Mathematics - Application of Derivative Question with Solution | TestHub

Let $f(x)={\log }_{e}x-(x-1)$.

$\Rightarrow f^{\prime }(x)=\frac{1}{x}-1=\frac{1-x}{x}<0$.

Since $f(x)$ is a decreasing function (because $x>1$),

$$x>1\Rightarrow f(x)<f(1)\Rightarrow {\log }_{e}x-(x-1)<0\Rightarrow (x-1)>{\log }_{e}x=y$$

Or $(x-1)>y$.

Now, let $g(x)={\log }_{e}x-(x^2-1)$.

$\Rightarrow g^{\prime }(x)=\frac{1}{x}-2x=\left(\frac{1-2x^2}{x}\right)<0$ (for $x>1$).

Therefore, $g(x)$ is a decreasing function.

$$\because x>1\Rightarrow g(x)<g(1)\Rightarrow {\log }_{e}x-(x^2-1)<0$$

Or $(x^2-1)>y$.

Again, let $h(x)=\frac{x-1}{x}-{\log }_{e}x$.

Therefore, $h^{\prime }(x)=0+\frac{1}{x^2}-\frac{1}{x}=\frac{1-x}{x^2}<0$ (for $x>1$).

Therefore, $h(x)$ is a decreasing function.

$$\because x>1\Rightarrow h(x)<h(1)\Rightarrow \frac{x-1}{x}-{\log }_{e}x<0\Rightarrow \frac{x-1}{x}<y$$