JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q6

← JEE Main 2022 · 26 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 6 of 90

If the maximum value of $a$ , for which the function $f_a(x)=\tan^{-1}2x-3ax+7$ is non-decreasing in $\left(-\dfrac\pi6,\dfrac\pi6\right)$ , is $\bar a$ , then $f_{\bar a}\left(\dfrac\pi8\right)$ is equal to

8-\dfrac{9\pi}{4\left(9+\pi^2\right)}

8-\dfrac{4\pi}{9\left(4+\pi^2\right)}

8\left(\dfrac{1+\pi^2}{9+\pi^2}\right)

8-\dfrac\pi4

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped in NTA's final key.

Why. $f_a'(x)=\frac2{1+4x^2}-3a\ge0$ on $\left(-\frac\pi6,\frac\pi6\right)$ needs $3a\le\frac{18}{9+\pi^2}$ , so $\bar a=\frac6{9+\pi^2}$ . Then $f_{\bar a}\left(\frac\pi8\right)=\tan^{-1}\frac\pi4+7-\frac{9\pi}{4(9+\pi^2)}$ , which is none of the options (option A would need $\tan^{-1}\frac\pi4=1$ ).

Topics:Monotonicity Inverse trigonometric functions

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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 6