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If a curve $y=a \sqrt{x}+b x$ passes through the point $(1,2)$ and the area bounded by the curve, line $x=4$ and $x$-axis is $8 \mathrm{sq}$. unit, then
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The correct answer is:
$a=3, b=-1$
Given curve $y=a \sqrt{x}+b x$. This curve passes through (1, 2), $\therefore 2=a+b...(i)$ and area bounded by this curve and line $x=4$ and $\quad x$-axis is 8 sq. unit, then $\int_0^4(a \sqrt{x}+b x) d x=8$
$\begin{aligned}& \Rightarrow \quad \frac{2 a}{3}\left[x^{3 / 2}\right]_0^4+\frac{b}{2}\left[x^2\right]_0^4=8, \frac{2 a}{3} \cdot 8+8 b=8 \\& \Rightarrow \quad 2 a+3 b=3 \\&\end{aligned}$
From equation (i) and (ii), we get $a=3, b=-1$
$\begin{aligned}& \Rightarrow \quad \frac{2 a}{3}\left[x^{3 / 2}\right]_0^4+\frac{b}{2}\left[x^2\right]_0^4=8, \frac{2 a}{3} \cdot 8+8 b=8 \\& \Rightarrow \quad 2 a+3 b=3 \\&\end{aligned}$
From equation (i) and (ii), we get $a=3, b=-1$
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