If f: \mathbbR arrow \mathbbR be a continuous function satisfying \int_0^\frac\pi2 f(\sin 2 x) \sin x d x+\alpha \int_0^\frac\pi4 f(\cos 2 x) \cos x d x=0 , then the value of \alpha is :

Definite Integration PYQ — JEE Main 2023 · 11 Apr · Shift 2 · Q32 with Solution

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Correct answer: C

Let

\qquad I_2=\int_{0}^{\pi/4} f(\cos 2x)\cos x\,dx.$$ We are given that $$I_1+\alpha I_2=0$$ for every continuous function $f$. We must find $\alpha$. ## 1. Simplify the first integral Since $\sin 2x=2\sin x\cos x$, split the interval: $$I_1=\int_0^{\pi/4} f(\sin 2x)\sin x\,dx+\int_{\pi/4}^{\pi/2} f(\sin 2x)\sin x\,dx.$$ In the second integral, put $$x=\frac{\pi}{2}-t \quad \Rightarrow \quad dx=-dt.$$ Then $$\sin 2x=\sin(\pi-2t)=\sin 2t, \qquad \sin x=\sin\left(\frac{\pi}{2}-t\right)=\cos t.$$ So $$\int_{\pi/4}^{\pi/2} f(\sin 2x)\sin x\,dx =\int_{\pi/4}^{0} f(\sin 2t)\cos t(-dt) =\int_0^{\pi/4} f(\sin 2t)\cos t\,dt.$$ Hence $$I_1=\int_0^{\pi/4} f(\sin 2x)(\sin x+\cos x)\,dx.$$ Now on $[0,\pi/4]$, let $$u=\sin 2x, \qquad du=2\cos 2x\,dx.$$ Also, $$(\sin x+\cos x)^2=1+\sin 2x=1+u,$$ and since $\sin x+\cos x>0$ on this interval, $$\sin x+\cos x=\sqrt{1+u}.$$ Further, $$\cos 2x=\sqrt{1-\sin^2 2x}=\sqrt{1-u^2}$$ because $2x\in[0,\pi/2]$, so $\cos 2x\ge 0$. Thus $$dx=\frac{du}{2\sqrt{1-u^2}},$$ and therefore $$I_1=\int_0^1 f(u)\,\frac{\sqrt{1+u}}{2\sqrt{1-u^2}}\,du.$$ Since $$\sqrt{1-u^2}=\sqrt{(1-u)(1+u)},$$ we get $$\frac{\sqrt{1+u}}{2\sqrt{1-u^2}}=\frac{1}{2\sqrt{1-u}}.$$ Hence $$I_1=\frac12\int_0^1 \frac{f(u)}{\sqrt{1-u}}\,du.$$ ## 2. Simplify the second integral Let $$v=\cos 2x, \qquad dv=-2\sin 2x\,dx=-4\sin x\cos x\,dx.$$ We want to express $\cos x\,dx$ in terms of $dv$. Since on $[0,\pi/4]$, $v=\cos 2x$ goes from $1$ to $0$, and $$1+v=1+\cos 2x=2\cos^2 x,$$ so $$\cos x=\sqrt{\frac{1+v}{2}}.$$ Also, $$\sin^2 x=\frac{1-v}{2} \Rightarrow \sin x=\sqrt{\frac{1-v}{2}}.$$ Thus $$dv=-4\sin x\cos x\,dx =-4\sqrt{\frac{1-v}{2}}\sqrt{\frac{1+v}{2}}\;dx =-2\sqrt{1-v^2}\,dx,$$ so $$dx=-\frac{dv}{2\sqrt{1-v^2}}.$$ Therefore $$I_2=\int_0^{\pi/4} f(\cos 2x)\cos x\,dx =\int_1^0 f(v)\sqrt{\frac{1+v}{2}}\left(-\frac{dv}{2\sqrt{1-v^2}}\right).$$ So $$I_2=\frac{1}{2\sqrt{2}}\int_0^1 f(v)\frac{\sqrt{1+v}}{\sqrt{1-v^2}}\,dv =\frac{1}{2\sqrt{2}}\int_0^1 \frac{f(v)}{\sqrt{1-v}}\,dv.$$ Hence $$I_2=\frac{1}{2\sqrt{2}}\int_0^1 \frac{f(t)}{\sqrt{1-t}}\,dt.$$ ## 3. Compare the two integrals From above, $$I_1=\frac12\int_0^1 \frac{f(t)}{\sqrt{1-t}}\,dt, \qquad I_2=\frac{1}{2\sqrt{2}}\int_0^1 \frac{f(t)}{\sqrt{1-t}}\,dt.$$ Therefore, $$I_1=\sqrt{2}\,I_2.$$ Given $$I_1+\alpha I_2=0,$$ we get $$\sqrt{2}I_2+\alpha I_2=0 \Rightarrow (\sqrt{2}+\alpha)I_2=0.$$ Since this must hold for every continuous $f$, we must have $$\alpha=-\sqrt{2}.$$ ## 4. Check options - A: $-\sqrt{3}$ - B: $\sqrt{2}$ - C: $-\sqrt{2}$ - D: $\sqrt{3}$ Hence the correct option is $$\boxed{\text{C }(-\sqrt{2})}.$$

