IIT JEE 1982 Maths - 'Adapted' - Subjective to Multi-Correct Q18

Calculus Level 2

If the following is an identity for some function f ( sin ⁡ x ) f(\sin x) defined on 0 < x < π 0<x<\pi , ∫ 0 π x f ( sin ⁡ x ) d x = a b π ∫ 0 π f ( sin ⁡ x ) d x , \large \int_0^{\pi}xf(\sin x) \ dx=\frac ab \pi \int_0^{\pi}f(\sin x) \, dx,

where a a and b b are integers, then the product a b ab can be

  • A: 2
  • B: − - 2
  • C: − - 4
  • D: 4

Enter your answer as a 4-digit string of 1s and 9s – 1 for correct option, 9 for wrong. For example: if statements A and B are correct, and C and D are incorrect, enter 1199. None, one or all may also be correct.

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The answer is 1999.

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

Chew-Seong Cheong
Dec 26, 2016

I = ∫ 0 π x f ( sin ⁡ x ) d x Using identity: ∫ a b f ( x ) d x = ∫ a ∗ b f ( a + b − x ) d x = 1 2 ∫ 0 π x f ( sin ⁡ x ) + ( π − x ) f ( sin ⁡ ( π − x ) ) d x Note that sin ⁡ ( π − x ) = sin ⁡ x = 1 2 ∫ 0 π x f ( sin ⁡ x ) + ( π − x ) f ( sin ⁡ x ) d x = 1 2 π ∫ 0 π f ( sin ⁡ x ) d x \begin{aligned} I & = \int_0^\pi x f(\sin x) \ dx & \small \color{#3D99F6} \text{Using identity: } \int_a^b f(x) \ dx = \int_a*b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\pi x f(\sin x) + (\pi - x) f({\color{#3D99F6}\sin (\pi -x)}) \ dx & \small \color{#3D99F6} \text{Note that }\sin (\pi - x) = \sin x \\ & = \frac 12 \int_0^\pi x f(\sin x) + (\pi - x) f({\color{#3D99F6}\sin x}) \ dx \\ & = \frac 12 \pi \int_0^\pi f(\sin x) \ dx \end{aligned}

⟹ a b = 2 \implies ab = 2 only statement A is true. Therefore, the answer is 1999 \boxed{1999} .

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