Question

Find the value of $k$if which the function f\left( x \right)=\left\\{ \begin{aligned} & {{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}},0 < x < \dfrac{\pi }{2} \\\ & k+\dfrac{2}{5}\,\,\,\,\,\,\,,x=\dfrac{\pi }{2}\, \\\ \end{aligned} \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

In this question, in order to the value of $k$ given the function f\left( x \right)=\left\\{ \begin{aligned} & {{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}},0 < x < \dfrac{\pi }{2} \\\ & k+\dfrac{2}{5}\,\,\,\,\,\,\,,x=\dfrac{\pi }{2}\, \\\ \end{aligned} \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{aligned} & {{\left( \dfrac{4}{5} \right)}^{\dfrac{\tan 4x}{\tan 5x}}},0 < x < \dfrac{\pi }{2} \\\ & k+\dfrac{2}{5}\,\,\,\,\,\,\,,x=\dfrac{\pi }{2}\, \\\ \end{aligned} \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

& \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{aligned}$$ Now we will evaluate the limit $$\displaystyle \lim_{x \to \dfrac{\pi }{2}}\left( \tan 4x\cot 5x \right)$$. Then we have $$\begin{aligned} & \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{aligned}$$ Now since the value of $$\tan 2\pi =0$$, therefore we have $$\begin{aligned} & \displaystyle \lim_{x \to \dfrac{\pi }{2}}\left( \tan 4x\tan 5x \right)=\tan 2\pi \cot \dfrac{5\pi }{4} \\\ & =0 \end{aligned}$$ Therefore using the above value in the limit in equation (3), we get $$\begin{aligned} & {{\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{aligned}$$ Now since $${{a}^{0}}=1$$ for all values of $$a$$, therefore from the above equation we have, $$\begin{aligned} & 1=k+\dfrac{2}{5} \\\ & \Rightarrow k=1-\dfrac{2}{5} \\\ & \Rightarrow k=\dfrac{5-2}{5} \\\ & \Rightarrow k=\dfrac{3}{5} \\\ \end{aligned}$$ Therefore we have that the value of $$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}}$$.