Continuity and Differentiability
NCERT Class 12
CBSE
Grade 12
Question:
If $f(x) = \begin{cases} \dfrac{1 - \cos 4x}{8x^2}, & x
eq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$, then $k$ is:
(a) $1$
(b) $2$
(c) $\dfrac{1}{2}$
(d) $4$
Step-by-Step Solution
$\lim_{x \to 0} \dfrac{2\sin^2 2x}{8x^2} = 1 \Rightarrow k = 1$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating limit $= 1$: 1.0 Mark
Correct Answer: $1$
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