Question

If the curve $y=a{x}^{\frac{1}{2}}+bx$ passes through the point (1, 2) and lies above the x-axis in $0\le x\le 9$ and the area enclosed by the curve, the x axis and the line x = 4 is 8 sq. units. Then

• A. a = 1
• B. b = 1
• C. a = 3
• D. b = - 1

Answer 1

Answer: (C), (D)

The correct options are
C

a = 3

D

b = - 1

Since the curve $y=a{x}^{\frac{1}{2}}+bx$ passes through the point (1, 2)
$\therefore$ 2 = a + b .... (1)
By observation the curve also passes through (0, 0). Therefore, the area enclosed by the curve, x-axis and x = 4 is given by
$A={\int }_0^4(a{x}^{\frac{1}{2}}+bx)dx=8\Rightarrow \frac{2a}{3}\times 8+\frac{b}{2}\times 16=8$
$\Rightarrow \frac{2a}{3}+b=1$ ..... (2)
On solving eq (1) and (2), we get a = 3, b = -1