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UPSC CDS 2 2021 Math Paper

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Question : 21 of 100

Marks: +1, -0

If

96-64a^{3}+\frac{8}{a^{6}}-\frac{48}{a^{3}}-t^{3}=0

a^{2}t + 4a^{3}

equal to ?

0
1
2
3

Solution:

Concept:

Rewriting the given equation as a perfect cube expression to find

t

in terms of

a

.

Explanation:

Given:

96 - 64a^{3} + \frac{8}{a^{6}} - \frac{48}{a^{3}} - t^{3} = 0

.

Rearrange:

96 - 64a^{3} + \frac{8}{a^{6}} - \frac{48}{a^{3}} = t^{3}

.

Observe that

96 = 3 \cdot \frac{2}{a^{2}} \cdot (4a)^{2}

and

\frac{-48}{a^{3}} = -3 \cdot \left(\frac{2}{a^{2}}\right)^{2} \cdot (4a)

.

Also

\frac{8}{a^{6}} = \left(\frac{2}{a^{2}}\right)^{3}

and

-64a^{3} = -(4a)^{3}

.

Thus the left side equals

\left(\frac{2}{a^{2}}\right)^{3} - (4a)^{3} - 3\left(\frac{2}{a^{2}}\right)^{2}(4a) + 3\left(\frac{2}{a^{2}}\right)(4a)^{2}

.

This matches the expansion of

\left(\frac{2}{a^{2}} - 4a\right)^{3}

.

Hence

t^{3} = \left(\frac{2}{a^{2}} - 4a\right)^{3}

\Rightarrow

t = \frac{2}{a^{2}} - 4a

.

Now compute

a^{2}t + 4a^{3}

:

a^{2}\left(\frac{2}{a^{2}} - 4a\right) + 4a^{3} = 2 - 4a^{3} + 4a^{3} = 2

.

Answer:

2

(Option C).

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