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UPSC NDA I 2022 Mathematics Solved Paper

Question : 59 of 120

Marks: +1, -0

f(x) = \sqrt{1 - (x-1)^2} ?

Solution:

Concept:

A square root function

f(x) = \sqrt{g(x)}

is defined only when

g(x) \ge 0

Explanation:

The given function is

f(x) = \sqrt{1 - (x-1)^2}

For it to be defined, the expression under the square root must be non‑negative:

1 - (x-1)^2 \ge 0

Rearrange:

(x-1)^2 \le 1

Taking square roots on both sides (since both sides are non‑negative):

-1 \le x-1 \le 1

Add

1

to all parts:

0 \le x \le 2

Thus the domain is the closed interval

[0, 2]

Answer:

[0, 2]

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