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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $x=\sin^{14}\theta+\cos ^{20}\theta$,then for all $\theta\in R$

$\begin{array}{1 1}(a)\;0< x <1&(b)\;0 < x \leq 1\\(c)\;0 \leq x< 1&(d)\;0 \leq x\leq 1\end{array}$

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We know that
$-1 \leq \sin\theta \leq 1$ and $-1 \leq \cos\theta \leq 1$ for all values of $\theta$
$\Rightarrow \sin^2\theta\leq 1$ & $\cos^2\theta \leq 1$
We also know that $a^n \leq a$ for all $n\in N,n> 1$ and 0 < a$\leq 1$
$\therefore \sin^{14}\theta=(\sin^2\theta)^7\leq \sin^2\theta$
$\cos^{20}\theta=(\cos^2\theta)^{10}\leq \cos^2\theta$ for all $\theta \in R$
$\Rightarrow \sin^{14}\theta+\cos ^{20}\theta\leq \sin^2\theta+\cos^2\theta$ for all $\theta\in R$
$\Rightarrow \sin^{14}\theta+\cos^{20}\theta \leq 1$ for all $\theta\in R$
$\Rightarrow x\leq 1$
Also $x=\sin^{14}\theta+\cos^{20}\theta >0$ for all $\theta\in R$
$\Rightarrow \theta < x\leq 1$
Hence (b) is the correct answer.
answered Oct 18, 2013 by sreemathi.v

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