JEE Main 2024 — Mathematics Trigonometry

Let,

$$f(x)={\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}2x$$f(x)=tan−1x+tan−12x

Now, differentiating the above function we get,

$$\Rightarrow {f}^{'}(x)=\frac{1}{1+{x}^{2}}+\frac{2}{1+4{x}^{2}}$$⇒f′(x)=1+x21​+1+4x22​

$$\Rightarrow {f}^{'}(x)>0;\forall x\in R$$⇒f′(x)>0;∀x∈R

$$\Rightarrow \underset{x\rightarrow \infty }{\mathrm{lim}}f(x)=\pi \text{and}\underset{x\rightarrow -\infty }{\mathrm{lim}}f(x)=-\pi$$⇒x→∞lim​f(x)=πandx→−∞lim​f(x)=−π

Hence, $f(x)$f(x) is monotonically increasing for $x\in R$x∈R and the range of $f(x)$f(x) is $(-\pi ,\pi )$(−π,π).

Hence, $f(x)=\frac{\pi }{4}$f(x)=4π​ has exactly one solution for $x>0$x>0.