Test Index

UPSC NDA I 2016 Mathematics Solved Paper

Show Para Hide Para

Question Numbers: 9-10

Direction: Consider the function

f(θ) = 4(sin^{2} θ + cos^{4} θ)

© examsnet.com

Question : 9 of 120

Marks: +1, -0

What is the maximum value of the function f(θ)?

1
2
3
4

Solution:

Concept:

The range of

\sin 2\theta

is used to find the maximum of

f(\theta)

.

Explanation:

Given

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

.

Rewrite

\sin^2 \theta = 1 - \cos^2 \theta

, so

f(\theta) = 4(1 - \cos^2 \theta + \cos^4 \theta)

.

This is quadratic in

\cos^2 \theta

. Simplify further:

f(\theta) = 4[1 - \cos^2 \theta(1 - \cos^2 \theta)] = 4(1 - \sin^2 \theta \cos^2 \theta)

.

Since

\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta

, we get

f(\theta) = 4 - (\sin 2\theta)^2 = 4 - \sin^2 2\theta

.

Note that

0 \le \sin^2 2\theta \le 1

because

-1 \le \sin 2\theta \le 1

.

Thus

3 \le 4 - \sin^2 2\theta \le 4

, so

3 \le f(\theta) \le 4

.

The maximum value is

4

.

Answer:

4

(Option D).

© examsnet.com

Go to Question:

Prev Question

Next Question