral be defined by the formula b
a
b af (x)dx (f (a) f(b)).
2
−= +∫ For
more accurate result for c (a,b)∈ , we can useb c b
a a cf(x)dx f (x)dx f (x)dx F(c)= + =∫ ∫ ∫ so
that forb
a
a b b ac , we get f (x)dx (f (a) f(b) 2f(c))
2 4
+ −= = + +∫
14. π/2
0∫ sin xdx is equal to
(a) π
(1 2)8
+ (b)π
(1 2)4
+ (c) π
8 2 (d)
π
4 2
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 10101010
15. If
x
a
3x a
x af (x)dx (f (x) f (a))
2lim 0, then f (x)
(x a)→
− − + =
−
∫is maximum degree
(a) 4 (b) 3 (c) 2 (d) 1
16. If f’’(x) 0 x (a,b)< ∀ ∈ and c is a point such that a < c < b, and (c, f (c)) is the point lying
on the curve for which F (c) is maximum, then f’ (c) is equal to
(a) f (b) f (a)
b a
−
− (b)
2(f (b) f (a))
b a
−
− (c)
2f(b) f(a)
2b a
−
− (d) 0
Sol.
14. (a) π/2
0
π0
π π2sin x dx sin 0 sin 2 sin
4 2 2
− = + +
∫
( )π1 2
8= +
15. (d).
x
a
3x a
x af (x) dx (f (x) f (a))
2lim 0
(x a)→
− − + =
−
∫
a h
a
3h 0
hf (x) dx (f (a h) f (a))
2lim 0h
+
→
− + +=
∫
2h 0
1 hf (a h) [f (a) f (a h)] ( f '(a h))
2 2lim3h→
+ − + + − +⇒ 0=
[ Using L’ Hospital’ s Rule]
2h 0
1 1 hf (a h) f (a) f '(a h)
2 2 2lim 03h→
+ − − +⇒ =
h 0
1 1 hf '(a h) f '(a h) f ''(a h)
2 2 2lim 06h→
+ − + − +⇒ =
[ Using L’ Hospital’ s Rule]
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 11111111
h 0
f ''(a h)lim 0
12→
− +⇒ =
f ''(a) 0, a R⇒ = ∀ ∈
⇒ f (x) must be of max. degree 1.
16. (b). f ''(x) 0, x (a,b),for c (a,b)< ∀ ∈ ∈
c a b cF(c) ( f (a) f (c)) ( f (b) f (c))
2 2
− −= + + +
b a c a b cf (c) f (a) f (b)
2 2 2
− − −= + +
b a 1 1F'(c) f '(c) f(a) f (b)
2 2 2
−⇒ = + −
1[(b a)f '(c) f(a) f (b)]
2= − + −
1F''(c) (b a)f ''(c) 0
2= − < [ f ''(x) 0, x (a,b) and b a]< ∀ ∈ >∵
∴ F (c) is max. at the point (c, f (c)) where
f (b) f (a)F'(c) 0 f '(c) 2
b a
−= ⇒ =
−
Paragraph for question 17 to 18
Integrals of the form 2R(x, ax bx c) dx+ +∫ are calculated with the aid of one of the
three Euler substitutions
1. 2ax bx c t x a if a 0;+ + = ± >
2. 2ax bx c tx c if c 0;+ + = ± >
3. 2ax bx c) (x α)t if a<0,c<0+ + = −
whereα is a real root of ax bx c a(x )(x )2 0+ + = = − α − β
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 12121212
17. Which of the following functions does not appear in the primitive of
2
1
1 x 2x 2+ + + if t is a function of x ?
(a) e
log t 1+ (b)e
log t 2+ (c)1
t 1+ (d) None of these
18.
( )3
2
xdx
7x 10 x− −∫ can be evaluated by substituting for x as
(a) 2
2
5 2tx
t 1
+=
+ (b)
2
2
5 tx
t 2
−=
+ (c)
2
2
2t 5x
3t 1
−=
− (d) None of these
Sol.
