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UPSC CDS I 2025 Mathematics Paper

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Question : 21 of 100

Marks: +1, -0

If

2 \sin^4 \alpha + 2 \cos^4 \alpha - 1 = 0

0 < \alpha < \frac{\pi}{2}

\sin 2\alpha + \cos 2\alpha

equal to ?

0
1
$\frac{\sqrt{3}+1}{2}$
$\frac{\sqrt{3}-1}{2}$

Solution:

Concept:

Use the identity

\sin^4\alpha + \cos^4\alpha = (\sin^2\alpha + \cos^2\alpha)^2 - 2\sin^2\alpha\cos^2\alpha

to simplify the given equation.

Explanation:

Start with

2\sin^4\alpha + 2\cos^4\alpha - 1 = 0

.

Divide by 2:

\sin^4\alpha + \cos^4\alpha = \frac12

.

Rewrite using the identity:

(\sin^2\alpha + \cos^2\alpha)^2 - 2\sin^2\alpha\cos^2\alpha = \frac12

.

Since

\sin^2\alpha + \cos^2\alpha = 1

, we get

1 - 2\sin^2\alpha\cos^2\alpha = \frac12

.

Thus

2\sin^2\alpha\cos^2\alpha = \frac12

, so

\sin^2\alpha\cos^2\alpha = \frac14

.

Take square root (positive because

0 < \alpha < \frac{\pi}{2}

):

\sin\alpha\cos\alpha = \frac12

.

Use

\sin2\alpha = 2\sin\alpha\cos\alpha = 2 \times \frac12 = 1

.

Hence

2\alpha = 90^\circ

, so

\alpha = 45^\circ

.

Then

\sin2\alpha = \sin90^\circ = 1

and

\cos2\alpha = \cos90^\circ = 0

.

Therefore

\sin2\alpha + \cos2\alpha = 1 + 0 = 1

.

Answer:

1

(Option B)

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