find all values of " a " for which the inequality \frac{1}{\sqrt{a}}\int_{1}^{a}\left( {\frac{3\sqrt{x}}{2}+1-\frac{1}{\sqrt{x}}\right)}<4 is satisfied.
ans-------->(0,4)
29\frac{1}{\sqrt{a}}\int_{1}^{a}{\frac{3}{2}\sqrt x+ 1 - \frac{1}{\sqrt{x}}} = a + \sqrt{a} - 2
now a + \sqrt{a} - 2 < 4
this cannot be solved by writing and squaring both sides
a - 6 < -√a
so we need to solve it like this
a + √a - 6 < 0
(√a - 2)(√a + 3) < 0...that means √a ε (-3, 2 )..since √a cannot be < 0...so then √a ε (0, 2 ) and so a ε ( 0 , 4)
thanks xYz for the hint
1\frac{1}{\sqrt{a}}\left[(a)^{3/2}+a-2\sqrt{a} -(1+1-2)\right]<4
\Rightarrow a+\sqrt{a}-2<4 \Rightarrow a+\sqrt{a}-6<0\Rightarrow a+3\sqrt{a}-2\sqrt{a}-6<0
\sqrt{a}(\sqrt{a}+3)-2(\sqrt{a}+3)\Rightarrow \sqrt{a}<2\Rightarrow a<4
since we squared and that √a is not less than 0 a\epsilon (0,4)
GUYS DON"T take me rong as i posted ans. to my question itself
had not seen that govind had already posted OOPS !