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MHTCET 23 Apr 2025 Shift 2 Paper

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Question : 102 of 150

Marks: +1, -0

If

\sin \theta=\frac{1}{2}\left(x+\frac{1}{x}\right)

\sin 3 \theta+\frac{1}{2}\left(x^{3}+\frac{1}{x^{3}}\right)=

0
1
$\frac{1}{4}$
2

Solution:

Concept:

Use the expansion of

\left(x+\frac{1}{x}\right)^3

and the identity

\sin 3\theta = 3\sin\theta - 4\sin^3\theta

.

Explanation:

Given

\sin\theta = \frac{1}{2}\left(x+\frac{1}{x}\right)

, we get:

x+\frac{1}{x}=2\sin\theta

Cubing both sides:

\left(x+\frac{1}{x}\right)^3 = 8\sin^3\theta

Expanding the left side:

x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=8\sin^3\theta

Substitute

x+\frac{1}{x}=2\sin\theta

:

x^3+\frac{1}{x^3}+6\sin\theta=8\sin^3\theta

Rearrange:

x^3+\frac{1}{x^3}+6\sin\theta-8\sin^3\theta=0

Factor the sine terms:

x^3+\frac{1}{x^3}+2\left(3\sin\theta-4\sin^3\theta\right)=0

Using

\sin 3\theta = 3\sin\theta-4\sin^3\theta

:

x^3+\frac{1}{x^3}+2\sin 3\theta=0

Divide by

2

:

\frac{1}{2}\left(x^3+\frac{1}{x^3}\right)+\sin 3\theta=0

This equals the required expression.

Answer:

The value is

0

, which is option A.

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