PAPER-A Integral calculus Sol - PioneerMathematics.Com

L.K.Gupta (Mathematic Classes) www.pioneermathematics.com. MOBILE: 9815527721,4617721

PAPER–A

IIT–JEE (2011) (Integral Calculus solutions) “TOWARDS IIT– IIT– JEE IS NOT A JOURNEY, IT’S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE” TIME: 60 MINS

MAX. MARKS: 84

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.

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

1


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 e1/n e2/n e(n−1)/n  1. lim  + + + ...... +  equals n →∞ n n n n   (a) 0

(b)

1

(c) e–1

(d) none of these

Sol:  1 + e1/n + (e1/n )2 + .......(e1/n )n−1  1.[(e1/n )n − 1] = lim lim   n →∞ n(e1/n − 1) n →∞ n   1 = (e − 1)× 1 = e − 1 n →∞  e1/n − 1   1/ n   

= (e − 1)lim

2.

c  c 2 1  sin2 t lim  ∫ e dt − ∫ esin t dt  is equal to(where c is a constant) x+y  x →0 x  y 

(a) esin

2

y

(b) sin2 yesin

2

y

(c) 0

(d) none of these

Sol: x+y x +y c c c   2 1  sin2 t 1  sin2 sin2 t lim  ∫ e dt − ∫ e dt  = lim  ∫ e dt + ∫ esin t dt  = lim   x →0 x →0 x  x →0 x  x+y c y  y 

∫

2

esin t dt

y

x

2

2 esin ( x + y ) − 0 = esin y = lim x →0 1

3. Area enclosed by the curve y = f (x) defined parametrically as x =

1 − t2 2t ,y= is 2 1 +t 1 + t2

equal to


2


(a) π sq. units

(b) π / 2 sq.units (c)

3π sq.units 4

(d)

3π sq.units 2

Sol: Clearly t can be any real number 1 − tan2 θ Let t = tan θ ⇒ x = 1 + tan2 θ ⇒ x = cos 2 θ, and

y=

2 tan θ = sin 2 θ 1 + tan2 θ

⇒ x2 + y 2 = 1

Thus, required area = π sq.units.

15

4. The value of

∫ sgn( {x} )dx, where{. } denotes the fractional part function, is

−1

(a) 8

(b) 16

(c)

24

(d) 0

Sol: 15

16

∫ sgn({x})dx = ∫ sgn({x − 1})dx

−1

(by property)

0

16

1

1

1

0

0

0

0

= ∫ sgn({x})dx = 16∫ sgn({x})dx = 16∫ sgn(x)dx = 16∫ 1.dx = 16

x2

5. The slope of the tangent to the curve y= ∫ cos−1 t 2dt at x = x

 4 8 3 (a)  −  π 2 4 

 4 8 1 (b)  −  π 3 4 

 4 8 1 (c)  −  π 3 4 

1 4 2 (d) none of these

Sol:


3

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∵ y = ∫ cos −1 t 2dt x

∴

dy = cos−1(x 4 ).2x − cos−1(x 2 ).1 dx

⇒

dy  1  1 2 | 1 = cos −1   4 − cos −1   dx x= 4 2 2 2  2

=

π 3/4 π .2 − 3 4

6.

 4 8 1 =  −  π 3 4 

(2x12 + 5x 9 ) ∫ (x5 + x3 + 1)3 dx is equal to.

(a)

x 2 + 2x +c (x5 + x3 + 1)2

(b)

(c)In | x5 + x3 + 1| + (2x7 + 5x 4 ) + c

x10 +c 2(x5 + 3x + 1)2

(d) none of the above

Sol: 2 5  x3 + x 6  2x + 5x  dx Let I = ∫ 5 3 dx = ∫  3 3 (x + x + 1) 1 1  1 + 2 + 5  x x   12

