If ex=1+t−1−t1+t+1−t and tan⁡y2=1−t1+t, then dydx at t=12 is

ex=1+t−1−t1+t+1−t⇒ex=1−1−t1+t1+1−t1+t=1−tan⁡y/21+tan⁡y/2⇒ex=tan⁡π4−y2⇒ex=sec2⁡π4−y2−12dydx⇒dydx=−2excos2⁡π4−y2t=1/2,tan⁡y/2=13⇒y2=π6dydx=−2ext=1/2⋅cos2⁡π4−π6=−23/2−1/23/2+1/2⋅12⋅32+12⋅122=−23−13+1⋅3+1222=−28(3−1)=−12

If ex=1+t−1−t1+t+1−t and tan⁡y2=1−t1+t, then dydx at t=12 is

ex=1+t−1−t1+t+1−t⇒ex=1−1−t1+t1+1−t1+t=1−tan⁡y/21+tan⁡y/2⇒ex=tan⁡π4−y2⇒ex=sec2⁡π4−y2−12dydx⇒dydx=−2excos2⁡π4−y2t=1/2,tan⁡y/2=13⇒y2=π6dydx=−2ext=1/2⋅cos2⁡π4−π6=−23/2−1/23/2+1/2⋅12⋅32+12⋅122=−23−13+1⋅3+1222=−28(3−1)=−12

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If $e^{x} = \frac{\sqrt{1 + t} - \sqrt{1 - t}}{\sqrt{1 + t} + \sqrt{1 - t}}$ and $\tan \frac{y}{2} = \sqrt{\frac{1 - t}{1 + t}},$ $then \frac{d y}{d x}$ at $t = \frac{1}{2}$ is

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$- \frac{1}{2}$

$\frac{1}{2}$

$0$

$2$

answer is A.

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Detailed Solution

$e^{x} = \frac{\sqrt{1 + t} - \sqrt{1 - t}}{\sqrt{1 + t} + \sqrt{1 - t}}$

$\begin{array}{l} \Rightarrow e^{x} = \frac{1 - \sqrt{\frac{1 - t}{1 + t}}}{1 + \sqrt{\frac{1 - t}{1 + t}}} = \frac{1 - \tan y / 2}{1 + \tan y / 2} \\ \Rightarrow e^{x} = \tan (\frac{π}{4} - \frac{y}{2}) \\ \Rightarrow e^{x} = \sec^{2} (\frac{π}{4} - \frac{y}{2}) (\frac{- 1}{2}) \frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x} = - 2 e^{x} \cos^{2} (\frac{π}{4} - \frac{y}{2}) \end{array}$

$t = 1 / 2, \tan y / 2 = \frac{1}{\sqrt{3}}$ $\Rightarrow \frac{y}{2} = \frac{π}{6}$

$\frac{d y}{d x} = - 2 {[e^{x}]}_{t = 1 / 2} \cdot \cos^{2} (\frac{π}{4} - \frac{π}{6})$

$\begin{matrix} = - 2 [\frac{\sqrt{3 / 2} - \sqrt{1 / 2}}{\sqrt{3 / 2} + \sqrt{1 / 2}}] \cdot {[\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}]}^{2} \\ = - 2 [\frac{\sqrt{3} - 1}{\sqrt{3} + 1}] \cdot {[\frac{\sqrt{3} + 1}{2 \sqrt{2}}]}^{2} = \frac{- 2}{8} (3 - 1) = \frac{- 1}{2} \end{matrix}$