17. (d)
Here a = 1 > 0, therefore we make the substitution 2x 2x 2 t x+ + = − . Squaring both
sides of this equality and reducing the similar terms, we get
2 22
2
t 2 t 2t 22x 2tx t 2 x dx dt ;
2(1 t) 2 (1 t)
− + ++ = − ⇒ = =
+ +
2 22 t 2 t 4t 4
1 x 2x 2 1 t2(1 t) 2 (1 t)
− + ++ + + = + − =
+ +
Substituting into the integral, we get
2 2
2 2 2
2(1 t)(t 2t 2) (t 2t 2)l dt dt
(t 4t 4)2 (1 t) (1 t) (t 2)
+ + + + += =
+ + + + +∫ ∫
Now let us expand the obtained proper rational fraction into per tail fractions :
2
2 2
t 2t 2 A B D
(t 1) (t 2) t 1 t 2 (t 2)
+ += +
+ + + + +.
18. (a) In this case a < 0 and c < 0, therefore neither the first, nor the second Euler
substitution is applicable. But the quadratic 7x − 10 − x2 has real roots α 2,β 5= = ,
therefore we use the third Euler substitution :
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 13131313
27x 10 x (x 2)(5 x) (x 2)t− − = − − = −
25 x (x 2)t ;⇒ − = −
2
2
5 2tx
t 1
+⇒ =
+
Section IV (Integer type)
This section contains TEN questions. The answer to each question is a single–digit
integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to
be bubbled.
19. 1
2
2x 2If sin dx
(4x 8x 13)
− + + +
∫1 22x 2
(x 1)tan λ ln(4x 8x 13) c,3
− + = + + + + +
then
the value of − 4 λ must be
Sol. 0003
1
2
2x 2I sin dx
(4x 8x 13)
− +
= + +
∫
1
2 2
2x 2I sin dx
(2x 2) 3
− +
= + +
∫
Put 2x 2 3 tanθ+ =
22dx 3 sec θ dθ∴ =
23 sec θ dθThen, I θ.
2= ∫
{ }3
θ. tan θ ln secθ c2
= − +
2
13 2x 2 2x 2 2x 2. tan ln 1 c
2 3 3 3
− + + +
= − + +
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 14141414
1 22x 2 3(x 1)tan ln(4x 8x 13) c
3 4
− + = + − + + +
3Here, λ
4= −
Then, 4λ 3− =
20. If π/2
8 4
0
14πsin xcos xdx
λ=∫ then the value of ( )λ 4090−
Sol. 0006
π/28 4
0
(7.5. 3. 1)(3. 1) πsin x cos x dx .
12.10. 8. 6. 4. 2 2=∫
(by Wallis’ formula)
7π 14π
2048 4096= =
λ 4096=
⇒ ( )λ 4090− = 4096 – 4090 = 6.
21. If 7 6 5 3 2
2
22
(2x 3x 10x 7x 12x x 1)dx
(x 2)−
+ − − − + +
+∫2
(5π λ)20
= − , then the value of
( )λ 60− must be
Sol. 0004
7 6 5 3 22
22
(2x 3x 10x 7x 12x x 1)Let I dx
(x 2)−
+ − − − + +=
+∫
7 5 32
22
(2x 10x 7x x)dx
(x 2)−
− − +=
+∫
6 22
22
(3x 12x 1)dx
(x 2)−
− ++
+∫
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 15151515
2 42
20
3x (x 4) 10 2
(x 2)
− += +
+∫ (by property)
22 2
20
12 3x (x 2) dx
x 2
= − +
+ ∫
25
3 1
0
3x 1 x2 2x tan
5 2 2
− = − +
12 2 1 π2 4 2 .
5 42
= − +
24 2 π8 2 2.
5 4= − +
2(5π 64)
20= −
λ 64∴ =
⇒ ( )λ 60− = 64 – 60 = 4.
