Question

Let $f\left(x\right)=lnmx(m>0)$ and $g\left(x\right)=px$. Then the equation $|f\left(x\right)|=g\left(x\right)$ has only one solution for

• A. $0<p<\frac{m}{e}$
• B. $p<\frac{e}{m}$
• C. $0<p<\frac{e}{m}$
• D. $p>\frac{m}{e}$

Answer 1

The correct option is D $p>\frac{m}{e}$

Let $P(h,k)$ be a point on $f\left(x\right)=\ln mx$
then, $k=\ln mh$
$\frac{dy}{dx}$ at point P $=\frac{1}{h}$
Equation of tangent at point P $L_1:y-k=\frac{(x-h)}{h}$
Since, it passes through origin
Therefore, $k=1$ & $h=e/m$
and $L_1:y=x/h$ $\Rightarrow L_1:y=mx/e$
From the figure: $|f\left(x\right)|=g\left(x\right)$ have one solution when slope of $g\left(x\right)$ is greater than slope of $L_1$
Therefore, $p>m/e$

Ans: D