Verified Solution • 2025–26 Session moderate

QJEEM23Ma8f55: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8f55 Solution: Let and . If is decreasing in the interval and increasing in…

Let g(x)=f(x)+f(1−x)\mathrm{g}(x)=f(x)+f(1-x) and f^{\prime \prime}(x) > 0, x \in(0,1) . If g\mathrm{g} is decreasing in the interval (0,a)(0, a) and increasing in the interval (α,1)(\alpha, 1) , then tan⁡−1(2α)+tan⁡−1(1α)+tan⁡−1(α+1α)\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) is equal to :

Multiple Choice Options

  • A 3π4\frac{3 \pi}{4}
  • B π\pi
  • C 5π4\frac{5 \pi}{4}
  • D 3π2\frac{3 \pi}{2}

Step-by-Step Verified Solution

We have, g(x)=f(x)+f(1−x)g(x)=f(x)+f(1-x)

Differentiating both side, we get

g′(x)=f′(x)−f′(1−x)g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)

As f^{\prime \prime}(x)>0, f^{\prime}(x) is an increasing function.

Also, g(x)=f(x)+f(2a−x)g(x)=f(x)+f(2 a-x) is always symmetric about x=ax=a

So, g(x)=f(x)+f(1−x)g(x)=f(x)+f(1-x) is also symmetric about x=1/2x=1 / 2

∴g\therefore g is decreasing in the interval (0,1/2)(0,1 / 2) and increasing in the interval (1/2,1)(1 / 2,1) .

Now,

amp;tan⁡−12α+tan⁡−1(12)+tan⁡−1(α+1α)amp;=tan⁡−11+tan⁡−1(2)+tan⁡−13=π\begin{aligned}& \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\& =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi\end{aligned}

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