Let k be a non - zero real number. If f x = ( e x − 1 ) 2 sin x k log 1 + x 4 , x ≠ 0 12 , x = 0 is a continuous function at x = 0 , then the value of k is

If the function is continuous then lim x → 0 f x = f 0 lim x → 0 ( e x − 1 ) 2 sin x k . log 1 + x 4 = 12 ⇒ lim x → 0 e x - 1 x 2 sin x k k . x k . log 1 + x 4 4 . x 4 = 12 ⇒ 1 1 k . 1 4 = 12 ⇒ 4 k = 12 ⇒ k = 3

Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityContinuity- MiscellaneousMedium2 minPYQ_2015

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Let $k$ be a non - zero real number. If $f (x) = \{\begin{cases} \frac{{(e^{x} - 1)}^{2}}{\sin (\frac{x}{k}) \log (1 + \frac{x}{4})} \begin{matrix} , & x \neq 0 \end{matrix} \\ \begin{matrix} 12, & x = 0 \end{matrix} \end{cases}$ is a continuous function at $x = 0$ , then the value of $k$ is

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Answer:

Solution:

If the function is continuous then

$\lim_{x \to 0} f (x) = f (0)$

$\lim_{x \to 0} \frac{{(e^{x} - 1)}^{2}}{\sin \frac{x}{k} . \log (1 + \frac{x}{4})} = 12$

$\Rightarrow$ $\lim_{x \to 0} \frac{{(\frac{e^{x} - 1}{x})}^{2}}{\frac{\sin \frac{x}{k}}{k . \frac{x}{k}} . \frac{\log (1 + \frac{x}{4})}{4 . \frac{x}{4}}}$ = 12

$\Rightarrow$ $\frac{1}{\frac{1}{k} . \frac{1}{4}} = 12$

$\Rightarrow$ $4 k = 12$

$\Rightarrow$ $k = 3$

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Continuity- Miscellaneous

⏱ 2mℹ️ Source: PYQ_2015

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