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UPSC CDS I 2024 Math Paper

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Question : 21 of 100

Marks: +1, -0

If

p = \sqrt[3]{(a+\sqrt{a^{2}+b^{3}})} + \sqrt[3]{(a-\sqrt{a^{2}+b^{3}})}

p^{3} + 3bp

equal to ?

-2a
a
2a
3a

Solution:

Concept:

Use simple substitution with convenient values to evaluate the expression.

Explanation:

Choose

a = 1

and

b = 0

.

Then

p = \sqrt[3]{(1+\sqrt{1^{2}+0^{3}})} + \sqrt[3]{(1-\sqrt{1^{2}+0^{3}})} = \sqrt[3]{(1+1)} + \sqrt[3]{(1-1)} = \sqrt[3]{2} + 0 = \sqrt[3]{2}

.

Now compute

p^{3} + 3bp

:

p^{3} = (\sqrt[3]{2})^{3} = 2

and

3bp = 3 \times 0 \times \sqrt[3]{2} = 0

.

So

p^{3} + 3bp = 2

.

Check the options with

a = 1

:

A:

-2a = -2

, B:

a = 1

, C:

2a = 2

, D:

3a = 3

.

Only option C gives

2

, which matches the computed value.

Answer:

Option C (2a)

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