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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 55-56

Consider the following for the next items that follow:

Let

f(x)=\sqrt{2-x}+\sqrt{2+x}

Question : 56 of 120

Marks: +1, -0

What is the greatest value of the function?

Solution:

Concept:

To find the greatest value of

f(x)=\sqrt{2-x}+\sqrt{2+x}

, we square and examine the expression under the square root.

Explanation:

Let

y=f(x)

. Squaring both sides gives:

y^2 = (2-x)+(2+x)+2\sqrt{(2-x)(2+x)} = 4 + 2\sqrt{4-x^2}

The value of

y^2

is maximum when

\sqrt{4-x^2}

is maximum, i.e., when

4-x^2

is largest.

Since the domain is

x\in[-2,2]

4-x^2

is greatest at

x=0

(where

4-x^2=4

Thus

y^2_{\text{max}} = 4 + 2\sqrt{4} = 4+4=8

, so

y_{\text{max}} = \sqrt{8}

Therefore, the greatest value of the function is

\sqrt{8}

Answer:

\sqrt{8}

(Option C)

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