If the function f defined on -1-3- 1-3 by f x - left matrix lend matrix right- 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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Byju's Answer

Standard XII

Mathematics

Continuity in an Interval

If the functi...

Question

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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Solution

As $f (x)$ is continuous,
$lim x \to 0 f (x) = f (0) = k$
$\Rightarrow lim x \to 0 \frac{1}{x} {log}_{e} (\frac{1 + 3 x}{1 - 2 x}) = k$

$\Rightarrow lim x \to 0 \frac{3 log (1 + 3 x)}{3 x} - lim x \to 0 \frac{(- 2) log (1 - 2 x)}{(- 2 x)} = k$
$\Rightarrow 3 + 2 = k \Rightarrow k = 5$

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Q. If the function

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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

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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}

is continuous, then k is equal to?

Q. The function

f (x)

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If the function f defined on (−13,13) by f(x)=⎧⎪⎨⎪⎩1xloge(1+3x1−2x), when x≠0k, when x=0 is continuous, then k is equal to

As f(x) is continuous, limx→0f(x)=f(0)=k ⇒limx→01xloge(1+3x1−2x)=k ⇒limx→03log(1+3x)3x−limx→0(−2)log(1−2x)(−2x)=k ⇒3+2=k⇒k=5

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

As $f (x)$ is continuous,
$lim x \to 0 f (x) = f (0) = k$
$\Rightarrow lim x \to 0 \frac{1}{x} {log}_{e} (\frac{1 + 3 x}{1 - 2 x}) = k$

$\Rightarrow lim x \to 0 \frac{3 log (1 + 3 x)}{3 x} - lim x \to 0 \frac{(- 2) log (1 - 2 x)}{(- 2 x)} = k$
$\Rightarrow 3 + 2 = k \Rightarrow k = 5$