JEE Main 2023 — Mathematics Calculus

Given,

$${\int }_{0}^{\frac{\pi }{2}}f(\mathrm{sin}2x)\mathrm{sin}xdx+\alpha {\int }_{0}^{\frac{\pi }{4}}f(\mathrm{cos}2x)\mathrm{cos}xdx=0$$∫02π​​f(sin2x)sinxdx+α∫04π​​f(cos2x)cosxdx=0

Now, let $\displaystyle I={\int }_{0}^{\frac{\pi }{2}}f(\mathrm{sin}2x).\mathrm{sin}xdx$I=∫02π​​f(sin2x).sinxdx

Now using property $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{c}f(x)dx+{\int }_{c}^{b}f(x)dx$∫ab​f(x)dx=∫ac​f(x)dx+∫cb​f(x)dx we get,

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{4}}f(\mathrm{sin}2x)\mathrm{sin}xdx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}f(\mathrm{sin}2x).\mathrm{sin}xdx$$⇒I=∫04π​​f(sin2x)sinxdx+∫4π​2π​​f(sin2x).sinxdx

Now using property $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx we get,

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{4}}f(\mathrm{cos}2x)\mathrm{sin}(\frac{\pi }{4}-x)dx+{\int }_{0}^{\frac{\pi }{4}}f(\mathrm{sin}2(\frac{\pi }{4}+x))\mathrm{sin}(\frac{\pi }{4}+x)dx$$⇒I=∫04π​​f(cos2x)sin(4π​−x)dx+∫04π​​f(sin2(4π​+x))sin(4π​+x)dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{4}}f(\mathrm{cos}2x)(\frac{1}{\sqrt{2}}\mathrm{cos}x-\frac{1}{\sqrt{2}}\mathrm{sin}x)dx+{\int }_{0}^{\frac{\pi }{4}}f(\mathrm{cos}2x)(\frac{1}{\sqrt{2}}\mathrm{cos}x+\frac{1}{\sqrt{2}}\mathrm{sin}x)dx$$⇒I=∫04π​​f(cos2x)(2​1​cosx−2​1​sinx)dx+∫04π​​f(cos2x)(2​1​cosx+2​1​sinx)dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{4}}f(\mathrm{cos}2x)(\sqrt{2}\mathrm{cos}x)dx$$⇒I=∫04π​​f(cos2x)(2​cosx)dx

So, the putting the value of $I$I in $\displaystyle {\int }_{0}^{\frac{\pi }{2}}f(\mathrm{sin}2x)\mathrm{sin}xdx+\alpha {\int }_{0}^{\frac{\pi }{4}}f(\mathrm{cos}2x)\mathrm{cos}xdx=0$∫02π​​f(sin2x)sinxdx+α∫04π​​f(cos2x)cosxdx=0 we get, $\alpha =-\sqrt{2}$α=−2​