if $f(x) = -1 + kx + k$ neither touches nor intercepts the curve $g(x)= \log x$,then the minimum value $k \in ? $
choose the correct option
a)$\left( \frac{1}{e}, \frac{1}{e^\frac{1}{2} }\right)$
b) $\left(e,e^2\right)$
c) $\left(\frac{1}{e^\frac{1}{2} },e\right)$
d) none of these
I was trying that $f'(x) = k$ and $k= \log x$ after that I get $x = e^k$
now here I can conclude that option b is correct answer
On
If you set $h(x)=kx+k-1-\log x$, the problem is to identify when the equation $h(x)=0$ has no solution.
By looking at limits at $0$ and $\infty$, we immediately disqualify $k\le0$, so we can assume $k>0$. Since $h'(x)=(kx-1)/x$, the function has an absolute minimum at $1/k$, where $$ h(1/k)=k+\log k $$ The equation $h(x)=0$ has no solution when $k+\log k>0$. Consider $\gamma(t)=t+\log t$. Then $\gamma'(t)=(t+1)/t$ is positive and there exists $k_0$ such that $\gamma(t)>0$ if and only $t>k_0$. Since $\gamma(e^{-1})=e^{-1}-1<0$ and $\gamma(e^{-1/2})=e^{-1/2}-1/2>0$, we know that $k_0\in(e^{-1},e^{-1/2})$.
The equation $h(x)=0$ has no solution if and only if $k>k_0$. However for $k=k_0$, the equation $h(x)=0$ has a solution. The correct option is (d), because there is no minimum.
Just for completeness, $e>1$ entails $e^{-1}<1$; $e<4$ entails $e^{1/2}<2$, so $e^{-1/2}>1/2$.
The value $k_0$ is where the line $f(x)=kx+k-1$ and the curve $g(x)=\log x$ touch each other. The problem states “neither touches nor intercepts”, so $k_0$ must be discarded.
HINT
Look for the value of k for the tangency condition.