If $\frac{1}{\sqrt{a}} \int\limits_1^a\left(\frac{3}{2} \sqrt{x}+1-\frac{1}{\sqrt{x}}\right) d x<4$, then 'a' may take values:
Answer & explanation
Correct answer: option 4
We have,
$\frac{1}{\sqrt{a}} \int\limits_1^a\left(\frac{3}{2} \sqrt{x}+1-\frac{1}{\sqrt{x}}\right) d x<4$
$\Rightarrow \frac{1}{\sqrt{a}}(a \sqrt{a}-1+a-1-2 \sqrt{a}+2)<4$
$\Rightarrow a+\sqrt{a}-2<4$
$\Rightarrow a+\sqrt{a}-6<0$
$\Rightarrow t^2+t-6<0$, where $t=\sqrt{a}$
$\Rightarrow t \in(-3,2) \Rightarrow \sqrt{a} \in(-3,2) \Rightarrow \sqrt{a} \in[0,2] \Rightarrow a \in[0,4)$
Clearly, the given inequality is defined for $a \neq 0$.
Hence, option (d) is correct.