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MHTCET 22 Apr 2025 Shift 2 Paper

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Question : 137 of 150

Marks: +1, -0

If

f(x)=\left\{\begin{array}{ccc}\frac{1-\cos 4 x}{x^{2}} & , \text { if } & x < 0 \\ \frac{a}{\sqrt{x}} & , \text { if } & x=0 \\ \frac{(16+\sqrt{x})^{\frac{1}{2}}-4}{16} & , \text { if } & x > 0\end{array}\right.

is continuous at

x=0

a=

4
8
-4
-8

Solution:

Concept:

For continuity at

x=0

, the left-hand limit, right-hand limit, and

f(0)

must all be equal.

Explanation:

Given

f(x)

is continuous at

x=0

, we have

\lim_{x \to 0^{-}} f(x) = f(0)

.

So,

\lim_{x \to 0} \frac{1-\cos 4x}{x^2} = a

.

Use the identity

1-\cos 4x = 2\sin^2 2x

.

Thus,

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

.

Rewrite as

8 \lim_{x \to 0} \left(\frac{\sin 2x}{2x}\right)^2 = a

.

Since

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

, we get

8(1)^2 = a

.

Therefore,

a = 8

.

Answer:

a = 8

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