Test Index

UPSC CDS 2 2021 Math Paper

© examsnet.com

Question : 50 of 100

Marks: +1, -0

What is the maximum value of

1 + 2\sin^{2}\theta \cos^{2}\theta - \sin^{4}\theta - \cos^{4}\theta

0^{\circ} < \theta < 90^{\circ}

?

1
2
3
4

Solution:

Concept:

The expression simplifies to

\sin^2 2\theta

, whose maximum value is

1

.

Explanation:

Let

a = \sin^2\theta

and

b = \cos^2\theta

. Then

a + b = 1

.

Rewrite the given expression:

1 + 2\sin^2\theta \cos^2\theta - \sin^4\theta - \cos^4\theta = 1 + 2ab - (a^2 + b^2)

.

Use the identity:

a^2 + b^2 = (a+b)^2 - 2ab = 1 - 2ab

.

Substitute:

1 + 2ab - (1 - 2ab) = 1 + 2ab - 1 + 2ab = 4ab

.

So

4\sin^2\theta \cos^2\theta = (2\sin\theta \cos\theta)^2 = (\sin 2\theta)^2 = \sin^2 2\theta

.

Since

0^{\circ} < \theta < 90^{\circ}

,

0^{\circ} < 2\theta < 180^{\circ}

. The maximum value of

\sin^2 2\theta

is

1

(occurs at

2\theta = 90^{\circ}

, i.e.,

\theta = 45^{\circ}

).

Thus the maximum value of the expression is

1

.

Answer:

A. 1

© examsnet.com

Go to Question:

Prev Question

Next Question