So far as I solved I not sure are
1 b
2 A Or C not sure
3 D
4 B
the rest is not yet solved I have to work on them to make sure
to #1:
$\displaystyle f'(x) = \int(6x-12)dx=3x^2-12x+C$ Since f has a stationary point at (0,0) you have:
$\displaystyle f'(x) = 3x^2-12x$
$\displaystyle f(x) = \int(3x^2-12x)dx=x^3-6x+C$ Since (0, 0) belongs to f you have:
$\displaystyle f(x) = x^3-6x$ So it is answer 1.c)
to #2:
$\displaystyle f(x)=\int(\frac1x+2)dx=\ln(x)+2x+C$
Therefore:
$\displaystyle f(e)-f(1)=(\ln(e)+2e+C) - (\ln(1)+2+C) = 1+2e+C-0-2-C = 2e-1$ . So it's 2.a)
to #3:
$\displaystyle \int_1^2\left(\frac a{x^2}+3\right)dx=\left. -\frac ax+3x\right|_1^2 = \left(-\frac12 a + 6\right) - \left(-a+3\right) = \frac12 a + 3$
According to your question you have to solve for a:
$\displaystyle \frac12 a + 3 = 4~\iff~\boxed{a = 2}$ Therefore it's answer 3.b)
$\displaystyle f(x)=\sqrt{x^2-2x}~,~x\leq 0~\vee~x\geq2$
$\displaystyle f'(x)=\frac12 \cdot (x^2-2x)^{-\frac12} \cdot (2x-2) = \frac{2x-2}{2\sqrt{x^2-2x}}$
$\displaystyle f'(x) = 0~\implies~x=1$ ..... But f is not define for x = 1 therefore f don't have any stationary points. So it's answer 5.d)
#6 Line has negative slope therefore minimum value occurs at right endpoint. So solve f(2a) = 8 for a ....
#7 Solve f'(a) = 1 for a. Then b = f(a) = .....
#8 I don't like any of the answers or the question for that matter. But they probably want the answer corresponding to $\displaystyle \frac{d^2 y}{dx^2}$ .....
#9 Evaluate $\displaystyle \int_2^3 f(x) \, dx$.
#10 Option (d) is wrong because it fails y(0) = 0. So test two of (a), (b) or (c). Either one of the two is correct, or both are wrong. Either way, you find the correct option.