JEE MAIN - Mathematics (2017 - 9th April Morning Slot - No. 16)
The value of k for which the function
$$f\left( x \right) = \left\{ {\matrix{ {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr } } \right.$$
is continuous at x = $${\pi \over 2},$$ is :
$$f\left( x \right) = \left\{ {\matrix{ {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr } } \right.$$
is continuous at x = $${\pi \over 2},$$ is :
$${{17} \over {20}}$$
$${{2} \over {5}}$$
$${{3} \over {5}}$$
$$-$$ $${{2} \over {5}}$$
Explanation
$f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}$
$\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$
$\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$
$\Rightarrow\left(\frac{4}{5}\right)^{\lim\limits_{x \rightarrow \frac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\frac{2}{5}$
$\Rightarrow\left(\frac{4}{5}\right)^{0 \times \cos \left(2 x+\frac{x}{2}\right)}=k+\frac{2}{5}$
$\Rightarrow\left(\frac{4}{5}\right)^0=k+\frac{2}{5}$
$\Rightarrow k+\frac{2}{5}=1 \Rightarrow k=1-\frac{2}{5} \Rightarrow k=\frac{3}{5}$
$\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$
$\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$
$\Rightarrow\left(\frac{4}{5}\right)^{\lim\limits_{x \rightarrow \frac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\frac{2}{5}$
$\Rightarrow\left(\frac{4}{5}\right)^{0 \times \cos \left(2 x+\frac{x}{2}\right)}=k+\frac{2}{5}$
$\Rightarrow\left(\frac{4}{5}\right)^0=k+\frac{2}{5}$
$\Rightarrow k+\frac{2}{5}=1 \Rightarrow k=1-\frac{2}{5} \Rightarrow k=\frac{3}{5}$
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