Sin^{2n}x + cos^{2n}x lies between

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Trigonometrical Equations

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$sin^{2n}x + cos^{2n}x$ lies between

A

$-1$ and $1$

B

$0$ and $1$

C

$1$ and $2$

D

None of these

Solution

Since $0 \leq \sin ^{2 \pi} x \leq \sin ^{2} x$

$0 \leq \cos ^{2 \pi} x \leq \cos ^{2} x$

$\left[\because \sin ^{4} x=\sin ^{2} x, \sin ^{2} x \leq \sin ^{2} x \cdot 1\right.$

$\left.\therefore \sin ^{4} x \leq \sin ^{2} x \text { etc. }\right]$

$\Rightarrow \quad 0<\sin ^{2 \pi} x+\cos ^{2 n} x \leq \sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \quad 0<\sin ^{2 n} x+\cos ^{2 n} x \leq 1$

Standard 11

Mathematics

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