Question

If A, B, C are respectively the minimum values of the functions
$\frac{x}{1+x^2},x^2+\frac{16}{x},\frac{x}{\log x}$ then the ascending order of A,B,C is

• A. A, B, C
• B. A, C, B
• C. B, A, C
• D. B, C, A

Answer 1

Answer: (B)

A, C, B

(A) $f\left(x\right)=\frac{x}{1+x^2}$
$f^{\prime }\left(x\right)=\frac{(x^2+1)-2x^2}{(1+x^2{)}^2}$
$f^{\prime }\left(x\right)=0$ $x=\pm 1$
$f(min{)}^{(-1)}=\frac{-1}{2}$
(B) Use AM & GM
$\frac{x^2+\frac{8}{x}+\frac{8}{x}}{3}\ge (64{)}^{\frac{1}{3}}$
$x^2+\frac{16}{x}\ge 3\times \left(4\right)\ge 12$
(C) $f\left(x\right)=\frac{x}{logx}$
$f^{\prime }\left(x\right)=\frac{1}{logx}-\frac{1}{\left(log\right(x){)}^2}$
$=\frac{log\left(x\right)-1}{\left(log\right(x){)}^2}$
$f^{\prime }\left(x\right)=0$ at $x=e$
$fmi{n}^{\left(e\right)}=e$
$A<C<B$