The function f(x) = \(\begin{cases} \frac{1}{x}log_e \Big(\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}\Big) & \quad, x < 0\\ k & \quad, x = 0 \\ \frac{cos^2x - sin^2 x-1}{\sqrt{x^2 + 1}-1} & \quad, x > 0 \end{cases}\)
is continuous at x = 0, then \(\frac{1}{a} +\frac{1}{b}+\frac{4}{k} \) is equal to
(1) -5
(2) 6
(3) -2
(4) 4
Correct answer: (1) -5
Explanation:
If ƒ(x) is continuous at x = 0, RHL = LHL = ƒ(0)
\(\lim\limits_{x \to 0^+} f(x)\) = \(\lim\limits_{x \to 0^+}\) \(\frac{cos^2x - sin^2 x - 1}{\sqrt{x^2 + 1}-1}\).\(\frac{\sqrt{x^2 + 1} + 1}{\sqrt{x^2 + 1}+1}\) (Rationalisation)
\(\lim\limits_{x \to 0^+}\)- \(\frac{2sin^2x}{x^2}\).\((\sqrt{x^2 + 1}+1)\) = -4
\(\lim\limits_{x \to 0^-} f(x)\) = \(\lim\limits_{x \to 0^-} \frac{1}{x} \ell n\) \(\Big(\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}\Big)\)
\(\lim\limits_{x \to 0^-}\) \(\frac{\ell n (1 + \frac{x}{a})}{(\frac{x}{a}).a} \) + \(\frac{\ell n (1 - \frac{x}{b})}{(-\frac{x}{b}).b} \)
= \(\frac{1}{a}\) + \(\frac{1}{b}\)
So, \(\frac{1}{a}\) + \(\frac{1}{b}\) = -4 = k
⇒ \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{4}{k}\)
= 4 - 1 = -5