Question

Let $f:R\to R$ is differentiable and strictly increasing function throughout its domain.
$S_1:$ If $|f\left(x\right)|$ is also strictly increasing function, then $f\left(x\right)=0$ has no real roots.
$S_2:$ At $\infty$ or $-\infty ,f\left(x\right)$ may approach to $0,$ but it can't be equal to zero

• A. $S_1$ is true, $S_2$ is true, $S_2$ is correct explanation of $S_1$
• B. $S_1$ is true, $S_2$ is true, $S_2$ is not correct explanation of $S_1$
• C. $S_1$ is true, $S_2$ is false.
• D. $S_1$ is false, $S_2$ is true.

Answer 1

The correct option is A $S_1$ is true, $S_2$ is true, $S_2$ is correct explanation of $S_1$

Suppose $f\left(x\right)=0$ has a real root say $x=a,$ then $f\left(x\right)<0$ for $x<a$.
Thus, $|f\left(x\right)|$ becomes strictly decreasing on $(-\infty ,a),$ which is the contradiction.
So $S_1$ is correct.Note that this case is possible if $f\left(x\right)$ lies above $x-$axis
$S_2$ is also correct as $f\left(x\right)$ is strictly increasing function and it is correct explanation of $S_1$