Test Index

UPSC NDA 2 2022 Mathematics Solved Paper

© examsnet.com

Question : 79 of 120

Marks: +1, -0

\text{What is the maximum value of } 3(\sin x - \cos x) + 4(\cos^{3} x - \sin^{3} x) \text{ ?}

1
$\sqrt{2}$
$\sqrt{3}$
2

Solution:

Concept:

Use triple-angle identities:

\sin 3x = 3\sin x - 4\sin^3 x

and

\cos 3x = 4\cos^3 x - 3\cos x

.

Explanation:

Let

f(x) = 3(\sin x - \cos x) + 4(\cos^3 x - \sin^3 x)

.

Expand:

f(x) = 3\sin x - 3\cos x + 4\cos^3 x - 4\sin^3 x

.

Group terms:

f(x) = (3\sin x - 4\sin^3 x) + (4\cos^3 x - 3\cos x)

.

Apply triple-angle identities:

(3\sin x - 4\sin^3 x) = \sin 3x

, and

(4\cos^3 x - 3\cos x) = \cos 3x

.

So

f(x) = \sin 3x + \cos 3x

.

To find maximum, differentiate:

f'(x) = 3\cos 3x - 3\sin 3x

.

Set

f'(x) = 0

:

3(\cos 3x - \sin 3x) = 0 \Rightarrow \cos 3x = \sin 3x \Rightarrow \tan 3x = 1

.

Thus

3x = \frac{\pi}{4}

(principal value for maximum).

Maximum value:

f_{\text{max}} = \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}

.

Answer:

The maximum value is

\sqrt{2}

. Hence the correct option is B.

© examsnet.com

Go to Question:

Prev Question

Next Question