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MHTCET 18 Apr 2026 Shift 2 Paper

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Question : 103 of 150

Marks: +1, -0

Let

f_{k}(x)=\frac{1}{k}\left(\cos ^{k} x+\sin ^{k} x\right)

k \in N

f_{6}(x)-f_{4}(x)=\ldots

$\frac{1}{6}$
$-\frac{1}{12}$
$-\frac{1}{6}$
$\frac{1}{12}$

Solution:

Concept:

Use trigonometric identities for

\sin^6 x + \cos^6 x

and

\sin^4 x + \cos^4 x

to simplify the difference.

Explanation:

Given:

f_k(x) = \frac{1}{k}\left(\cos^k x + \sin^k x\right)

So,

f_6(x) = \frac{1}{6}\left(\sin^6 x + \cos^6 x\right)

And

f_4(x) = \frac{1}{4}\left(\sin^4 x + \cos^4 x\right)

Use

\sin^2 x + \cos^2 x = 1

.

\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x(\sin^2 x + \cos^2 x)

= 1 - 3\sin^2 x \cos^2 x

\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x

= 1 - 2\sin^2 x \cos^2 x

Now substitute:

f_6(x) - f_4(x) = \frac{1}{6}(1 - 3\sin^2 x \cos^2 x) - \frac{1}{4}(1 - 2\sin^2 x \cos^2 x)

= \frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x - \frac{1}{4} + \frac{1}{2}\sin^2 x \cos^2 x

The

\sin^2 x \cos^2 x

terms cancel:

f_6(x) - f_4(x) = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12}

Answer:

f_6(x) - f_4(x) = -\frac{1}{12}

Correct option: B.

-\frac{1}{12}

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