InterviewSolution
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InterviewSolution
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If f(x) = \(\begin{cases} \frac{log(1+ax)-log(1-bx)}{x} &, \quad x ≠{0}\\ k&, \quad \text x =0 \end{cases} \)and f(x) is continuous at x = 0, then the value of k is :A. a – b B. a + b C. log a + log b D. none of these |
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Answer» Option : (b) Formula : - (i) standard limit \(\lim\limits_{x \to 0}\frac{log(1-x)}{x}\) = 1 (ii) A function f(x) is said to be continuous at a point x=a of its domain, if \(\lim\limits_{x \to a}f(x)\) = f(a) \(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a) (iii) \(\lim\limits_{x \to a}{\{f(x)±g(x)}\}\) = 1 ± m, Where \(\lim\limits_{x \to a}f(x)\) = 1, \(\lim\limits_{x \to a}g(x)\) = m Given :- f(x) =\(\begin{cases} \frac{log(1+ax)-log(1-bx)}{x} &, \quad x ≠{0}\\ k&, \quad \text x =0 \end{cases} \) And, f(x) is continuous at x = 0 \(\lim\limits_{x \to 0}\frac{log(1+ax)-log(1-bx)}{x} \) = k Using formula (ii), \(a\lim\limits_{x \to 0}\frac{log(1+ax)}{ax} \) - \(b\lim\limits_{x \to 0}\frac{log(1-bx)}{bx} \) = k Using formula (i), a + b = k |
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