Put

1+

9

1 1 + =t x 2 x5

Then I = − ∫

 2 5 ∴  − 3 − 6  dx = dt  x x 

dt 1 1 x10 = = + c = + c +c 2 5 3 2 2(x + x + 1) t 3 2t 2 1 1  2 1 + 2 + 5  x x  


4


7. If ∫ f(x)cos xdx =

1 2 {f(x)} + c then f(x) is 2

(a) x + c

(b) sin x + c

(c) cos x + c

(d) c

Sol: 1 f(x)cos x = .2f(x)f '(x) 2 Then f '(x) = cos x ∴ f(x) = sin x + c

π/2

8. If I1 =

π/2

∫ cos(sin x)dx; I

2

0

=

π/2

∫ sin(cos x)dx and I

3

0

(a) I1 > I2 > I3

=

∫ cos xdx, then. 0

(b) I2 > I3 > I1

(c) I3 > I1 > I2

(d) I1 > I3 > I2

Sol: ∵ x > 0 ∴sin x < x ⇒ cos(sin x) > cos x ....(1) Also 0 < x <

π 2

∴ 1 > cos x > 0 , sin(cosx)
cos (sin x) > cos x >sin (cos x) Or

π/2

(2)

π/2

∫ cos(sin x)dx > ∫ cos x dx > ∫ sin(cos x)dx 0

0

0

⇒ I1 > I3 > I2

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). f (x,y) =

x−y x2 + y 2


5

L.K.Gupta (Mathematic Classes) www.pioneermathematics.com. MOBILE: 9815527721,4617721 1 3

(b) f (x,y) = x y

−

2 3

tan −1

x y

(c) f (x,y) = x (ln x 2 + y 2 − ln y) + yex/y

 2x 2 + y 2  x + 2y (d) f (x,y) = x ln − ln(x + y) + y 2 tan x 3x − y  

Sol: (a). f (λx,λy) =

λ(x − y) = λ−1 f (x,y) 2 2 2 λ (x + y )

⇒ homogeneous of degree (−1).

(b). f (λx, λy) = (λx)1/3 (λ y)−2/3 tan −1 = λ−1/3 x1/3 y −2/3 tan −1 =λ

−

1 3

x y

x y

f (x,y)

⇒ homogeneous

(

)

(c). f (λx, λy) = λx ln λ2 (x2 + y 2 ) − ln λy + λ yex/y   λ (x2 + y 2 )   = λx ln    + λyex/y  λy    

(

)

= λ  x ln x2 + y 2 − ln y + yex/y    = λ f (x,y) ⇒ homogeneous  2λ2 x2 + λ2 y 2  x + 2y (d). f (λx,λy) = λx ln + λ2 x2 tan  3x − y  λ x λ(x + y) 


6


 2x 2 + y 2  2 2 x + 2y = λx ln + λ x tan  3x − y  x(x + 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 ⇒ 9 = 3 + 4 + c1 ⇒ c1 = 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). f (x,y) = ey/x + tan

y is homogeneous of degree zero. x

y y2 y (b). x ln dx + sin −1 dy = 0 is homogeneous differential equation. x x x (c). f (x,y) = x2 + sin x cos y is not homogeneous. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

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(d). (x 2 + y 2 )dx − (xy 2 − y 3 )dy = 0 is a homogeneous differential equation.

Sol: (a), (b), (c)

12.

xdx + ydy 1 − x2 − y 2 The solution of = is xdy − ydx x2 + y 2

(a).

x2 + y 2 = sin tan −1(y / x) + C (b).

x2 + y 2 = cos (tan −1 y / x) + C

(c).

x2 + y 2 = (tan (sin −1 y / x) + C)

(d). y = x tan c + sin −1 x 2 + y 2

{

}

{

(

}

)

Sol: (a), (b) The D. E can be re-written as x dx + y dy

(

1 − x2 + y 2

)

=

x dy − y dx x2 + y 2

Since d tan −1 ( y / x) =

x dy − ydx ,and d x2 + y 2 2 2 x +y

(

)

= 2(x dx + y dy), 1 d x2 + y 2 x dy − ydx 2 = x2 + y 2 x2 + y 2 1 − x 2 + y 2

(

∴ we have

)

(

{

)

}

= d tan −1 (y / x)

Put x 2 + y 2 = t 2 in t he L.H.S and get

{

= d tan −1 (y / x)

t dt t 1− t2

}

Integrating both sides, we get sin −1 t = tan −1 (y / x) + c

i.e.,sin −1

(x

2

)

+ y 2 = tan −1 (y / x) + c


8

L.K.Gupta (Mathematic Classes) www.pioneermathematics.com. MOBILE: 9815527721,4617721 x

Let f(x) =

13.

