For all $x∈ (0,1)$
Answer & explanation
Correct answer: option 2
$e^x=1+x+\frac{x^2}{2!}+.....>1+x⇒x>\log_e(1+x)$ i.e., $\log_e(1+x)<x$
For all $x∈ (0,1)$
Correct answer: option 2
$e^x=1+x+\frac{x^2}{2!}+.....>1+x⇒x>\log_e(1+x)$ i.e., $\log_e(1+x)<x$