Question

Find the value of \[k\]if which the function \[f\left( x \right)=\left\{ \begin{align}
  & {{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}},0 < x < \dfrac{\pi }{2} \\
 & k+\dfrac{2}{5}\,\,\,\,\,\,\,,x=\dfrac{\pi }{2}\, \\
\end{align} \right.\]​​ is continuous at \[x=\dfrac{\pi }{2}\]​.
(a) \[\dfrac{17}{20}\]
(b) \[\dfrac{2}{5}\]
(c) \[\dfrac{3}{5}\]
(d) \[-\dfrac{2}{5}\]

Answer 1

Hint: In this question, in order to the value of \[k\] given the function \[f\left( x \right)=\left\{ \begin{align}
  & {{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}},0 < x < \dfrac{\pi }{2} \\
 & k+\dfrac{2}{5}\,\,\,\,\,\,\,,x=\dfrac{\pi }{2}\, \\
\end{align} \right.\] ​​ which is continuous at \[x=\dfrac{\pi }{2}\]​. Now we know that if a function \[f\left( x \right)\] is continuous at \[x=a\], then we have \[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]. Therefore here we should have \[\displaystyle \lim_{x \to \dfrac{\pi }{2}}f\left( x \right)=f\left( \dfrac{\pi }{2} \right)\]. This is true because of the fact that if a \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( \displaystyle \lim_{x \to a}x \right)\]. Using these properties for continuous functions we will get our desired answer.

Complete step by step answer:
Let the function \[f\] is defined by \[f\left( x \right)=\left\{ \begin{align}
  & {{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}},0 < x < \dfrac{\pi }{2} \\
 & k+\dfrac{2}{5}\,\,\,\,\,\,\,,x=\dfrac{\pi }{2}\, \\
\end{align} \right.\].
Now for the values of \[x\] such that \[0 < x < \dfrac{\pi }{2}\], we have that
\[f\left( x \right)={{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}}\]
Since we know that if a function \[f\left( x \right)\] is continuous at \[x=a\], then we have \[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\].
Therefore here we should have \[\displaystyle \lim_{x \to \dfrac{\pi }{2}}f\left( x \right)=f\left( \dfrac{\pi }{2} \right)\].
Now since \[f\left( x \right)={{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}}\] and \[f\left( \dfrac{\pi }{2} \right)=k+\dfrac{2}{5}\], therefore we have that
\[\displaystyle \lim_{x \to \dfrac{\pi }{2}}{{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}}=k+\dfrac{2}{5}............(1)\]
Now since we have the property of trigonometric function that \[\dfrac{1}{\tan x}=\cot x\], therefore we have
\[\dfrac{\tan 4x}{\tan 5x}=\tan 4x\cot 5x...........(2)\]
Substituting the value in equation (2) in (1), we get
\[\displaystyle \lim_{x \to \dfrac{\pi }{2}}{{\left( \dfrac{4}{5} \right)}^{\tan 4x\cot 5x}}=k+\dfrac{2}{5}\]
Since we know that if a \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( \displaystyle \lim_{x \to a}x \right)\].
Therefore using this, we get that
\[\begin{align}
  & \displaystyle \lim_{x \to \dfrac{\pi }{2}}{{\left( \dfrac{4}{5} \right)}^{\tan 4x\cot 5x}}=k+\dfrac{2}{5} \\
 & \Rightarrow {{\left( \dfrac{4}{5} \right)}^{\displaystyle \lim_{x \to \dfrac{\pi }{2}}\left( \tan 4x\cot 5x \right)}}=k+\dfrac{2}{5}........(3)
\end{align}\]
Now we will evaluate the limit \[\displaystyle \lim_{x \to \dfrac{\pi }{2}}\left( \tan 4x\cot 5x \right)\].
Then we have
\[\begin{align}
  & \displaystyle \lim_{x \to \dfrac{\pi }{2}}\left( \tan 4x\tan 5x \right)=\tan 4\left( \dfrac{\pi }{2} \right)\cot 5\left( \dfrac{\pi }{2} \right) \\
 & =\tan 2\pi \cot \dfrac{5\pi }{4}
\end{align}\]
Now since the value of \[\tan 2\pi =0\], therefore we have
\[\begin{align}
  & \displaystyle \lim_{x \to \dfrac{\pi }{2}}\left( \tan 4x\tan 5x \right)=\tan 2\pi \cot \dfrac{5\pi }{4} \\
 & =0
\end{align}\]
Therefore using the above value in the limit in equation (3), we get
\[\begin{align}
  & {{\left( \dfrac{4}{5} \right)}^{\displaystyle \lim_{x \to \dfrac{\pi }{2}}\left( \tan 4x\cot 5x \right)}}=k+\dfrac{2}{5} \\
 & \Rightarrow {{\dfrac{4}{5}}^{0}}=k+\dfrac{2}{5} \\
\end{align}\]
Now since \[{{a}^{0}}=1\] for all values of \[a\], therefore from the above equation we have,
\[\begin{align}
  & 1=k+\dfrac{2}{5} \\
 & \Rightarrow k=1-\dfrac{2}{5} \\
 & \Rightarrow k=\dfrac{5-2}{5} \\
 & \Rightarrow k=\dfrac{3}{5} \\
\end{align}\]
Therefore we have that the value of \[k\] is equal to \[\dfrac{3}{5}\].

So, the correct answer is “Option C”.

Note: In this problem, we know that if a function \[f\left( x \right)\] is continuous at \[x=a\], then we have \[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]. Therefore here we should have \[\displaystyle \lim_{x \to \dfrac{\pi }{2}}f\left( x \right)=f\left( \dfrac{\pi }{2} \right)\]. This is true because of the fact that if a \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( \displaystyle \lim_{x \to a}x \right)\].we are using this same property while evaluating the limit \[\displaystyle \lim_{x \to \dfrac{\pi }{2}}{{\left( \dfrac{4}{5} \right)}^{\tan 4x\cot 5x}}\].