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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\sin^{-1}\big(x-\large\frac{x^2}{2}+\frac{x^3}{4}-........)$$+\cos^{-1}(x^2-\large\frac{x^4}{2}+\frac{x^6}{4}.......)=\large\frac{\pi}{2}$.For $0<\mid x\mid<\sqrt 2$ then $x$ equals

$\begin{array}{1 1}(a)\;\large\frac{1}{2}&(b)\;1\\(c)\;-\large\frac{1}{2}&(d)\;-1\end{array}$

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1 Answer

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  • Sum of infinite G.P.$=\large\frac{a}{1-r}$ where $a=1^{st}\:term$ and $r=common\: ratio.$
Given: $\sin^{-1}\big(x-\large\frac{x^2}{2}+\frac{x^3}{4}-........)$$+\cos^{-1}(x^2-\large\frac{x^4}{2}+\frac{x^6}{4}.......)=\large\frac{\pi}{2}$
$x-\large\frac{x^2}{2}+\frac{x^3}{4}-.......$ is an infinite G.P.
Sum of this series $=\large\frac{x}{1+x/2}$$=\large\frac{2x}{2+x}$
Similarly $x^2-\large\frac{x^4}{2}+\frac{x^6}{4}.......$ is also an infinite G.P.
Sum of this series $=\large\frac{x^2}{1+x^2/2}$$=\large\frac{2x^2}{2+x^2}$
Substituting these values in the given equation, we get
$\sin^{-1}\big(\large\frac{x}{1+x/2}\big)$$+\cos^{-1}\big(\large\frac{x^2}{1+x^2/2}\big)=\frac{\pi}{2}$
$\Rightarrow\:\sin^{-1}\big(\large\frac{2x}{2+x}\big)=\frac{\pi}{2}$$-\cos^{-1}\big(\large\frac{2x^2}{2+x^2}\big)$
$\qquad\quad\quad\quad\;\;=\sin^{-1}\big(\large\frac{2x^2}{2+x^2}\big)$
$\Rightarrow\:\large\frac{2x}{2+x}=\frac{2x^2}{2+x^2}$
$\Rightarrow 2x\big[\large\frac{1}{2+x}-\frac{x}{2+x^2}\big]$$=0$
$\Rightarrow\:x=0$ or $2+x^2=2x+x^2$
$\Rightarrow x=0$ or $x=1$
But $0 < \mid x\mid <\sqrt 2$
$\Rightarrow x=1$
Hence (b) is the correct answer.
answered Nov 18, 2013 by sreemathi.v
edited Mar 25, 2014 by rvidyagovindarajan_1
 

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