If the function f defined on -13-13 by fx-1xloge 1-3x1-2x - when x - 0 k- when x-0-is continuous- then k is equal to

As f(x) is continuous, x → 0limf(x)=f(0)=k ⇒x → 0lim1xlog_e ( 1+3x1 2x )=k ⇒x → 0lim3log(1+3x)3x x → 0lim( 2)log(1 2x)( 2x)=k ⇒ 3+2=k ⇒ k=5 ...

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Continuity in an Interval

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If the function $f$ defined on $(- \frac{1}{3}, \frac{1}{3})$ by $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{1}{x} {log}_{e} (\frac{1 + 3 x}{1 - 2 x}), when x \neq 0 k, when x = 0 \end{matrix}$
is continuous, then $k$ is equal to

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{x} {log}_{e} ⎛ ⎜ ⎜ ⎝ \frac{1 + \frac{x}{a}}{1 - \frac{x}{b}} ⎞ ⎟ ⎟ ⎠, & x < 0 k, & x = 0 \frac{{cos}^{2} x - {sin}^{2} x - 1}{\sqrt{x^{2} + 1} - 1}, & x > 0 \end{matrix}

x = 0,

\frac{1}{a} + \frac{1}{b} + \frac{4}{k}

f

(- \frac{1}{3}, \frac{1}{3})

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{1}{x} {log}_{e} (\frac{1 + 3 x}{1 - 2 x}), when x \neq 0 k, when x = 0 \end{matrix}

k

f

f (x) = \frac{1}{x} - \frac{k - 1}{e^{2 x} - 1}

x \neq 0

x = 0

(k, f (0))

f : (- 1, \infty) \to R

f (0) = 1

f (x) = \frac{1}{x} {log}_{e} (1 + x), x \neq 0

f

(\frac{π}{6}, \frac{π}{3})

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{2} cos x - 1}{cot x - 1}, & x \neq \frac{π}{4} k, & x = \frac{π}{4} \end{matrix}

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