Question

If $f\left(x\right)=k^{3}x+k^3-2$ cuts the curve $g\left(x\right)=\frac{1}{2}\ln x^2$ at exactly two points then ${}^{\prime }{k}^{\prime }$ may lie in the interval

• A. $(\frac{1}{\sqrt{e}},e)$
• B. $(\frac{1}{e},\frac{1}{\sqrt{e}})$
• C. $(\frac{1}{e^2},\frac{1}{\sqrt{e}})$
• D. None of these

Answer 1

The correct option is A $(\frac{1}{\sqrt{e}},e)$

The curves may cut exactly at one point, two points or three points. Since, the curves are cut exactly at two points is possible only when the curve $f\left(x\right)$ is tangent to the curve $g\left(x\right)$.
Let the point of contact be $(t,\ln t)$

$g\left(x\right)=\frac{1}{2}\ln x^2=\ln x(\text{when,}x>0)g^{\prime }\left(x\right)=\frac{1}{x}\text{Slope of the tangent at}t=\frac{1}{t}\text{Equation of the tangent is,}y-\ln t=\frac{1}{t}(x-t)\Rightarrow y=\frac{x}{t}+\ln t-1....\left(1\right)y=k^{3}x+k^3-2....\left(2\right)\text{Comparing eqn(1) and (2), we get,}\Rightarrow k^3=\frac{1}{t}\&k^3-2=\ln t-1\Rightarrow k^3-2=\ln \left(\frac{1}{k^3}\right)-1\Rightarrow k^3+3\ln k-1=0\Rightarrow f\left(k\right)=k^3+3\ln k-1\Rightarrow f^{\prime }\left(k\right)=3k^2+\frac{3}{k}>0\text{Using intermediate value theorem, we have}f\left(\frac{1}{\sqrt{e}}\right).f\left(e\right)<0\text{Hence,}k\text{may lie in the interval}(\frac{1}{\sqrt{e}},e)$