If the function \(f(x) = \begin{cases}\displaystyle\frac{1 - \cos 6x}{x^{2}}, & x \lt 0\\ a, & x = 0\\ \displaystyle\frac{\sqrt{x}}{\sqrt{81 + \sqrt{x}} - 9}, & x \gt 0\end{cases}\) is continuous at \(x = 0\), then the value of a is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 18
For continuity at 0, both one-sided limits must equal a.
Left-hand limit, using \(1 - \cos 2\theta = 2\sin^{2}\theta\) with \(\theta = 3x\):
\( \displaystyle \frac{1 - \cos 6x}{x^{2}} = \frac{2\sin^{2} 3x}{x^{2}} = 18\left(\frac{\sin 3x}{3x}\right)^{2} \)
\( \displaystyle \lim_{x \to 0^-} f(x) = 18 \times 1^{2} = 18 \)
Right-hand limit, rationalising the denominator:
\( \displaystyle \frac{\sqrt{x}}{\sqrt{81 + \sqrt{x}} - 9} = \frac{\sqrt{x}\left(\sqrt{81 + \sqrt{x}} + 9\right)}{(81 + \sqrt{x}) - 81} \)
\( \displaystyle = \sqrt{81 + \sqrt{x}} + 9 \)
\( \displaystyle \lim_{x \to 0^+} f(x) = 9 + 9 = 18 \)
Both limits are 18, so a = 18.