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 :

0 4 9 $\frac{13±\sqrt{313}}{2}$

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$\frac{1}{\sqrt{a}}\int\limits_1^a\left(\frac{3}{2}\sqrt{x}+1-\frac{1}{\sqrt{x}}\right)dx < 4$ $⇒\frac{1}{\sqrt{a}}(a\sqrt{a}-1+a-1-2\sqrt{a}+2)<4$ $⇒a+\sqrt{a}-6<0⇒(\sqrt{a})^2+\sqrt{a}-6<0$ $\sqrt{a}∈(–3, 2)$