JEE Main 2021 — Mathematics Calculus

Put $x={\mathrm{sin}}^{2}\theta ,0<x<1$x=sin2θ,0<x<1

$$\mathrm{sin}\theta =\sqrt{x}$$sinθ=x​

$$\Rightarrow f(x)=\mathrm{cos}(2{\mathrm{tan}}^{-1}\mathrm{sin}({\mathrm{cot}}^{-1}\sqrt{\frac{1-{\mathrm{sin}}^{2}\theta }{{\mathrm{sin}}^{2}\theta }}))$$⇒f(x)=cos(2tan−1sin(cot−1sin2θ1−sin2θ​​))

$$\Rightarrow f(x)=\mathrm{cos}(2{\mathrm{tan}}^{-1}(\mathrm{sin}\theta ))$$⇒f(x)=cos(2tan−1(sinθ))

$$\Rightarrow f(x)=\mathrm{cos}(2{\mathrm{tan}}^{-1}\sqrt{x})$$⇒f(x)=cos(2tan−1x​)

$$=\frac{1-{\mathrm{tan}}^{2}({\mathrm{tan}}^{-1}\sqrt{x})}{1+{\mathrm{tan}}^{2}({\mathrm{tan}}^{-1}\sqrt{x})}$$=1+tan2(tan−1x​)1−tan2(tan−1x​)​

$$\Rightarrow f(x)=\frac{1-x}{1+x}$$⇒f(x)=1+x1−x​

$$\Rightarrow {f}^{'}(x)=\frac{(1+x)(-1)-(1-x)\cdot 1}{{(1+x)}^{2}}$$⇒f′(x)=(1+x)2(1+x)(−1)−(1−x)⋅1​

$$\Rightarrow {f}^{'}(x)=\frac{-2}{{(1+x)}^{2}}$$⇒f′(x)=(1+x)2−2​

Multiply, ${(1-x)}^{2}$(1−x)2 on both sides

$$\Rightarrow {(1-x)}^{2}{f}^{'}(x)=\frac{-2{(1-x)}^{2}}{{(1+x)}^{2}}$$⇒(1−x)2f′(x)=(1+x)2−2(1−x)2​

Now, $2{(f(x))}^{2}=\frac{2{(1-x)}^{2}}{{(1+x)}^{2}}$2(f(x))2=(1+x)22(1−x)2​

$\therefore {(1-x)}^{2}{f}^{'}(x)+2{(f(x))}^{2}=0$∴(1−x)2f′(x)+2(f(x))2=0 option $(1)$(1) satisfied