Continuity and Differentiability Question - CBSE

Question

If $f(x) = \begin{cases} \dfrac{1-\cos 4x}{x^2}, & x < 0 \ a, & x = 0 \ \dfrac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a$ is:
(a) $8$
(b) $4$
(c) $2$
(d) $16$

Accepted Answer

\lim_{x o 0-} (1-cos 4x)/x^2 = 8 \implies a = 8. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Selecting $8$: 1.0 Mark

Question:

Step-by-Step Solution

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