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

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Question

(A) Given the curve $y = a\sqrt{x} + bx$. Since the curve passes through $(1, 2)$,we have $2 = a(1)^{1/2} + b(1)$,which implies $a + b = 2$ ... $(i)$.

Vedclass · Accepted Answer

$a = 3, b = -1$

Vedclass · Answer

(A) Given the curve $y = a\sqrt{x} + bx$. Since the curve passes through $(1, 2)$,we have $2 = a(1)^{1/2} + b(1)$,which implies $a + b = 2$ ... $(i)$. The area bounded by the curve,the line $x = 4$,and the $x$-axis is given by $\int_{0}^{4} (a\sqrt{x} + bx) dx = 8$. Evaluating the integral: $\left[ \frac{2a}{3}x^{3/2} + \frac{b}{2}x^2 \right]_{0}^{4} = 8$. Substituting the limits: $\frac{2a}{3}(8) + \frac{b}{2}(16) = 8$. This simplifies to $\frac{16a}{3} + 8b = 8$. Dividing by $8$,we get $\frac{2a}{3} + b = 1$,or $2a + 3b = 3$ ... $(ii)$. From $(i)$,$b = 2 - a$. Substituting into $(ii)$: $2a + 3(2 - a) = 3$. $2a + 6 - 3a = 3$,which gives $-a = -3$,so $a = 3$. Then $b = 2 - 3 = -1$. 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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