- k - - X-1lend cases is continuous at x -1- then the value of k isA- eB- 1C- 0D- -1

The correct option is B 1 nbsp; nbsp; nbsp;Since f is continuous at x=1, lim_x → 1 f(x)= f(1) ⇒lim_x → 1log_ex/x 1=k ⇒lim_x → 11/x=k [By L Hopital' ...

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Standard XII

Mathematics

Number of Elements in a Cartesian Product

∼∼ k ∼∼∼∼∼∼∼∼...

Question

$If f (x) = {\begin{matrix} \frac{{log}_{e} x}{x - 1}; x \neq 1 k; x = 1 \end{matrix}$ is continuous at $x = 1$ , then the value of $k$ is

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Solution

The correct option is B $1$
Since $f$ is continuous at $x = 1$ ,
$lim x \to 1 f (x) = f (1)$
$\Rightarrow lim x \to 1 \frac{{log}_{e} x}{x - 1} = k$
$\Rightarrow lim x \to 1 \frac{1}{x} = k [By L Hopital's Rule]$
$\Rightarrow k = 1$

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If f(x)={logexx−1 ;x≠1 k ;x=1 is continuous at x=1, then the value of k is

The correct option is B 1 Since f is continuous at x=1, limx→1f(x)=f(1) ⇒limx→1logexx−1=k ⇒limx→11x=k [By L Hopital's Rule] ⇒k=1

$If f (x) = {\begin{matrix} \frac{{log}_{e} x}{x - 1}; x \neq 1 k; x = 1 \end{matrix}$ is continuous at $x = 1$ , then the value of $k$ is

The correct option is B $1$
Since $f$ is continuous at $x = 1$ ,
$lim x \to 1 f (x) = f (1)$
$\Rightarrow lim x \to 1 \frac{{log}_{e} x}{x - 1} = k$
$\Rightarrow lim x \to 1 \frac{1}{x} = k [By L Hopital's Rule]$
$\Rightarrow k = 1$