JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $f : \mathbb{R} \to \mathbb{R}$ be a continuous function satisfying $\int_{0}^{π /2} f (sin 2 x) sin x d x + α \int_{0}^{π /4} f (cos 2 x) cos x d x = 0,$ then the value of $\alpha$ is:

Choose the correct answer:

Explanation

1. Pehle integral ( $I_1$ ) ko simplify karte hain:

I_1 = \int_{0}^{\pi/2} f(\sin 2x) \sin x \, dx

Isme property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$ use karte hain:

I_1 = \int_{0}^{\pi/2} f(\sin 2(\pi/2 - x)) \sin(\pi/2 - x) \, dx

I_1 = \int_{0}^{\pi/2} f(\sin(\pi - 2x)) \cos x \, dx

I_1 = \int_{0}^{\pi/2} f(\sin 2x) \cos x \, dx

2. $I_1$ ke dono forms ko add karte hain:

2I_1 = \int_{0}^{\pi/2} f(\sin 2x) (\sin x + \cos x) \, dx

3. Substitution:

Maante hain $x = \frac{\pi}{4} + t$ , toh $dx = dt$ .

Limits badal jayengi: Jab $x=0, t=-\pi/4$ ; jab $x=\pi/2, t=\pi/4$ .

Also, $\sin 2x = \sin(2(\pi/4 + t)) = \sin(\pi/2 + 2t) = \cos 2t$ .

Aur $(\sin x + \cos x) = \sqrt{2} \sin(x + \pi/4) = \sqrt{2} \sin(t + \pi/2) = \sqrt{2} \cos t$ .

Ab equation hogi:

2I_1 = \int_{-\pi/4}^{\pi/4} f(\cos 2t) (\sqrt{2} \cos t) \, dt

Chunki $f(\cos 2t) \cos t$ ek even function hai, hum limits ko $2 \int_{0}^{\pi/4}$ likh sakte hain:

2I_1 = 2\sqrt{2} \int_{0}^{\pi/4} f(\cos 2t) \cos t \, dt

I_1 = \sqrt{2} \int_{0}^{\pi/4} f(\cos 2x) \cos x \, dx

4. $\alpha$ ki value nikalna:

Sawal mein diya gaya hai:

I_1 + \alpha \int_{0}^{\pi/4} f(\cos 2x) \cos x \, dx = 0

Substitute $I_1$ :

\sqrt{2} \int_{0}^{\pi/4} f(\cos 2x) \cos x \, dx + \alpha \int_{0}^{\pi/4} f(\cos 2x) \cos x \, dx = 0

(\sqrt{2} + \alpha) \int_{0}^{\pi/4} f(\cos 2x) \cos x \, dx = 0

Isse humein milta hai:

\alpha = -\sqrt{2}

Correct Option: (3) $-\sqrt{2}$

Choose the correct answer:

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