If a curve y = asqrt{x} + bx passes through t | vedclass

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(A) The given curve is $y = a\sqrt{x} + bx$. Since it passes through $(1, 2)$,we have $2 = a(1) + b(1)$,which gives $a + b = 2$ ...$(i)$.The area boun

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$a = 3, b = -1$

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(A) The given curve is $y = a\sqrt{x} + bx$. Since it passes through $(1, 2)$,we have $2 = a(1) + b(1)$,which gives $a + b = 2$ ...$(i)$.The area bounded by the curve,$x = 4$,and the $X$-axis is given by $\int_0^4 (a\sqrt{x} + bx) dx = 8$.Evaluating the integral:$\int_0^4 (ax^{1/2} + bx) dx = \left[ a \cdot \frac{x^{3/2}}{3/2} + b \cdot \frac{x^2}{2} \right]_0^4 = 8$.Substituting the limits:$\left( \frac{2a}{3} \cdot 4^{3/2} + \frac{b}{2} \cdot 4^2 \right) = 8$.$\frac{2a}{3} \cdot 8 + \frac{b}{2} \cdot 16 = 8$.$\frac{16a}{3} + 8b = 8$.Dividing by $8$,we get $\frac{2a}{3} + b = 1$,which simplifies to $2a + 3b = 3$ ...(ii).Solving equations $(i)$ and (ii):From $(i)$,$b = 2 - a$. Substituting into (ii):$2a + 3(2 - a) = 3$.$2a + 6 - 3a = 3$.$-a = -3 \Rightarrow a = 3$.Substituting $a = 3$ into $(i)$:$3 + b = 2 \Rightarrow b = -1$.Thus,$a = 3$ and $b = -1$.

If a curve $y = a\sqrt{x} + bx$ passes through the point $(1, 2)$ and the area bounded by the curve,the line $x = 4$,and the $X$-axis is $8$ sq. units,then:

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