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UPSC NDA I 2025 Mathematics Paper

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Question : 22 of 120

Marks: +1, -0

If

k<(\sqrt{2}+1)^{3} < k+2,

where k is a natural number, the value of

k

?

11
13
15
17

Solution:

Concept:

Simplify the expression

(\sqrt{2}+1)^3

and approximate its value to find the integer

k

that satisfies the inequality

k < (\sqrt{2}+1)^3 < k+2

.

Explanation:

First, expand

(\sqrt{2}+1)^3

using

(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

.

(\sqrt{2}+1)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(1) + 3(\sqrt{2})(1)^2 + 1^3 = 2\sqrt{2} + 3\cdot 2 \cdot 1 + 3\sqrt{2} + 1 = 2\sqrt{2} + 6 + 3\sqrt{2} + 1 = 5\sqrt{2} + 7

.

Approximate

\sqrt{2} \approx 1.414

:

5 \times 1.414 + 7 = 7.07 + 7 = 14.07

.

The inequality becomes

k < 14.07 < k+2

.

From

k < 14.07

, we have

k \le 13

. From

14.07 < k+2

, we get

k > 12.07

, so

k \ge 13

.

The only natural number satisfying both is

k = 13

.

Answer:

k = 13

(Option B).

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