Q104 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

If $f(x)=\sqrt{\log \left(x^2+x+1\right)+\sqrt{\cosh (2 x-3)}}$, then $f^{\prime}(0)=$

  • A
    $\frac{1}{2 \sqrt{\sqrt{\cosh (3)}}}\left(1+\frac{\sinh (3)}{\sqrt{\cosh (3)}}\right)$
  • B
    $\frac{1}{2 \sqrt{\sqrt{\cosh (3)}}}\left(\log 3-\frac{\sinh (3)}{\sqrt{\cosh (3)}}\right)$
  • C
    $\frac{\log 3 \sqrt{\cosh (3)}-\sinh (3)}{2(\cosh (3))^{\frac{3}{4}}}$
  • Correct answer
    D
    $\frac{\sqrt{\cosh (3)}-\sinh (3)}{2(\cosh (3))^{\frac{3}{4}}}$

Solution

$\begin{aligned} & \text {} f(x)=\sqrt{\log \left(x^2+x+1\right)+\sqrt{\cosh (2 x-3)}} \\ & \Rightarrow f^{\prime}(x)=\frac{1 \times \frac{2 x+1}{x^2+x+1}+\frac{2 \sinh (2 x-3)}{2 \sqrt{\cos h(2 x-3)}}}{2 \sqrt{\log \left(x^2+x+1\right)+\sqrt{\cosh (2 x-3)}}} \\ & \Rightarrow f^{\prime}(0)=\frac{1+\frac{\sin h(-3)}{\sqrt{\cosh (-3)}}}{2 \sqrt{\sqrt{\cos h(-3)}}} \\ & \Rightarrow f^{\prime}(0)=\frac{1-\frac{\sin h(3)}{\sqrt{\cosh (3)}}}{2(\operatorname{cosh(3))})^{1 / 4}} \\ & \Rightarrow f^{\prime}(0)=\frac{\sqrt{\cosh (3)}-\sin h(3)}{2(\cosh (3))^{3 / 4}}\end{aligned}$

Hence, the correct answer is option: 'D'.

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