PYQs / GATE CS / 2015 / Set 3 / Q54 GATE CS 2015 Set 3 — Question 54 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Integration Calculus Engineering Mathematics
Engineering Mathematics → Calculus → Integration
Last updated 5 September 2026
Question If for non-zero
x x x ,
a f ( x ) + b f ( 1 x ) = 1 x − 25 af(x) + bf(\frac{1}{x}) = \frac{1}{x} - 25 a f ( x ) + b f ( x 1 ) = x 1 − 25 where
a ≠ b a \neq b a = b then
∫ 1 2 f ( x ) d x \int_1^2 f(x)dx ∫ 1 2 f ( x ) d x is
Correct answer (A) (1)/(a² - b²) [ a(ln 2 - 25) + (47b)/(2) ]
Solution Given:
(1)
a f ( x ) + b f ( 1 / x ) = 1 x − 25 a f(x) + b f(1/x) = \frac{1}{x} - 25 a f ( x ) + b f ( 1/ x ) = x 1 − 25 Replace
x x x with
1 / x 1/x 1/ x :
(2)
a f ( 1 / x ) + b f ( x ) = x − 25 a f(1/x) + b f(x) = x - 25 a f ( 1/ x ) + b f ( x ) = x − 25 Multiply (1) by
a a a and (2) by
b b b :
a 2 f ( x ) + a b f ( 1 / x ) = a x − 25 a a^2 f(x) + ab f(1/x) = \frac{a}{x} - 25a a 2 f ( x ) + ab f ( 1/ x ) = x a − 25 a b 2 f ( x ) + a b f ( 1 / x ) = b x − 25 b b^2 f(x) + ab f(1/x) = bx - 25b b 2 f ( x ) + ab f ( 1/ x ) = b x − 25 b Subtract the second from the first:
( a 2 − b 2 ) f ( x ) = a x − b x − 25 a + 25 b (a^2 - b^2) f(x) = \frac{a}{x} - bx - 25a + 25b ( a 2 − b 2 ) f ( x ) = x a − b x − 25 a + 25 b f ( x ) = 1 a 2 − b 2 [ a x − b x − 25 ( a − b ) ] f(x) = \frac{1}{a^2 - b^2} \left[ \frac{a}{x} - bx - 25(a - b) \right] f ( x ) = a 2 − b 2 1 [ x a − b x − 25 ( a − b ) ] Integrate from 1 to 2:
∫ 1 2 f ( x ) d x = 1 a 2 − b 2 [ ∫ 1 2 a x d x − ∫ 1 2 b x d x − ∫ 1 2 25 ( a − b ) d x ] \int_1^2 f(x) dx = \frac{1}{a^2 - b^2} \left[ \int_1^2 \frac{a}{x} dx - \int_1^2 bx dx - \int_1^2 25(a-b) dx \right] ∫ 1 2 f ( x ) d x = a 2 − b 2 1 [ ∫ 1 2 x a d x − ∫ 1 2 b x d x − ∫ 1 2 25 ( a − b ) d x ] = 1 a 2 − b 2 [ a [ ln x ] 1 2 − b [ x 2 2 ] 1 2 − 25 ( a − b ) [ x ] 1 2 ] = \frac{1}{a^2 - b^2} \left[ a[\ln x]_1^2 - b[\frac{x^2}{2}]_1^2 - 25(a-b)[x]_1^2 \right] = a 2 − b 2 1 [ a [ ln x ] 1 2 − b [ 2 x 2 ] 1 2 − 25 ( a − b ) [ x ] 1 2 ] = 1 a 2 − b 2 [ a ( ln 2 ) − b ( 2 − 0.5 ) − 25 ( a − b ) ( 1 ) ] = \frac{1}{a^2 - b^2} \left[ a(\ln 2) - b(2 - 0.5) - 25(a-b)(1) \right] = a 2 − b 2 1 [ a ( ln 2 ) − b ( 2 − 0.5 ) − 25 ( a − b ) ( 1 ) ] = 1 a 2 − b 2 [ a ln 2 − 1.5 b − 25 a + 25 b ] = \frac{1}{a^2 - b^2} \left[ a \ln 2 - 1.5b - 25a + 25b \right] = a 2 − b 2 1 [ a ln 2 − 1.5 b − 25 a + 25 b ] = 1 a 2 − b 2 [ a ( ln 2 − 25 ) + 23.5 b ] = \frac{1}{a^2 - b^2} \left[ a(\ln 2 - 25) + 23.5b \right] = a 2 − b 2 1 [ a ( ln 2 − 25 ) + 23.5 b ] = 1 a 2 − b 2 [ a ( ln 2 − 25 ) + 47 b 2 ] = \frac{1}{a^2 - b^2} \left[ a(\ln 2 - 25) + \frac{47b}{2} \right] = a 2 − b 2 1 [ a ( ln 2 − 25 ) + 2 47 b ] Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer (A) (1)/(a² - b²) [ a(ln 2 - 25) + (47b)/(2) ]
