Logo
Jeeify
MCQMediumJEE 2023Definite Integrals

JEE Mathematics 2023 Question with Solution

Let f(x)=x+a(π2−4)sin⁡x+b(π2−4)cos⁡x,  x∈Rf(x) = x + \frac{a}{\left(\frac{\pi}{2} - 4\right)} \sin x + \frac{b}{\left(\frac{\pi}{2} - 4\right)} \cos x, \; x \in \mathbb{R} be a function which satisfies f(x)=x+∫0π/2sin⁡(x+y)f(y) dy.f(x) = x + \int_{0}^{\pi/2} \sin(x + y)f(y) \, dy. Then (a+b)(a+b) is equal to:

  • A

    −π(π+2)-\pi(\pi + 2)

  • B

    −2π(π+2)-2\pi(\pi + 2)

  • C

    −2π(π−2)-2\pi(\pi - 2)

  • D

    −π(π−2)-\pi(\pi - 2)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

f(x)=x+∫0π/2sin⁡(x+y)f(y) dyf(x) = x + \int_{0}^{\pi/2} \sin(x+y)f(y) \, dy

and

f(x)=x+a(π2−4)sin⁡x+b(π2−4)cos⁡xf(x) = x + \frac{a}{\left(\frac{\pi}{2} - 4\right)} \sin x + \frac{b}{\left(\frac{\pi}{2} - 4\right)} \cos x

Find: (a+b)(a+b).

Using

sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x+y) = \sin x \cos y + \cos x \sin y

we get

f(x)=x+∫0π/2(cos⁡ysin⁡x+sin⁡ycos⁡x)f(y) dyf(x) = x + \int_{0}^{\pi/2} \left(\cos y \sin x + \sin y \cos x\right) f(y) \, dy

so

f(x)=x+(∫0π/2cos⁡y f(y) dy)sin⁡x+(∫0π/2sin⁡y f(y) dy)cos⁡x.f(x) = x + \left(\int_{0}^{\pi/2} \cos y \, f(y) \, dy\right) \sin x + \left(\int_{0}^{\pi/2} \sin y \, f(y) \, dy\right) \cos x.

Comparing coefficients with the given form used in the solution,

aπ2/4=∫0π/2cos⁡y f(y) dy,bπ2/4=∫0π/2sin⁡y f(y) dy\frac{a}{\pi^2/4} = \int_{0}^{\pi/2} \cos y \, f(y) \, dy, \qquad \frac{b}{\pi^2/4} = \int_{0}^{\pi/2} \sin y \, f(y) \, dy

therefore

a=π24∫0π/2cos⁡y f(y) dy,b=π24∫0π/2sin⁡y f(y) dy.a = \frac{\pi^2}{4} \int_{0}^{\pi/2} \cos y \, f(y) \, dy, \qquad b = \frac{\pi^2}{4} \int_{0}^{\pi/2} \sin y \, f(y) \, dy.

Adding,

a+b=π24(∫0π/2cos⁡y f(y) dy+∫0π/2sin⁡y f(y) dy)a+b = \frac{\pi^2}{4} \left( \int_{0}^{\pi/2} \cos y \, f(y) \, dy + \int_{0}^{\pi/2} \sin y \, f(y) \, dy \right)

that is,

a+b=π24∫0π/2(cos⁡y+sin⁡y)f(y) dy.a+b = \frac{\pi^2}{4} \int_{0}^{\pi/2} (\cos y + \sin y) f(y) \, dy.

From the final simplification shown in the solution,

a+b=−2π(π+2).a+b = -2\pi(\pi+2).

Hence the value of (a+b)(a+b) is −2π(π+2)-2\pi(\pi+2).

The solution marks C, but the computed value −2π(π+2)-2\pi(\pi+2) matches option B in the listed options. Therefore, the defensible correct option from the working is B.

Answer discrepancy note

The solution contains an internal inconsistency: it states "The Correct Option is C" and later says "the correct answer is option (3)", but the numerical result obtained there is −2π(π+2)-2\pi(\pi+2), which corresponds to option B in the provided option list. By the stated working, the answer should be taken as B.

Common mistakes

  • Using sin⁡(x+y)\sin(x+y) incorrectly. The identity is sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x+y)=\sin x\cos y+\cos x\sin y, not a product of separate sines or cosines. Expand first before comparing coefficients.

  • Comparing coefficients with the wrong form of f(x)f(x). The coefficient of sin⁡x\sin x must be matched with the integral containing cos⁡y f(y)\cos y\,f(y), and the coefficient of cos⁡x\cos x with sin⁡y f(y)\sin y\,f(y).

  • Trusting the marked option label without checking the computed value. Here the solution's option label conflicts with its own final expression. Always match the final expression with the listed options.

Practice more Definite Integrals questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Definite Integrals

Related questions