Consider the function defined by . If and be respectively the number of points at which is not continuous and is not differentiable, then is

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Solution

Q.
Consider the function $f : (0, \infty) \to R$ defined by $f (x) = e^{- | \log_{e} x |}$ . If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is [2024]

1 1

2 0

3 2

4 3

Ans.
(1)

Given, $f (x) = e^{- | \log_{e} x |}$

$\Rightarrow$ $f (x) =$ ${\begin{matrix} 1 / e^{- \log_{e} x}, & 0 < x < 1 \\ 1 / e^{\log_{e} x}, & x \geq 1 \end{matrix} = {\begin{matrix} x, & 0 < x < 1 \\ 1 / x, & x \geq 1 \end{matrix}$

$∴$ $f (x)$ is continuous everywhere for $x > 0$ but not differentiable at $x = 1 .$

Thus, $m = 0, n = 1$

Hence, $m + n = 1$

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