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QJEEM22Ma2eee: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2eee Solution: If the maximum value of , for which the function is…

If the maximum value of aa , for which the function fa(x)=tan⁡−12x−3ax+7f_{a}(x)=\tan ^{-1} 2 x-3 a x+7 is non-decreasing in (−π6,π6)\left(-\frac{\pi}{6}, \frac{\pi}{6}\right) , is aˉ\bar{a} , then faˉ(π8)f_{\bar{a}}\left(\frac{\pi}{8}\right) is equal to :

Multiple Choice Options

  • A 8−9π4(9+π2)8-\frac{9 \pi}{4\left(9+\pi^{2}\right)}
  • B 8−4π9(4+π2)8-\frac{4 \pi}{9\left(4+\pi^{2}\right)}
  • C 8(1+π29+π2)8\left(\frac{1+\pi^{2}}{9+\pi^{2}}\right)
  • D 8−π48-\frac{\pi}{4}

Step-by-Step Verified Solution

Given, \text {Given, }

fa(x)amp;=tan⁡−12x−3ax+7fa′(x)amp;=21+4x2−3a\begin{aligned}f_a(x) & =\tan ^{-1} 2 x-3 a x+7 \\\\f_a^{\prime}(x) & =\frac{2}{1+4 x^2}-3 a\end{aligned}

As the function fa′(x)f_a^{\prime}(x) is non-decreasing

amp; in (−π6,π6),amp;fa′(x)≥0\begin{aligned}& \text { in }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \\\\& f_a^{\prime}(x) \geq 0\end{aligned}

amp;⇒21+4x2−3a≥0⇒21+4x2≥3aamp;⇒a≤23(1+4x2), when x∈(−π6,π6)amp;∵a is maximum when x2=π236,\begin{aligned}& \Rightarrow \frac{2}{1+4 x^2}-3 a \geq 0 \Rightarrow \frac{2}{1+4 x^2} \geq 3 a \\\\& \Rightarrow a \leq \frac{2}{3\left(1+4 x^2\right)}, \text { when } x \in\left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \\\\& \because a \text { is maximum when } x^2=\frac{\pi^2}{36},\end{aligned}

amax⁡amp;=23(1+4π236)=2×1236+4π2amp;=69+π2∴aˉamp;=69+π2\begin{aligned}a_{\max } & =\frac{2}{3\left(1+\frac{4 \pi^2}{36}\right)}=\frac{2 \times 12}{36+4 \pi^2} \\\\& =\frac{6}{9+\pi^2} \\\\\therefore \bar{a} & =\frac{6}{9+\pi^2}\end{aligned}

amp;∴fa(x)=tan⁡−12x−189+π2x+7amp;fπ(π8)=tan⁡−12(π8)−189+π2×π8+7amp;=tan⁡−1π4−9π36+4π2+7amp;=8−9π36+4π2=8−9π4(9+π2)\begin{aligned}& \therefore f_a(x)=\tan ^{-1} 2 x-\frac{18}{9+\pi^2} x+7 \\\\& f_\pi\left(\frac{\pi}{8}\right)=\tan ^{-1} 2\left(\frac{\pi}{8}\right)-\frac{18}{9+\pi^2} \times \frac{\pi}{8}+7 \\\\& =\tan ^{-1} \frac{\pi}{4}-\frac{9 \pi}{36+4 \pi^2}+7 \\\\& =8-\frac{9 \pi}{36+4 \pi^2}=8-\frac{9 \pi}{4\left(9+\pi^2\right)}\end{aligned}

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