∫ 1

3t dt, where x > 0, then 1 + t2

(a). for0 < α < β,f(α)< f(β)

(b). for0 < α < β,f(α) > f(β)

(c). f(x) + π / 4 < tan −1 x, ∀ x ≥ 1

(d). f(x) + π / 4 > tan −1 x, ∀ x ≥ 1

Sol. 3x 3x 1 f '(x) = > 0 ∀ x > 0 ⇒ f '(x) = > , ∀ x ≥1 2 2 1+x 1 + x 1 + x2 x

x

1

1

⇒ ∫ f '(x) dx > ∫

1 dx 1 + x2

⇒ f(x) > tan −1 x − tan −1 1 ⇒ f(x) + π / 4 > tan −1 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

f (x)dx =

b−a (f (a) + f(b)). For 2

b

c

b

a

a

c

more accurate result for c ∈ (a,b) , we can use ∫ f(x)dx = ∫ f (x)dx + ∫ f (x)dx = F(c) so that for c = π/2

a+b , we get 2

14.

∫

(a)

π (1 + 2) 8

0

∫

b

a

f (x)dx =

b−a (f (a) + f(b) + 2f(c)) 4

sin xdx is equal to π (b) (1 + 2) 4

(c)

π 8 2

(d)

π 4 2


9


∫ 15. If lim

x

a

x →a

x −a f (x)dx −  ( f (x) + f (a)) 2   = 0, then f (x) is maximum degree (x − a)3

(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 f (b) − f (a) b−a

(a)

(b)

2(f (b) − f (a)) b−a

(c)

2f(b) − f(a) 2b − a

(d) 0

Sol. π  − 0  π/2 2   sin 0 + sin π + 2 sin π  14. (a) ∫ sin x dx =  0 4  2 2  =

π 1+ 2 8

(

)

x −a f (x) dx −  2 ( f (x) + f (a)) ∫a   =0 15. (d). lim x →a (x − a)3 x

lim

h →0

∫

a +h

a

h f (x) dx − (f (a + h) + f (a)) 2 =0 h3

1 h f (a + h) − [f (a) + f (a + h)]− ( f '(a + h)) 2 2 =0 ⇒ lim 2 h→0 3h [ Using L’ Hospital’ s Rule] 1 1 h f (a + h) − f (a) − f '(a + h) 2 2 ⇒ lim 2 =0 2 h →0 3h 1 1 h f '(a + h) − f '(a + h) − f ''(a + h) 2 2 ⇒ lim 2 =0 h→0 6h [ Using L’ Hospital’ s Rule] PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

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⇒ lim

h →0

− f ''(a + h) =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) F(c) = =

c−a b−c ( f (a) + f (c)) + ( f (b) + f (c)) 2 2

b−a c−a b−c f (c) + f (a) + f (b) 2 2 2

⇒ F'(c) = =

b−a 1 1 f '(c) + f(a) − f (b) 2 2 2

1 [(b − a)f '(c) + f(a) − f (b)] 2

1 F''(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

∫ R(x,

ax2 + bx + c) dx are calculated with the aid of one of the

three Euler substitutions 1.

ax2 + bx + c = t ± x a if a > 0;

2. ax2 + bx + c = tx ± c if c > 0; 3. ax 2 + bx + c) = (x − α)t if a<0,c<0

where α is a real root of ax 2 + bx + c = 0 = a(x − α)(x − β) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

11


17.

Which of the following functions does not appear in the primitive of 1

if t is a function of x ?

2

1 + x + 2x + 2 (a) log e t + 1

18.