Solution Given:
(1)
a f ( x ) + b f ( 1 / x ) = 1 x − 25 a f(x) + b f(1/x) = \frac{1}{x} - 25 a f ( x ) + b f ( 1/ x ) = x 1 − 25 Replace
x x x with
1 / x 1/x 1/ x :
(2)
a f ( 1 / x ) + b f ( x ) = x − 25 a f(1/x) + b f(x) = x - 25 a f ( 1/ x ) + b f ( x ) = x − 25 Multiply (1) by
a a a and (2) by
b b b :
a 2 f ( x ) + a b f ( 1 / x ) = a x − 25 a a^2 f(x) + ab f(1/x) = \frac{a}{x} - 25a a 2 f ( x ) + ab f ( 1/ x ) = x a − 25 a b 2 f ( x ) + a b f ( 1 / x ) = b x − 25 b b^2 f(x) + ab f(1/x) = bx - 25b b 2 f ( x ) + ab f ( 1/ x ) = b x − 25 b Subtract the second from the first:
( a 2 − b 2 ) f ( x ) = a x − b x − 25 a + 25 b (a^2 - b^2) f(x) = \frac{a}{x} - bx - 25a + 25b ( a 2 − b 2 ) f ( x ) = x a − b x − 25 a + 25 b f ( x ) = 1 a 2 − b 2 [ a x − b x − 25 ( a − b ) ] f(x) = \frac{1}{a^2 - b^2} \left[ \frac{a}{x} - bx - 25(a - b) \right] f ( x ) = a 2 − b 2 1 [ x a − b x − 25 ( a − b ) ] Integrate from 1 to 2:
∫ 1 2 f ( x ) d x = 1 a 2 − b 2 [ ∫ 1 2 a x d x − ∫ 1 2 b x d x − ∫ 1 2 25 ( a − b ) d x ] \int_1^2 f(x) dx = \frac{1}{a^2 - b^2} \left[ \int_1^2 \frac{a}{x} dx - \int_1^2 bx dx - \int_1^2 25(a-b) dx \right] ∫ 1 2 f ( x ) d x = a 2 − b 2 1 [ ∫ 1 2 x a d x − ∫ 1 2 b x d x − ∫ 1 2 25 ( a − b ) d x ] = 1 a 2 − b 2 [ a [ ln x ] 1 2 − b [ x 2 2 ] 1 2 − 25 ( a − b ) [ x ] 1 2 ] = \frac{1}{a^2 - b^2} \left[ a[\ln x]_1^2 - b[\frac{x^2}{2}]_1^2 - 25(a-b)[x]_1^2 \right] = a 2 − b 2 1 [ a [ ln x ] 1 2 − b [ 2 x 2 ] 1 2 − 25 ( a − b ) [ x ] 1 2 ] = 1 a 2 − b 2 [ a ( ln 2 ) − b ( 2 − 0.5 ) − 25 ( a − b ) ( 1 ) ] = \frac{1}{a^2 - b^2} \left[ a(\ln 2) - b(2 - 0.5) - 25(a-b)(1) \right] = a 2 − b 2 1 [ a ( ln 2 ) − b ( 2 − 0.5 ) − 25 ( a − b ) ( 1 ) ] = 1 a 2 − b 2 [ a ln 2 − 1.5 b − 25 a + 25 b ] = \frac{1}{a^2 - b^2} \left[ a \ln 2 - 1.5b - 25a + 25b \right] = a 2 − b 2 1 [ a ln 2 − 1.5 b − 25 a + 25 b ] = 1 a 2 − b 2 [ a ( ln 2 − 25 ) + 23.5 b ] = \frac{1}{a^2 - b^2} \left[ a(\ln 2 - 25) + 23.5b \right] = a 2 − b 2 1 [ a ( ln 2 − 25 ) + 23.5 b ] = 1 a 2 − b 2 [ a ( ln 2 − 25 ) + 47 b 2 ] = \frac{1}{a^2 - b^2} \left[ a(\ln 2 - 25) + \frac{47b}{2} \right] = a 2 − b 2 1 [ a ( ln 2 − 25 ) + 2 47 b ] Understand the concept, then try another question Revisit Engineering Mathematics with concept notes, common mistakes and an original worked example before your next attempt.
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