[Solved] The function f(x) = {1/x loge ((1+x/a)/(1x/b)), x < 0 f(x) = k, x = 0, f(x) = (cos^2  sin^2  1)/(√(x^2 +1)1, x > 0) is continuous at x = 0, then 1/a + 1/b + 4/k is equal to
The function f(x) = \(\begin{cases} \frac{1}{x}log_e \Big(\frac{1 + \frac{x}{a}}{1  \frac{x}{b}}\Big) & \quad, x < 0\\ k & \quad, x = 0 \\ \frac{cos^2x  sin^2 x1}{\sqrt{x^2 + 1}1} & \quad, x > 0 \end{cases}\)
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

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