Question

$y=\log x$ satisfies for $x>1$, the equality

• A. $x-1>y$
• B. $x^2-1>y$
• C. $y>x-1$
• D. $\frac{x-1}{x}<y$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $x-1>y$
B $x^2-1>y$
D $\frac{x-1}{x}<y$

Consider $f\left(x\right)=\log x-(x-1)\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}-1$

So $f^{\prime }\left(x\right)<0$ for $x>1$.

Thus $f\left(1\right)>f\left(x\right)$ for $x>1$

$\Rightarrow 0>\log x-(x-1)\Rightarrow x-1>\log x$

Consider $g\left(x\right)=\frac{x-1}{x}-\log x\Rightarrow g^{\prime }\left(x\right)=\frac{{(x-\frac{1}{2})}^2+\frac{1}{4}}{x^2}>0$

for all $x$.

Hence $g\left(x\right)>g\left(1\right),x>1\Rightarrow \frac{x-1}{x}>\log x$

$x^2-1>\log x$ is true (we can check by taking $x=2$)