JEE Main 2023 — Mathematics Calculus

Given,

$g(x)=f(x)+f(1-x)$g(x)=f(x)+f(1−x) and ${f}^{"}(x)>0,x\in (0,1)$f"(x)>0,x∈(0,1). If $g$g is decreasing in the interval $(0,\alpha )$(0,α) and increasing in the interval $(\alpha ,1)$(α,1),

Now solving,

$$g(x)=f(x)+f(1-x)$$g(x)=f(x)+f(1−x)

Now differentiating both side we get,

$${g}^{'}(x)={f}^{'}(x)-{f}^{'}(1-x)$$g′(x)=f′(x)−f′(1−x)

Differentiating again we get,

$${g}^{"}(x)={f}^{"}(x)+{f}^{"}(1-x)>0$$g"(x)=f"(x)+f"(1−x)>0

So, ${g}^{'}(x)$g′(x) is increasing as given ${f}^{"}(x)>0$f"(x)>0

$$\Rightarrow {g}^{'}(0)<{g}^{'}(1)$$⇒g′(0)<g′(1)

$$\Rightarrow {f}^{'}(0)-{f}^{'}(1)<{f}^{'}(1)-{f}^{'}(0)$$⇒f′(0)−f′(1)<f′(1)−f′(0)

$$\Rightarrow {f}^{'}(0)<{f}^{'}(1)$$⇒f′(0)<f′(1)

Now finding ${g}^{'}(x)=0$g′(x)=0 we get,

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

$$\Rightarrow x=1-x$$⇒x=1−x

$$\Rightarrow x=\frac{1}{2}$$⇒x=21​

Now ${g}^{'}(x)$g′(x) is positive for $x\in (0,\frac{1}{2})$x∈(0,21​)

And ${g}^{'}(x)$g′(x) is negative for $x\in (\frac{1}{2},1)$x∈(21​,1)

$$\therefore \alpha =\frac{1}{2}$$∴α=21​

So,

$${\mathrm{tan}}^{-1}2\alpha +{\mathrm{tan}}^{-1}(\frac{1}{\alpha })+{\mathrm{tan}}^{-1}(\frac{\alpha +1}{\alpha })$$tan−12α+tan−1(α1​)+tan−1(αα+1​)

$$={\mathrm{tan}}^{-1}1+{\mathrm{tan}}^{-1}2+{\mathrm{tan}}^{-1}3$$=tan−11+tan−12+tan−13

$$=\frac{\pi }{4}+{\mathrm{tan}}^{-1}(\frac{2+3}{1-6})$$=4π​+tan−1(1−62+3​)

$$=\frac{\pi }{4}+{\mathrm{tan}}^{-1}(-1)$$=4π​+tan−1(−1)

$$=\frac{\pi }{4}+\frac{3\pi }{4}=\pi$$=4π​+43π​=π