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If $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right)$
$+\cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$ and $0 < \mathrm{x} < \sqrt{2}$,
then $x$ is equal to
Options:
$+\cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$ and $0 < \mathrm{x} < \sqrt{2}$,
then $x$ is equal to
Solution:
1141 Upvotes
Verified Answer
The correct answer is:
$1$
Given that, $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^2}{4}-\ldots \infty\right)$
$\cos ^{-1}\left(x-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$
Here, $\quad x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty$
Forms G.P. so its sum can be given as $\frac{x}{1-\frac{x}{2}}$ Similarly, for series
$x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty \Rightarrow \text { sum }=\frac{x^2}{1-\frac{x^2}{2}}$
$\begin{array}{ll}\sin ^{-1}\left[\frac{x}{1-\frac{x}{2}}\right]+\cos ^{-1}\left[\frac{x^2}{1-\frac{x^2}{2}}\right]=\frac{\pi}{2} & \\ \sin ^{-1}\left[\frac{2 x}{2-x}\right]+\cos ^{-1}\left[\frac{2 x^2}{2-x^2}\right]=\frac{\pi}{2} & \\ & \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]\end{array}$
$\begin{aligned} & \frac{2 x}{2-x}=\frac{2 x^2}{2-x^2} \\ & \frac{1}{2-x}=\frac{x}{2-x^2} \\ & \left(2-x^2\right)=x(2-x) \\ & 2-x^2=2 x-x^2 \\ & 2 x=2 \Rightarrow x=1\end{aligned}$
$\cos ^{-1}\left(x-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$
Here, $\quad x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty$
Forms G.P. so its sum can be given as $\frac{x}{1-\frac{x}{2}}$ Similarly, for series
$x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty \Rightarrow \text { sum }=\frac{x^2}{1-\frac{x^2}{2}}$
$\begin{array}{ll}\sin ^{-1}\left[\frac{x}{1-\frac{x}{2}}\right]+\cos ^{-1}\left[\frac{x^2}{1-\frac{x^2}{2}}\right]=\frac{\pi}{2} & \\ \sin ^{-1}\left[\frac{2 x}{2-x}\right]+\cos ^{-1}\left[\frac{2 x^2}{2-x^2}\right]=\frac{\pi}{2} & \\ & \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]\end{array}$
$\begin{aligned} & \frac{2 x}{2-x}=\frac{2 x^2}{2-x^2} \\ & \frac{1}{2-x}=\frac{x}{2-x^2} \\ & \left(2-x^2\right)=x(2-x) \\ & 2-x^2=2 x-x^2 \\ & 2 x=2 \Rightarrow x=1\end{aligned}$
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