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UPSC NDA 2 2021 Mathematics Solved Paper

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Question : 28 of 120

Marks: +1, -0

Let

\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}

0 \le x, y z \le 1

. What is the value of

x^{1000} + y^{1001} + z^{1002}

?

0
1
3
6

Solution:

Concept:

The maximum value of

\sin^{-1} \theta

for

\theta

in

[0,1]

is

\frac{\pi}{2}

.

The sum of three such terms reaches

\frac{3\pi}{2}

only when each term attains its maximum.

Explanation:

Given

\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}

where

0 \le x,y,z \le 1

.

Each

\sin^{-1}

lies between

0

and

\frac{\pi}{2}

, so the maximum possible sum is

\frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2}

.

For the sum to equal

\frac{3\pi}{2}

, each term must be

\frac{\pi}{2}

.

Thus

\sin^{-1}x = \sin^{-1}y = \sin^{-1}z = \frac{\pi}{2}

.

Taking sine on both sides gives

x = y = z = \sin(\frac{\pi}{2}) = 1

.

Now compute

x^{1000} + y^{1001} + z^{1002} = 1^{1000} + 1^{1001} + 1^{1002} = 1 + 1 + 1 = 3

.

Answer:

3

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