∫

(b) log e t + 2 xdx

(

7x − 10 − x

5 + 2t 2 (a) x = 2 t +1

2

3

)

(c)

1 t +1

(d) None of these

can be evaluated by substituting for x as

5 − t2 (b) x = 2 t +2

2t 2 − 5 (c) x = 2 3t − 1

(d) None of these

Sol. 17. (d) Here a = 1 > 0, therefore we make the substitution

x2 + 2x + 2 = t − x . Squaring both

sides of this equality and reducing the similar terms, we get t2 − 2 t 2 + 2t + 2 2x + 2tx = t − 2 ⇒ x = dx = dt ; 2(1 + t) 2 (1 + t)2 2

t2 − 2 t 2 + 4t + 4 1 + x + 2x + 2 = 1 + t − = 2(1 + t) 2 (1 + t) 2

Substituting into the integral, we get 2(1 + t)(t 2 + 2t + 2) (t 2 + 2t + 2) l =∫ 2 dt = ∫ dt (t + 4t + 4)2 (1 + t)2 (1 + t) (t + 2)2 Now let us expand the obtained proper rational fraction into per tail fractions : t 2 + 2t + 2 A B D = + . 2 (t + 1) (t + 2) t + 1 t + 2 (t + 2)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 : PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

12


7x − 10 − x2 = (x − 2)(5 − x) = (x − 2)t

⇒ 5 − x = (x − 2)t 2 ; 5 + 2t 2 ⇒x= 2 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.

  2x + 2  2x + 2  If ∫ sin −1   dx = (x + 1)tan −1  + λ ln(4x 2 + 8x + 13) + c, then   (4x 2 + 8x + 13)   3   

the value of − 4 λ must be

Sol. 0003   2x + 2 I = ∫ sin −1   dx  (4x2 + 8x + 13)     2x + 2 I = ∫ sin −1   (2x + 2)2 + 32 

  dx  

Put 2x + 2 = 3 tanθ 2dx = 3 sec2 θ dθ

∴

Then, I = ∫ θ. =

3 sec2 θ dθ 2

3 {θ. tan θ − ln secθ} + c 2

3 = 2

2  2x + 2    2x + 2  −1  2x + 2    . tan  3  − ln 1 +  3  3       

 +c 


13


 2x + 2  3 = (x + 1)tan −1  − ln(4x 2 + 8x + 13) + c   3  4 3 4

Here, λ = −

Then, − 4λ = 3

∫

If

20.

π/2

0

sin8 xcos4 xdx =

14π then the value of ( λ − 4090) λ

Sol. 0006

∫

π/2

0

sin8 x cos 4 x dx =

(7.5. 3. 1)(3. 1) π . 12.10. 8. 6. 4. 2 2 (by Wallis’ formula)

=

7π 14π = 2048 4096

λ = 4096 ⇒ ( λ − 4090) = 4096 – 4090 = 6.

∫

If

21.

−

(2x7 + 3x 6 − 10x5 − 7x3 − 12x 2 + x + 1) 2 dx = (5π − λ) , then the value of 2 2 (x + 2) 20

2

( λ − 60) must be Sol. 0004 Let I = ∫

2

− 2

=∫

2

+∫

2

− 2

−

(2x7 + 3x 6 − 10x5 − 7x3 − 12x 2 + x + 1) dx (x 2 + 2)

(2x7 − 10x5 − 7x3 + x) dx (x 2 + 2)

(3x 6 − 12x 2 + 1) dx 2 (x 2 + 2)


14


= 0 + 2∫

2

0

= 2∫

0

2

3x2 (x 4 − 4) + 1 (x2 + 2)

(by property)

 2 2 1   3x (x − 2) + 2  dx x +2  

 3x5 1  x  =2 − 2x3 + tan −1   2  2  0  5

2

1 π   12 2 = 2 −4 2 + .  5 4  2  =

24 2 π − 8 2 + 2. 5 4

=

2 (5π − 64) 20

∴ λ = 64 ⇒ ( λ − 60) = 64 – 60 = 4.

1

If I = ∫ x (1 − x)49 dx, then the value of

22.

0

1 − 2545 = must be I

Sol. 0005 1

∵ I = ∫ x(1 − x)49 dx 0

1

= ∫ (1 − x)(1 − (1 − x))49 dx 0

1

1

0

0

[By property]

= ∫ (1 − x)x 49 dx = ∫ (x 49 − x50 )dx 1

 x50 x51  = − 50 51 0 


15


=

1 1 1 − = 50 51 2550

∴

1 = 2550 I

⇒

1 − 2545 = 2550 – 2545 = 5 I

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

1

Required area λ = ∫ (xex − xe − x ) dx 0

{ {

1

}

= x (ex + e− x ) −(ex + e− x )

