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UPSC NDA I 2019 Mathematics Solved Paper

Question Numbers: 33-35

Direction: Consider the following for the next 03 (three) items:

p = X\cos\theta - Y\sin\theta

q = X\sin\theta + Y\cos\theta

and

p^2 + 4pq + q^2 = AX^2 + BY^2

0 \leq \theta \leq \frac{\pi}{2}

Question : 35 of 120

Marks: +1, -0

What is the value of B?

Solution:

Concept:

Use algebraic expansion and comparison of coefficients to find

B

Explanation:

Given

p = X\cos\theta - Y\sin\theta

and

q = X\sin\theta + Y\cos\theta

Compute

p^2 = X^2\cos^2\theta + Y^2\sin^2\theta - 2XY\cos\theta\sin\theta

Compute

q^2 = X^2\sin^2\theta + Y^2\cos^2\theta + 2XY\cos\theta\sin\theta

Compute

4pq = 4\left[ X^2\cos\theta\sin\theta + XY\cos^2\theta - XY\sin^2\theta - Y^2\sin\theta\cos\theta \right]

Add

p^2 + q^2 + 4pq

= X^2(\cos^2\theta + \sin^2\theta + 4\cos\theta\sin\theta) + Y^2(\sin^2\theta + \cos^2\theta - 4\sin\theta\cos\theta) + 4XY(\cos^2\theta - \sin^2\theta)

Simplify using identities:

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

2\cos\theta\sin\theta = \sin 2\theta

, and

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

Thus,

p^2 + 4pq + q^2 = X^2(1 + 2\sin 2\theta) + Y^2(1 - 2\sin 2\theta) + 4XY\cos 2\theta

This must equal

AX^2 + BY^2

for all

X

Y

. Hence the

XY

term must be zero:

\cos 2\theta = 0

Then

B = 1 - 2\sin 2\theta

. With

\cos 2\theta = 0

\theta = 45^\circ

(or

\pi/4

), so

\sin 2\theta = 1

Therefore

B = 1 - 2(1) = -1

Answer:

B = -1

(Option A).

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