Aspire Faculty ID #16442 · Topic: NIMCET 2009 · Just now
NIMCET 2009

If $A = \cos^2\theta + \sin^4\theta$, then for all values of $\theta$:

Solution

$A = \cos^2\theta + \sin^4\theta$ 
Let $\sin^2\theta = t$, $0 \le t \le 1$ 
 Then $A = (1 - t) + t^2 = t^2 - t + 1$ 
 This is a quadratic in $t$. 
 Minimum value at $t = \dfrac{1}{2}$: 
$A_{\min} = \left(\dfrac{1}{2}\right)^2 - \dfrac{1}{2} + 1 = \dfrac{3}{4}$ 
 Maximum value at endpoints $t=0$ or $t=1$: 
$A_{\max} = 1$ 
 Thus: ${\dfrac{3}{4} \le A \le 1}$

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