Question

If $f\left(x\right)=\{\begin{array}{c}{\left(\frac{6}{5}\right)}^{\frac{tan6x}{tan5x}};0<x<\frac{\pi }{2}\\ a+2;x=\frac{\pi }{2}\\ (1+|cotx|{)}^{\frac{b|tanx|}{a}};\frac{\pi }{2}<x<\pi \end{array}$ is continuous at $x=\frac{\pi }{2}$. Then the values of $a$ and $b$ are

• A. $a=0:b=0$
• B. $a=-1:b=0$
• C. $a=0:b=-1$
• D. $a=0:b=1$

Answer 1

The correct option is B $a=-1:b=0$

For function to be continuous at $\frac{\pi }{2}$:

$\underset{x\to (\frac{\pi }{2}{)}^{-}}{lim}f\left(x\right)$=$\underset{x\to (\pi /2{)}^{+}}{lim}f\left(x\right)$=$f\left(\frac{\pi }{2}\right)$

$\underset{x\to (\frac{\pi }{2}{)}^{-}}{lim}f\left(x\right)$=$\underset{x\to (\frac{\pi }{2}{)}^{-}}{lim}{\left(\frac{6}{5}\right)}^{\frac{tan6x}{tan5x}}={\frac{6}{5}}^{0/\infty }=1$..........[1]

$\underset{x\to (\frac{\pi }{2}{)}^{+}}{lim}f\left(x\right)$=$\underset{x\to (\frac{\pi }{2}{)}^{+}}{lim}(1+|cotx|{)}^{\frac{b|tanx|}{a}}$........ ( $1^{\infty }form$)

$\underset{x\to (\frac{\pi }{2}{)}^{+}}{lim}(1+|cotx|{)}^{\frac{b|tanx|}{a}}=e^k$ where

$k=\underset{x\to \frac{\pi }{2}}{lim}(1+|cotx|-1)\left(\frac{b|tanx|}{a}\right)=\frac{b}{a}$............[2]

$\underset{x\to (\frac{\pi }{2}{)}^{+}}{lim}(1+|cotx|{)}^{\frac{b|tanx|}{a}}=e^{\frac{b}{a}}$

$f\left(\frac{\pi }{2}\right)=a+2$.........[3]

Hence, from [1] and [3]:

$a+2=1⟹a=-1$.......[4]

And from [1],[2] and [4]:

$e^{\frac{b}{a}}=1⟹\frac{b}{a}=0⟹b=0$