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View written solutionFree

Correct answer: C

Let

\qquad I_2=\int_{0}^{\pi/4} f(\cos 2x)\cos x\,dx.$$ We are given that $$I_1+\alpha I_2=0$$ for every continuous function $f$. We must find $\alpha$. ## 1. Simplify the first integral Since $\sin 2x=2\sin x\cos x$, split the interval: $$I_1=\int_0^{\pi/4} f(\sin 2x)\sin x\,dx+\int_{\pi/4}^{\pi/2} f(\sin 2x)\sin x\,dx.$$ In the second integral, put $$x=\frac{\pi}{2}-t \quad \Rightarrow \quad dx=-dt.$$ Then $$\sin 2x=\sin(\pi-2t)=\sin 2t, \qquad \sin x=\sin\left(\frac{\pi}{2}-t\right)=\cos t.$$ So $$\int_{\pi/4}^{\pi/2} f(\sin 2x)\sin x\,dx =\int_{\pi/4}^{0} f(\sin 2t)\cos t(-dt) =\int_0^{\pi/4} f(\sin 2t)\cos t\,dt.$$ Hence $$I_1=\int_0^{\pi/4} f(\sin 2x)(\sin x+\cos x)\,dx.$$ Now on $[0,\pi/4]$, let $$u=\sin 2x, \qquad du=2\cos 2x\,dx.$$ Also, $$(\sin x+\cos x)^2=1+\sin 2x=1+u,$$ and since $\sin x+\cos x>0$ on this interval, $$\sin x+\cos x=\sqrt{1+u}.$$ Further, $$\cos 2x=\sqrt{1-\sin^2 2x}=\sqrt{1-u^2}$$ because $2x\in[0,\pi/2]$, so $\cos 2x\ge 0$. Thus $$dx=\frac{du}{2\sqrt{1-u^2}},$$ and therefore $$I_1=\int_0^1 f(u)\,\frac{\sqrt{1+u}}{2\sqrt{1-u^2}}\,du.$$ Since $$\sqrt{1-u^2}=\sqrt{(1-u)(1+u)},$$ we get $$\frac{\sqrt{1+u}}{2\sqrt{1-u^2}}=\frac{1}{2\sqrt{1-u}}.$$ Hence $$I_1=\frac12\int_0^1 \frac{f(u)}{\sqrt{1-u}}\,du.$$ ## 2. Simplify the second integral Let $$v=\cos 2x, \qquad dv=-2\sin 2x\,dx=-4\sin x\cos x\,dx.$$ We want to express $\cos x\,dx$ in terms of $dv$. Since on $[0,\pi/4]$, $v=\cos 2x$ goes from $1$ to $0$, and $$1+v=1+\cos 2x=2\cos^2 x,$$ so $$\cos x=\sqrt{\frac{1+v}{2}}.$$ Also, $$\sin^2 x=\frac{1-v}{2} \Rightarrow \sin x=\sqrt{\frac{1-v}{2}}.$$ Thus $$dv=-4\sin x\cos x\,dx =-4\sqrt{\frac{1-v}{2}}\sqrt{\frac{1+v}{2}}\;dx =-2\sqrt{1-v^2}\,dx,$$ so $$dx=-\frac{dv}{2\sqrt{1-v^2}}.$$ Therefore $$I_2=\int_0^{\pi/4} f(\cos 2x)\cos x\,dx =\int_1^0 f(v)\sqrt{\frac{1+v}{2}}\left(-\frac{dv}{2\sqrt{1-v^2}}\right).$$ So $$I_2=\frac{1}{2\sqrt{2}}\int_0^1 f(v)\frac{\sqrt{1+v}}{\sqrt{1-v^2}}\,dv =\frac{1}{2\sqrt{2}}\int_0^1 \frac{f(v)}{\sqrt{1-v}}\,dv.$$ Hence $$I_2=\frac{1}{2\sqrt{2}}\int_0^1 \frac{f(t)}{\sqrt{1-t}}\,dt.$$ ## 3. Compare the two integrals From above, $$I_1=\frac12\int_0^1 \frac{f(t)}{\sqrt{1-t}}\,dt, \qquad I_2=\frac{1}{2\sqrt{2}}\int_0^1 \frac{f(t)}{\sqrt{1-t}}\,dt.$$ Therefore, $$I_1=\sqrt{2}\,I_2.$$ Given $$I_1+\alpha I_2=0,$$ we get $$\sqrt{2}I_2+\alpha I_2=0 \Rightarrow (\sqrt{2}+\alpha)I_2=0.$$ Since this must hold for every continuous $f$, we must have $$\alpha=-\sqrt{2}.$$ ## 4. Check options - A: $-\sqrt{3}$ - B: $\sqrt{2}$ - C: $-\sqrt{2}$ - D: $\sqrt{3}$ Hence the correct option is $$\boxed{\text{C }(-\sqrt{2})}.$$

Definite Integration question

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