Question

If ݂f(x) = $\left \{ {{\frac{4^x - 3^x}{sin 2x}, x !=(not-equals-to) 0} \atop {k, x=0}} \right.$ is continuous, then k equals.

Answer 1

${ \huge{ \star}} \: \: \: \: \: \Large\bf{ \underline{Overview :}}$

$\natural \: \tt \: \lim_{x \to \:a {}^{ + } } \: f( x) = f(a) =\lim_{x \to \:a {}^{ - } } \: f( x) \: \: for \: \: any \: \: function \: \: to \: \: be \: \: continuous \: \: any \: \: x = a \\ \tt$

${ \huge{ \star}} \: \: \: \: \: \Large\bf{ \underline{Solution :}}$

$\dashrightarrow \tt \: f(x) = \begin{cases} \tt\frac{ {4}^{x} - {3}^{x} }{ \sin 2x} \: \: \: ;& \tt \: x \not = 0 \\ \tt\: k \: \: \: ;& \tt x = 0\end{cases} \\ \\ \\ \circ \sf \: \: A/C ,\: \tt \: \: \: function \: \: f(x) \: \: is \: \: continuous \\ \\ \tt \: hence, \\ \tt \: \implies \: \lim_{x \to \:a {}^{ + } } \: f( x) = f(x) = \lim_{x \to \:a {}^{ - } } \: f( x) \\ \\ \implies \tt \lim_{x \to \:0 {}^{ + } } \: \frac{ {4}^{x} - {3}^{x} }{ \sin \: 2x} = k = \lim_{x \to 0 {}^{ - } } \: \frac{ {4}^{x} - {3}^{x} }{ \sin \: 2x} \\ \\ \implies \tt\lim_{x \to 0{}^{ + } } \frac{ \frac{ {4}^{x} - {3}^{x} }{2x} }{ \frac{ \sin \: 2x}{2x} } = k= \tt\lim_{x \to 0{}^{ - } } \frac{ \frac{ {4}^{x} - {3}^{x} }{2x} }{ \frac{ \sin \: 2x}{2x} } \\ \\ \tt \:\implies \tt\lim_{x \to 0{}^{ + } } { \frac{ {4}^{x} - {3}^{x} }{2x} } = k = \tt\lim_{x \to 0{}^{ - } } { \frac{ {4}^{x} - {3}^{x} }{2x} } \: \: \: [ \because \lim_{x \to 0 } \frac{\sin x}{x} = 1 ] \\ \\ \tt \implies\lim_{x \to \: 0 {}^{ + } } \frac{ \frac{d}{dx} ( {4}^{x} - {3}^{x}) }{ \frac{d}{dx}(2x) } = k = \lim_{x \to \: 0 {}^{ - } } \frac{ \frac{d}{dx} ( {4}^{x} - {3}^{x}) }{ \frac{d}{dx}(2x) } \\ \\ \tt \implies\lim_{x \to \: 0 {}^{ + } } \frac{ ln(2) \times {2}^{(2x + 1) } - ln(3) \times {3}^{x}}{ 2 } = k= \lim_{x \to \: 0 {}^{ - } } \frac{ ln(2) \times {2}^{(2x + 1) } - ln(3) \times {3}^{x}}{ 2 } \\ \\ \implies \: {\tt\frac{ ln( \frac{4}{3} ) }{2} = k = \frac{ ln( \frac{4}{3} ) }{2} } \\ \\ \\ \therefore \: \: \: \large{ \green{ { \underline{\boxed{ { \green{\tt{ k = { \green{\frac{ ln( \frac{4}{3} ) }{2} }}}}}}}}}} \\ \\ \\ \tt \Box\: \:\:\:We \: \: know; \: \: ln(x) = 2.303 \times log(x) \\ \\ \tt \: putting \: \: value \: \\ \\ \tt \: k = 2.303 \times \frac{ log( \frac{4}{3} ) }{2} \\ \\ \implies\tt \: k \approx \: log( \frac{4}{3} ) \\ \\ \tt \therefore \: \green{ \boxed{ \tt \: b) \: \: log( \frac{4}{3} ) }}$

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