The maximum value of ^2 (2 3 +x+ ^2 (2 3 -x ) is: - Vidyadip

MCQ

The maximum value of $\sin^2\Big(\frac{2\pi}{3}+\text{x}+\sin^2\Big(\frac{2\pi}{3}-\text{x}\Big)$ is:

A
$\frac12$
✓
$\frac32$
C
$\frac14$
D
$\frac34$

✓

Answer

Correct option: B.

$\frac32$

$\frac{2\pi}{3}=120^\circ$
Let $\text{f(x)}=\sin^2(90+30+\text{x})+\sin^2(90+30-\text{x})$
$=[\cos(30+\text{x})]^2+[\cos(30=\text{x})]^2$ $[\text{Using }\sin(90+\text{A})=\cos\text{A}]$
$=\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2+\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2$
$=\frac{\sqrt{3}}{2}\cos^2\text{x}-\frac14\sin^2\text{x}-\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}+\frac34\cos^2\text{x}+\frac14\sin^2\text{x}+\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}$
$=\frac{3}{2}\cos^2\text{x}-\frac12\sin^2\text{x}$
$=\frac{3}{2}(1-\sin^2\text{x})+\frac12\sin^2\text{x}$
$=\frac32-\frac{3}{2}\sin^2\text{x}+\frac12\sin^2\text{x}$
$=\frac32-\sin^2\text{x}$.
For $f(x)$ to be maximum, $\sin^2\text{x}$ must have minimum value, which is $0.$
$\therefore\frac32$ is the maximum value of $f(x).$

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