Question

Which of the following statements is/are correct in stating the number of solutions of the equation
$x\log x+x=\lambda ,$ where $\lambda \in R.$

• A. If $\lambda =0$, then the equation has exactly two solutions in $x$.
• B. If $\lambda \in (0,\frac{1}{e^2})$, then the equation has exactly one solution in $x$.
• C. If $\lambda \in (\frac{-1}{e^2},0)$, then equation has at least one solution in $x$.
• D. the equation has at least one solution in $[1,\lambda ]$

Answer 1

Answer: (B), (C), (D)

the equation has at least one solution in $[1,\lambda ]$

$x\log x+x=\lambda \left(i\right)$
Let $f\left(x\right)=x\log x+x=0x\log x+x=0\Rightarrow x(\log x+1)=0$
$\Rightarrow x=0$ or $x=\frac{1}{e}$
$x=0$ is not a solution as $\log$ of $0$ is not defined.
So, the equation has exactly one solution if $\lambda =0$

$\left(ii\right)$ Exactly $1$ solution in $x$ if $\lambda \in (0,\frac{1}{e^2})$
$\left(iii\right)$ At least $1$ solution in $x$ if $\lambda \in (\frac{-1}{e^2},0)$
${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}(x\log x+x)={lim}_{x\to 0^{+}}\frac{\log x+1}{\frac{1}{x}}$
Using L'hospitals' rule, we get
${lim}_{x\to 0^{+}}\frac{\frac{1}{x}}{\frac{-1}{x^2}}={lim}_{x\to 0^{+}}(-x)=0$

$f^{\prime }\left(x\right)=x\times \frac{1}{x}+\log x+1=2+\log x{f}^{\prime }\left(x\right)=0\Rightarrow \log x=-2\Rightarrow x=\frac{1}{e^2}$
$f^{\prime \prime }\left(x\right)=\frac{1}{x}>0$ at $x=\frac{1}{e^2}$
Hence, $x=\frac{1}{e^2}$ is a point of local minima.

$f^{\prime }\left(1\right)=2(>0)$
$f^{\prime }\left(\frac{1}{e^3}\right)=-1(<0)$

If $\lambda >0\to$ Only $1$ solution
If $\lambda <0\to$ $2$ solutions.

$\left(iv\right)$ At least one solution in $[1,\lambda ]$
$f\left(x\right)=x\log x+x=x(1+\log x)\Rightarrow f\left(x\right)\ge x,\forall x\in [1,\lambda ]$