Test Index

MHTCET 11 May 2024 Shift 2 Paper

© examsnet.com

Question : 131 of 150

Marks: +1, -0

Let

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

\mathrm{f}(x)

is continuous at

x=0

, then the value of

a

is

8
4
$\frac{1}{2}$
2

Solution:

Concept:

For a function to be continuous at

x=0

, the value of the function at

x=0

must equal its limit as

x

approaches

0

.

Explanation:

Since

f(x)

is continuous at

x=0

, we write:

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

Substitute the given values:

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

Use the identity

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

:

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

Rewrite to apply the standard limit

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

:

a = 2 \lim_{x \to 0^{-}} \left( \frac{\sin 2x}{2x} \right)^2 \times 4

a = 2 \times (1)^2 \times 4 = 8

Answer:

a = 8

, so the correct option is A. 8.

© examsnet.com

Go to Question:

Prev Question

Next Question