If a curve $y=\mathrm{a} \sqrt{x}+\mathrm{b} x$ passes through the point $(1,2)$ and the area bounded by the…
If a curve $y=\mathrm{a} \sqrt{x}+\mathrm{b} x$ passes through the point $(1,2)$ and the area bounded by the curve, line $x=4$ and $\mathrm{X}$-axis is 8 sq. units, then
$\mathrm{a}=3, \mathrm{~b}=-1$
$\mathrm{a}=3, \mathrm{~b}=1$
$\mathrm{a}=-3, \mathrm{~b}=1$
$\mathrm{a}=-3, \mathrm{~b}=-1$
Solution
The given curve passes through $(1,2)$.
$\therefore \quad 2=a+b$
According to the given condition,
$\int_0^4(a \sqrt{x}+b x) d x=8$ ...(i)
$\Rightarrow \frac{2 \mathrm{a}}{3}\left[x^{3 / 2}\right]_0^4+\frac{\mathrm{b}}{2}\left[x^2\right]_0^4=8 \Rightarrow \frac{2 \mathrm{a}}{3} \cdot 8+8 \mathrm{~b}=8$
$\Rightarrow 2 a+3 b=3$ ...(ii)
From (i) and (ii), we get
$\mathrm{a}=3, \mathrm{~b}=-1$