Question 1

Let f(x) = ( 16x + 24 x^2 + 2x - 15 ) dx. If f(4) = 14 _e(3) and f(7) = _e(2^ 3^ ), , N , then + is equal to:

Accepted Answer

The given integral is f(x) = ( 16x + 24 x^2 + 2x - 15 ) dx. Factorizing the denominator, we get x^2 + 2x - 15 = (x + 5)(x - 3). Using partial fractions, we can write: 16x + 24 (x + 5)(x - 3) = A x + 5 + B x - 3 16x + 24 = A(x - 3) + B(x + 5) Substituting x = 3, we get 72 = 8B B = 9. Substituting x = -5, we get -56 = -8A A = 7. Therefore, the integral becomes: f(x) = ( 7 x + 5 + 9 x - 3 ) dx f(x) = 7 _e|x + 5| + 9 _e|x - 3| + C Given f(4) = 14 _e(3), we substitute x = 4: f(4) = 7 _e(9) + 9 _e(1)

Question 2

Let f(x)= d x x^ ( 2 3 ) +2 x^ ( 1 2 ) be such that f(0)=-26+24 _ e (2). If f(1)= a + b _ e (3), where a , b Z , then a + b is equal to :

Accepted Answer

Let u = x^ 1/6 , so x = u^6 and dx = 6u^5 du. Then x^ 2/3 = u^4, x^ 1/2 = u^3. f(x) = 6u^5 du u^4 + 2u^3 = 6u^2 du u+2 . By polynomial division: u^2 u+2 = u - 2 + 4 u+2 . f(x) = 6 ( u^2 2 - 2u + 4 |u+2| ) + C = 3u^2 - 12u + 24 (u+2) + C. Substituting back: f(x) = 3x^ 1/3 - 12x^ 1/6 + 24 (x^ 1/6 +2) + C. From f(0) = -26 + 24 (2): 24 (2) + C = -26 + 24 (2), so C = -26. At x = 1: f(1) = 3 - 12 + 24 (3) - 26 = -35 + 24 (3). Thus a = -35, b = 24, and a + b = -11.

Question 3

Let f(t)= ( 1- ( _ e t ) 1- ( _ e t ) ) d t, t>1. If f (e^ / 2 )=-e^ / 2 and f (e^ / 4 )= e^ / 4 , then equals

Accepted Answer

Substituting u = t with dt = e^u du: f(t) = 1 - u 1 - u e^u du Express as: 1 - u 1 - u = 1 2 ^2(u/2) - (u/2) Using integration by parts and standard techniques: f(t) = e^ t ( t 2 ) + C = t ( t 2 ) + C From f(e^ /2 ) = -e^ /2 : e^ /2 ( /4) + C = -e^ /2 → e^ /2 + C = -e^ /2 → C = -2e^ /2 For f(e^ /4 ): f(e^ /4 ) = e^ /4 ( /8) - 2e^ /2 With ( /8) = 1 + 2 : = -1 - 2

Question 4

Let I (x)= 3 d x (4 x+6) ( 4 x^ 2 +8 x+3 ) and I (0)= 3 4 +20. If I ( 1 2 )= a 2 b + c , where a, b, c N , gcd (a, b)=1, then a+b+c is equal to

Accepted Answer

4x^2+8x+3 = (2x+1)(2x+3) and 4x+6 = 2(2x+3). I(x) = 3 4 du u^ 3/2 u-2 where u = 2x+3. Substituting u = 2 ^2 : I(x) = 3 4 + C = 3 4 2x+1 2x+3 + C. I(0) = 3 4 1 3 + C = 3 4 + C = 3 4 + 20 C = 20. I ( 1 2 ) = 3 4 2 4 + 20 = 3 4 2 + 20 = 3 2 8 + 20. a = 3, b = 8, c = 20, (3,8) = 1. a+b+c = 31.

Question 5

Let f(x)= (2-x^ 2 ) e ^ x ( 1+x )(1-x)^ 3 / 2 ~d x. If f(0)=0, then f ( 1 2 ) is equal to:

Accepted Answer

Write 2 - x^2 = (1-x)(1+x) + 1, so the integrand splits as e^x ( 1+x 1-x + 1 1+x (1-x)^ 3/2 ). Let g(x) = 1+x 1-x . Then g'(x) = 1 1+x (1-x)^ 3/2 . Using e^x(g + g')dx = e^x g + C: f(x) = e^x 1+x 1-x + C. f(0) = 1 + C = 0 C = -1. f(1/2) = e^ 1/2 3/2 1/2 - 1 = e 3 - 1 = 3e - 1.

Question 6

If ( x)^ -11 2 ( x)^ -5 2 d x= - p_ 1 q_ 1 ( x)^ 9 2 - p_ 2 q_ 2 ( x)^ 5 2 - p_ 3 q_ 3 ( x)^ 1 2 + p_ 4 q_ 4 ( x)^ -3 2 + C , where p_ i and q_ i are positive integers with gcd (p_ i , q_ i )=1 for i=

Accepted Answer

^ -11/2 x ^ -5/2 x\,dx. Since m+n = -8 (negative even), substitute t = x, dx = -dt 1+t^2 . Integral becomes - (1+t^2)^3 t^ 5/2 dt = - (t^ -5/2 +3t^ -1/2 +3t^ 3/2 +t^ 7/2 )dt. = 2 3 t^ -3/2 - 6t^ 1/2 - 6 5 t^ 5/2 - 2 9 t^ 9/2 + C. Comparing: p_1 q_1 = 2 9 , p_2 q_2 = 6 5 , p_3 q_3 = 6 1 , p_4 q_4 = 2 3 . 15p_1p_2p_3p_4 q_1q_2q_3q_4 = 15 2 6 6 2 9 5 1 3 = 2160 135 = 16.

Indefinite Integration — Previous Year Questions & Solutions