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MHTCET 25 Apr 2025 Shift 1 Paper

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Question : 128 of 150

Marks: +1, -0

If

\mathrm{f}(x)=\frac{\sin \left(\pi \cos ^{2} x\right)}{3 x^{2}}, x \neq 0

is continuous at

x=0

\mathrm{f}(0)=

0
$\frac{\pi}{3}$
$\frac{-\pi}{3}$
$\frac{3}{\pi}$

Solution:

Concept:

For continuity at

x=0

,

f(0)

must equal

\lim_{x \to 0} f(x)

.

Explanation:

Since

f(x)

is continuous at

x=0

, we get:

f(0)=\lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{3x^2}

Rewrite

\pi \cos^2 x

as

\pi-\pi\sin^2 x

and use

\sin(\pi-\theta)=\sin\theta

:

\sin(\pi \cos^2 x)=\sin(\pi-\pi\sin^2 x)=\sin(\pi\sin^2 x)

So the limit becomes:

f(0)=\frac{1}{3}\lim_{x \to 0}\frac{\sin(\pi\sin^2 x)}{x^2}

Multiply and divide by

\pi\sin^2 x

to apply the standard limit

\lim_{t \to 0}\frac{\sin t}{t}=1

:

f(0)=\frac{1}{3}\lim_{x \to 0}\frac{\sin(\pi\sin^2 x)}{\pi\sin^2 x}\cdot\frac{\pi\sin^2 x}{x^2}

Now,

\lim_{x \to 0}\frac{\sin(\pi\sin^2 x)}{\pi\sin^2 x}=1

and

\lim_{x \to 0}\frac{\sin^2 x}{x^2}=1

.

Hence:

f(0)=\frac{1}{3}\times 1 \times \pi=\frac{\pi}{3}

Answer:

\frac{\pi}{3}

, i.e. option B.

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