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If $A=\sin ^{2}x+\cos ^{4}x,$ then for all real $x$

Option 1)
$1\leq A\leq 2\;$

Option 2)
$\; \frac{3}{4}\leq A\leq \frac{13}{16}\;$

Option 3)
$\; \frac{3}{4}\leq A\leq 1\;$

Option 4)
$\; \frac{13}{16}\leq A\leq 1$

Answers (1)

best_answer

As we learnt in

Trigonometric Identities -

$\sin ^{2}\Theta + \cos ^{2}\Theta = 1$

$1 + \tan ^{2}\Theta = \sec ^{2}\Theta$

$1 + \cot ^{2}\Theta = cosec ^{2}\Theta$

- wherein

They are true for all real values of $\Theta$

$A = \sin ^{2}x+ \cos^{4}x =1 - \cos^{2}x +\cos^{4}x = \cos^{4}x - 2 \cos^{2}x \times \frac{1}{2}+\frac{1}{4}+\frac{3}{4}$

$A = (\cos^{2}x - \frac{1}{2})^{2}+\frac{3}{4}$

Range of A depends on $\cos^{2}x$ . Minimum value occurs at $cos^{2}x=\frac{1}{2},\ \: \: \: \: \: \: \: \: A=\frac{3}{4}$

Maximum value occurs at $cos^{2}x=\frac{1}{2},\ \, \, \, \, \, \, \, A = \frac{1}{4}\: \: \: \: \frac{1-3}{4} = 1$

For all other values of $\cos^{2}x$ , A varies between these two values[ function is continues]

$A \varepsilon \left [ \frac{3}{4} , 1\right ]$

Option 1)

$1\leq A\leq 2\;$

This option is incorrect.

Option 2)

$\; \frac{3}{4}\leq A\leq \frac{13}{16}\;$

This option is incorrect.

Option 3)

$\; \frac{3}{4}\leq A\leq 1\;$

This option is correct.

Option 4)

$\; \frac{13}{16}\leq A\leq 1$

This option is incorrect.

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