Question

If equation of tangent drawn to the curve $y=\left|{\log }_2\right|x||$ at $x=\frac{-1}{3}$ is $px+y\ln (q)+\ln (3e)=0$ then $(p+q)$ is equal to

• A. 3
• B. $\frac{5}{2}$
• C. 5
• D. $\frac{7}{2}$

Answer 1

Answer: (D)

$\text{ at }x=-1/3,y={\log }_{2}3$
$y=\left|{\log }_2\right|x||=-{\log }_2(-x)$$-1<x<0$
$\frac{dy}{dx}=\frac{-1}{\ln 2}\cdot \frac{(-1)}{(-x)}=\frac{-1}{x\ln 2}$
${\frac{dy}{dx}|}_{x=-1/3}=\frac{3}{\ln 2}$
$\therefore \text{ equation of tangent at }\left(-1/3,{\log }_{2}3\right)$
$y-{\log }_{2}3=\frac{3}{\ln 2}\left(x+\frac{1}{3}\right)$
$y\ln 2-\ln 3=3x+1$
$3x-y\ln 2+1+\ln 3=0$
$px+y\ln (q)+\ln (3e)=0$
$\therefore p=3,q=\frac{1}{2}\Rightarrow p+q=\frac{7}{2}$