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  4. If equation of tangent drawn to the curve y=| log 2| x|| at x=(-1/3) is p x+y ln (q)+ ln (3 e)=0 then (p+q) is equal to

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

Application of Derivatives

Solution:

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