22. If 1
49
0I x (1 x) dx,= −∫ then the value of
12545
I− = must be
Sol. 0005
149
0I x(1 x) dx= −∫∵
149
0(1 x)(1 (1 x)) dx= − − −∫ [By property]
1 149 49 50
0 0(1 x)x dx (x x )dx= − = −∫ ∫
150 51
0
x x
50 51
= −
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1 1 1
50 51 2550= − =
12550
I∴ =
⇒ 1
2545I
− = 2550 – 2545 = 5
23. If area between the curves y = xex and y = xe−x and the line x = 1 is λ sq unit, then
the value of e( λ ) = 2 must be
Sol. 0002
Required area 1
x x
0λ (xe xe ) dx−= −∫
{ }1
x x x x
0x (e e ) (e e )− −= + − +
{ } { }1 1(e e ) (e e ) 0 (1 1)− −= + − − − − −
2
e=
⇒2
128eλ 128e 256e
= × =
⇒ e( λ ) = 2
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24. If area enclosed between the curves y = ln(x + e) and x = ln1
y
and the axis of is x
is λ sq unit, then the value of 2λ6 1292−
Sol. 0004
1y ln(x e) and x ln
y
= + =
x x1e y e
y
−= ⇒ =
0x
1 e 0Required area ln (x e)dx e dx
∞−
−∴ = + +∫ ∫
ex
1 0ln x dx e dx
∞−= +∫ ∫ ( by property)
{ } { }e x
1 0x ln x x e
∞−= − −
{ } { }(e e) (0 1) 0 1= − − − − −
2 sq unit=
λ 2∴ =
2λ 4Then, 6 6 1296= =
⇒ 2λ6 1292− = 4.
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25. The equation of a curve whose slope at any point is thrice its abscissa and which
passes through 2( 1, 3)is 2y λ(x 3)− − = − , then the value of λ must be
Sol. 0003
dy3x
dx=∵
dy 3x dx⇒ =
23xint ergrating, we get y c
2= +
Since, it passes through (− 1, − 3)
then 3
3 c2
− = +
9c
2∴ = −
223x 9
y 2y 3(x 3)2 2
∴ = − ⇒ = −
λ 3∴ =
26. If the solution of the differential equation 2 3dysec y 2x tan y x
dx+ = is
2 tan y = 22 xλ(x 1) ce ,−− + c is arbitrary constant, then the numerical value of λ must be
Sol. 0001
2 3dysec y 2x tan y x
dx+ =∵ …..(i)
let tan y v=
2 dy dvsec y
dx dx∴ =
Then from Eq. (i),
3dv2vx x
dx+ =
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2x
e
IF 2xdx e∴ = =∫
Solution is∴
3v. (IF) x . (IF)dx c= +∫
2 2x 3 xtan y . e x .e dx c⇒ = +∫
2 dtPut x t xdx
2= ⇒ =
2x t1tan y . e te dt c
2∴ = +∫
t t1( te e ) c
2= − +
2x1tan y (t 1) ce
2
−⇒ = − +
22 x1or tan y (x 1) ce
2
−= − +
22 xor 2 tan y (x 1) 2ce−= − +
22 x(x 1) ce (Replacing 2c by c)−= − +
Hence, λ 1=
27. A particle moves in a straight line with a velocity given by dx
x 1dt
= + (x is the
distance travelled). If the time taken by a particle to traverse a distance of 99 m is
λ , then the value of ( )e
1020λ log 35− must be
Sol. 0005
dx dxx 1 dt
dt x 1= + ⇒ =
+
ln(x 1) t c⇒ + = +
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Putting t 0, x 0, we get c 0= = =
t ln(x 1)⇒ = +
efor x 99, t ln 100 2 log 10= = =
10 e 1020λ log e 20 2 log 10 log e= × ×
40=
⇒ ( )e
1020λ log 35− = 40 – 35 = 5.
28. The differential equation whose solution represents the family
3x 5xy ae be= + is 2
2
d y dy8 λy 0
dx dx− + = , then the value of λ 8− must be
Sol. 0007
3x 5x 3x 5xy ae be or ae be y 0= + + − =∵ … (i)
3x 5xdy3ae 5be
dx∴ = +
3x 5x dyor 3ae 5be 0
dx+ − = …. (ii)
Again differentiating both sides w.r.t. x, then
23x 5x
2
d y9ae 25be 0
dx+ − = …. (iii)
From Eqns. (i), (ii) and (iii), we get
2
2
1 1 y
dy3 5 0
dx
d y9 25
dx
=
Expanding w.r.t. R 1, then
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2 2
2 2
d y dy d y dy5 25 1 3 9 y (75 45) 0
dx dx dx dx
− − − + − =
⇒2
2
d y dy2 16 30y 0
dx dx− + =
∴ 2
2
d y dy8 15y 0
dx dx− + =
∴ λ 15=