If f(x) = […] is continuous at x = 0, then the value of k is:

The correct answer is Option (1) → 8

f is continuous at x = 0 when the limit of f(x) as x → 0 equals f(0) = k.

Using \(1 - \cos 2\theta = 2\sin^2\theta\) with \(\theta = 2x\):

    \( \displaystyle \lim_{x \to 0}\frac{1-\cos 4x}{x^2} = \lim_{x \to 0}\frac{2\sin^2 2x}{x^2} \)

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

Since \(\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1\):

    \( = 8 \times 1 = 8 \)

So k = 8.

NCERT Exemplar:
Continuity and Differentiability