Question

If $\frac{4\alpha }{{\alpha }^2+1}\ge 1$ and $\alpha +\frac{1}{\alpha }$ is an odd integer then number of possible values of $\alpha$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

We know that,

$A.M.\ge G.M.⟹\frac{a+b}{2}\ge \sqrt{ab}$

Let $a=1$ and $b={\alpha }^2$

$⟹\frac{1+{\alpha }^2}{2}\ge \sqrt{1\ast {\alpha }^2}$

$⟹\frac{1+{\alpha }^2}{2}\ge \alpha$

$⟹\frac{1+{\alpha }^2}{2\alpha }\ge 1$

$⟹\frac{2\alpha }{1+{\alpha }^2}\le 1$

multiplying 2 on both sides

$⟹\frac{4\alpha }{1+{\alpha }^2}\le 2$

given that $\frac{4\alpha }{1+{\alpha }^2}\ge 1$

and $\alpha +\frac{1}{\alpha }=\frac{1+{\alpha }^2}{\alpha }$ is odd integer

$⟹\frac{4}{odd}\ge 1$ and $\frac{4}{odd}\le 2$

that odd integer can only be $3$ satisfying above conditions.

Therefore, $\alpha$ can take only one value .