Question

Let $M=\left[\begin{array}{ccc}x& 2x& 3x\\ f\left(x\right)& g\left(x\right)& h\left(x\right)\\ 0& 1& 1\end{array}\right],x\ne 0$ be a singular matrix. If $f\left(x\right)=\ln (e^x+1)$ and $g\left(x\right)=\ln (e^x-1),$ then the value of $h^{\prime }\left(\ln 3\right)$ is

• A. $\frac{9}{8}$
• B. $\frac{9}{4}$
• C. $3$
• D. $6$

Answer 1

The correct option is B $\frac{9}{4}$

$M=\left[\begin{array}{ccc}x& 2x& 3x\\ f\left(x\right)& g\left(x\right)& h\left(x\right)\\ 0& 1& 1\end{array}\right]$

Applying $C_2\to C_2-C_3$
$\Rightarrow M=\left[\begin{array}{ccc}x& -x& 3x\\ f\left(x\right)& g\left(x\right)-h\left(x\right)& h\left(x\right)\\ 0& 0& 1\end{array}\right]$

Since $M$ is singular matrix,
$\therefore |M|=0$
$\Rightarrow x\left[g\right(x)-h(x)+f(x\left)\right]=0$
$\Rightarrow h\left(x\right)=f\left(x\right)+g\left(x\right)$
$\Rightarrow h\left(x\right)=\ln (e^x+1)+\ln (e^x-1)$
$\Rightarrow h\left(x\right)=\ln (e^{2x}-1)$
Differentiating w.r.t. $x,$ we get
$h^{\prime }\left(x\right)=\frac{2e^{2x}}{e^{2x}-1}$
For $x=\ln 3,$ we have
$h^{\prime }\left(\ln 3\right)=\frac{2e^{2\ln 3}}{e^{2\ln 3}-1}=\frac{9}{4}$