QJEEM17Mad305: JEE Main 2017 Mathematics - Class 12 JEEM17Mad305 Solution: The value of k for which the function f ( x ) = { ( 4 5 ) tan 4 …

Question

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The value of k for which the function

f (x) = {\begin{matrix} {(\frac{4}{5})}^{\frac{\tan 4 x}{\tan 5 x}}, & 0 < x < \frac{π}{2} \\ k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}

is continuous at x =

\frac{π}{2},

is :

Accepted Answer

f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}

\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}

\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}

\Rightarrow\left(\frac{4}{5}\right)^{\lim\limits_{x \rightarrow \frac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\frac{2}{5}

\Rightarrow\left(\frac{4}{5}\right)^{0 \times \cos \left(2 x+\frac{x}{2}\right)}=k+\frac{2}{5}

\Rightarrow\left(\frac{4}{5}\right)^0=k+\frac{2}{5}

\Rightarrow k+\frac{2}{5}=1 \Rightarrow k=1-\frac{2}{5} \Rightarrow k=\frac{3}{5}

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