Question

The area bounded by the lines $y=2,y=1,x=a$ and the curve $y=f\left(x\right)$, which cuts the last two lines above the first line for all $a\ge 1$, is equal to $[{\left(2a\right)}^{3/2}-3a+3-2\sqrt{2}]$. Then $f\left(x\right)=$

• A. $2\sqrt{2x},x\ge 1$
• B. $\sqrt{2x},x\ge 1$
• C. $2\sqrt{x},x\ge 1$
• D. none of these

Answer 1

The correct option is A $2\sqrt{2x},x\ge 1$

We are given

${\int }_1^a[f\left(x\right)-2]dx=\frac{2}{3}[{\left(2a\right)}^{3/2}-3a+3-2\sqrt{2}]$

Differentiating w.r.t $a$, we get

$f\left(a\right)-2=\frac{2}{3}[\frac{3}{2}\sqrt{2a}.2-3]\Rightarrow f\left(a\right)=2\sqrt{2a},a\ge 1\therefore f\left(x\right)=2\sqrt{2x},x\ge 1$