If , then for all values of :

Explanation: To find the range of A , we simplify the expression: 1. Simplify the Expression A = cos 2 θ + sin 4 θ Substitute cos 2 θ = 1 − sin 2 θ : A = ( 1 − sin 2 θ ) + sin 4 θ A = sin 4 θ − sin 2 θ + 1 2. Perfect Square Method Let t = sin 2 θ . Since 0 ≤ sin 2 θ ≤ 1 , we have 0 ≤ t ≤ 1 . The expression becomes: A = t 2 − t + 1 To find the minimum and maximum, we complete the square: A = ( t 2 − t + 4 1 ) − 4 1 + 1 A = ( t − 2 1 ) 2 + 4 3 3. Finding the Maximum and Minimum Values Minimum Value: Occurs when ( t − 2 1 ) 2 = 0 , i.e., t = 2 1 . A m i n = 0 + 4 3 = 4 3 Maximum Value: Occurs at the boundaries of t ∈ [ 0 , 1 ] . If t = 0 : A = ( 0 − 2 1 ) 2

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NIMCET 2009 — Mathematics PYQ

NIMCET | Mathematics | 2009

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

Choose the correct answer:

B.

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

C.

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

D.

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

(Correct Answer)

Explanation

To find the range of A, we simplify the expression:

1. Simplify the Expression

A = \cos^2 \theta + \sin^4 \theta

Substitute $cos 2 θ = 1 - sin 2 θ$ :

A = (1 - \sin^2 \theta) + \sin^4 \theta

A = \sin^4 \theta - \sin^2 \theta + 1

2. Perfect Square Method Let $t = sin 2 θ$ . Since $0 \leq sin 2 θ \leq 1$ , we have $0 \leq t \leq 1$ . The expression becomes:

A = t^2 - t + 1

To find the minimum and maximum, we complete the square:

A = \left( t^2 - t + \frac{1}{4} \right) - \frac{1}{4} + 1

A = \left( t - \frac{1}{2} \right)^2 + \frac{3}{4}

3. Finding the Maximum and Minimum Values

Minimum Value: Occurs when $(t - 21) 2 = 0$ , i.e., $t = 21$ .

$A_{\min} = 0 + \frac{3}{4} = \frac{3}{4}$
Maximum Value: Occurs at the boundaries of $t \in [0, 1]$ . If $t = 0$ : $A = (0 - 21) 2 + 43 = 41 + 43 = 1$ . If $t = 1$ : $A = (1 - 21) 2 + 43 = 41 + 43 = 1$ .

$A_{\max} = 1$

Conclusion: The range of A is $43 \leq A \leq 1$ .

Correct Option: (d) $43 \leq A \leq 1$