If $\frac{1}{\sqrt{a}}\int\limits_1^a\left(\frac{3}{2}\sqrt{x}+1-\frac{1}{\sqrt{x}}\right)dx < 4$, then ‘a’ may take values :
Answer & explanation
Correct answer: option 1
$\frac{1}{\sqrt{a}}\int\limits_1^a\left(\frac{3}{2}\sqrt{x}+1-\frac{1}{\sqrt{x}}\right)dx < 4$
$⇒\frac{1}{\sqrt{a}}(a\sqrt{a}-1+a-1-2\sqrt{a}+2)<4$
$⇒a+\sqrt{a}-6<0⇒(\sqrt{a})^2+\sqrt{a}-6<0$
$\sqrt{a}∈(–3, 2)$