Question

If $\frac{1}{\sqrt{a}}{\int }_1^a(\frac{3}{2}\sqrt{x}+1-\frac{1}{\sqrt{x}})dx<4$, then ${}^{\prime }{a}^{\prime }$ may take the value

• A. $0$
• B. $4$
• C. $9$
• D. none of these

Answer 1

The correct option is A $0$

We have, $\frac{1}{\sqrt{a}}{\int }_1^a(\frac{3}{2}\sqrt{x}+1-\frac{1}{\sqrt{x}})dx<4$

$\Rightarrow \frac{1}{\sqrt{a}}(a\sqrt{a}-1+a-1-2\sqrt{a}+2)<4$

$\Rightarrow t^2+t-6<0$, where $t=\sqrt{a}$

$\Rightarrow (t+3)(t-2)<0$

$\Rightarrow t\in (-3,2)\Rightarrow \sqrt{a}\in (-3,2)\Rightarrow \sqrt{a}\in [0,2)\Rightarrow a\in (0,4)$

$[$ The given inequality is defined for $a\ne 0]$