JEE Main 2021 — Mathematics Calculus

Given $f(x)=\mathrm{sin}({\mathrm{cos}}^{-1}(\frac{1-{2}^{2x}}{1+{2}^{2x}}))\text{ at }x=1;{2}^{2x}={({2}^{x})}^{2}$f(x)=sin(cos−1(1+22x1−22x​)) at x=1;22x=(2x)2

For $\mathrm{sin}({\mathrm{cos}}^{-1}(\frac{1-{({2}^{x})}^{2}}{1+{({2}^{x})}^{2}}))$sin(cos−1(1+(2x)21−(2x)2​))

Let ${\mathrm{tan}}^{-1}{2}^{x}=\theta ;\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$tan−12x=θ;θ∈(−2π​,2π​)

$$\therefore \mathrm{sin}({\mathrm{cos}}^{-1}\mathrm{cos}2\theta )=\mathrm{sin}2\theta$$∴sin(cos−1cos2θ)=sin2θ

$${\begin{matrix}\text{ If } & x>1\Rightarrow \frac{\pi }{2}>\theta >\frac{\pi }{4} \\ \therefore & \pi >2\theta >\frac{\pi }{2}\end{matrix}}$$ If ∴​x>1⇒2π​>θ>4π​π>2θ>2π​​

$$=2\mathrm{sin}\theta \mathrm{cos}\theta =\frac{2\mathrm{tan}\theta }{1+{\mathrm{tan}}^{2}\theta }$$=2sinθcosθ=1+tan2θ2tanθ​

Hence, $f(x)=\frac{2\cdot {2}^{x}}{1+{2}^{2x}}$f(x)=1+22x2⋅2x​

$$\therefore {f}^{'}(x)=\frac{(1+{2}^{2x})(2.{2}^{x}\mathrm{ln}2)-{2}^{2x}\cdot 2\cdot \mathrm{ln}2\cdot 2\cdot {2}^{x}}{(1+{2}^{2x})}$$∴f′(x)=(1+22x)(1+22x)(2.2xln2)−22x⋅2⋅ln2⋅2⋅2x​

$$\therefore {f}^{'}(1)=\frac{20\mathrm{ln}2-32\mathrm{ln}2}{25}=-\frac{12}{25}\mathrm{ln}2$$∴f′(1)=2520ln2−32ln2​=−2512​ln2

So, $a=25,b=12$a=25,b=12.

Then, $|{a}^{2}-{b}^{2}|={25}^{2}-{12}^{2}$∣a2−b2∣=252−122

$$=625-144$$=625−144

$$=481$$=481