0

}

= (e + e−1 ) − (e − e−1 ) − {0 − (1 − 1)} =

2 e

⇒ 128eλ = 128e ×

2 = 256 e

⇒ e( λ ) = 2


16


1 If area enclosed between the curves y = ln(x + e) and x = ln   and the axis of is x y

24.

is λ sq unit, then the value of 62λ − 1292

Sol. 0004 1 y = ln(x + e) and x = ln   y 1 = ex ⇒ y = e − x y

∴ Required area = ∫

0

1−e

e

= ∫ ln x dx + 1

e

∫

∞

0

ln (x + e)dx +

e− x dx

∫

∞

0

e− x dx

( by property)

∞

{ }

= {x ln x − x}1 − e − x

0

= {(e − e) − (0 − 1)} − {0 − 1} = 2 sq unit ∴λ =2 Then, 62λ = 64 = 1296

⇒ 62λ − 1292 = 4. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

17


25.

The equation of a curve whose slope at any point is thrice its abscissa and which

passes through ( −1, − 3)is 2y = λ(x 2 − 3) , then the value of λ must be

Sol. 0003 ∵

dy = 3x dx

⇒

dy = 3x dx

3x 2 int ergrating, we get y = +c 2 Since, it passes through (− 1, − 3) then −3 = ∴c=−

3 +c 2

9 2

3x 2 9 ∴y = − ⇒ 2y = 3(x 2 − 3) 2 2 ∴ λ =3

26.

If the solution of the differential equation sec2 y

dy + 2x tan y = x3 is dx

2

2 tan y = λ(x2 − 1) + ce− x , c is arbitrary constant, then the numerical value of λ must be

Sol. 0001 ∵ sec2 y

dy + 2x tan y = x3 dx

…..(i)

let tan y = v ∴ sec2 y

dy dv = dx dx

Then from Eq. (i), dv + 2vx = x3 dx PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

18


∴ IF = ∫ 2xdx = ex

2

e

∴ Solution is v . (IF) = ∫ x3 . (IF)dx + c ⇒ tan y . ex = ∫ x3 .ex dx + c 2

2

Put x 2 = t ⇒ xdx = 2

∴ tan y . ex = =

1 2

∫

dt 2

tet dt + c

1 ( tet − et ) + c 2

⇒ tan y =

2 1 (t − 1) + ce− x 2

or tan y =

2 1 2 (x − 1) + ce− x 2

or 2 tan y = (x 2 − 1) + 2ce− x

2

2

= (x2 − 1) + ce− x (Replacing 2c by c) Hence, λ = 1

27.

A particle moves in a straight line with a velocity given by

dx = x + 1 (x is the dt

distance travelled). If the time taken by a particle to traverse a distance of 99 m is

(

)

e λ , then the value of 20λ log10 − 35 must be

Sol. 0005 dx dx = x +1⇒ = dt dt x+1 ⇒ ln(x + 1) = t + c PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

19


Putting t = 0, x = 0, we get c = 0 ⇒ t = ln(x + 1)

for x = 99, t = ln 100 = 2 log e 10 20λ log10 e = 20 × 2 log e 10 × log10 e = 40

(

)

e ⇒ 20λ log10 − 35 = 40 – 35 = 5.

The differential equation whose solution represents the family

28.

d2 y dy y = ae + be is 2 − 8 + λy = 0 , then the value of λ − 8 must be dx dx 3x

5x

Sol. 0007 ∵ y = ae3x + be5x or ae3x + be5x − y = 0 ∴

… (i)

dy = 3ae3x + 5be5x dx

or 3ae3x + 5be5x −

dy =0 dx

…. (ii)

Again differentiating both sides w.r.t. x, then d2 y 9ae + 25be − 2 = 0 dx 3x

5x

…. (iii)

From Eqns. (i), (ii) and (iii), we get 1

1

y

dy =0 dx d2 y 9 25 dx2

3

5

Expanding w.r.t. R 1, then


20


 d2 y  d2 y dy  dy   5 2 − 25  − 1  3 2 − 9  + y (75 − 45) = 0 dx  dx   dx  dx ⇒2

d2 y dy − 16 + 30y = 0 dx2 dx

d2 y dy + 15y = 0 ∴ 2 −8 dx dx ∴ λ = 15


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