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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
The derivative of the function $\cot^{-1}\Big|(\cos2\text{x})^{\frac{1}{2}}\Big|$ at $\text{x}=\frac{\pi}{6}$ is:
  • ✓
    $\Big(\frac{2}{3}\Big)^\frac{1}{2}$
  • B
    $\Big(\frac{1}{3}\Big)^\frac{1}{2}$
  • C
    $3^\frac{1}{2}$
  • D
    $6^\frac{1}{2}$
✓

Answer

Correct option: A.
$\Big(\frac{2}{3}\Big)^\frac{1}{2}$
We have$, \text{y}=\cot^{-1}\Big(\sqrt{\cos2\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{1+\cot2\text{x}}\frac{\text{d}}{\text{dx}}\sqrt{\cos2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\cos^2\text{x}}\times\frac{1}{2\sqrt{\cos2\text{x}}}\frac{\text{d}}{\text{dx}}(\cos2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\cos^2\text{x}}\times\frac{1}{2\sqrt{\cos2\text{x}}}\times-2\sin2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin2\text{x}}{\cos^2\text{x}\times2\sqrt{\cos2\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\sin\text{x}\cos\text{x}}{\cos^2\text{x}\times2\sqrt{\cos2\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\tan\text{x}}{\sqrt{\cos2\text{x}}}$
So, at $\text{x}=\frac{\pi}{6},$ we get
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{6}}=\frac{\tan\big(\frac{\pi}{6}\big)}{\sqrt{\cos2\big(\frac{\pi}{6}\big)}}=\frac{\big(\frac{1}{\sqrt{3}}\big)}{\sqrt{\frac{1}{2}}}=\Big(\frac{2}{3}\Big)^\frac{1}{2}$

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