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UPSC CDS 1 2021 Math Paper

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Question : 98 of 100

Marks: +1, -0

If

p + \frac{1}{p} = 4

then what is the value of

p^{2} + \frac{1}{p^{2}}

?

10
14
21
9

Solution:

Concept:

Square the given equation and use the identity

(a+b)^2 = a^2 + b^2 + 2ab

to find the required expression.

Explanation:

We are given

p + \frac{1}{p} = 4

.

Square both sides:

\left(p + \frac{1}{p}\right)^2 = 4^2

.

Using the identity:

p^2 + 2 \cdot p \cdot \frac{1}{p} + \frac{1}{p^2} = 16

.

Simplify:

p^2 + 2 + \frac{1}{p^2} = 16

.

Subtract 2 from both sides:

p^2 + \frac{1}{p^2} = 14

.

Thus the required value is 14.

(Note: This matches the shortcut: if

x + \frac{1}{x} = k

, then

x^2 + \frac{1}{x^2} = k^2 - 2

; here

k=4

, so

4^2 - 2 = 14

.)

Answer:

p^2 + \frac{1}{p^2} = 14

(Option B).

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