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# [Solved] The function f(x) = {1/x loge ((1+x/a)/(1-x/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

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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

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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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