1. Multi questions

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

2. Originally Posted by soso love
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:
$f'(x) = \int(6x-12)dx=3x^2-12x+C$ Since f has a stationary point at (0,0) you have:
$f'(x) = 3x^2-12x$

$f(x) = \int(3x^2-12x)dx=x^3-6x+C$ Since (0, 0) belongs to f you have:
$f(x) = x^3-6x$ So it is answer 1.c)

to #2:
$f(x)=\int(\frac1x+2)dx=\ln(x)+2x+C$

Therefore:

$f(e)-f(1)=(\ln(e)+2e+C) - (\ln(1)+2+C) = 1+2e+C-0-2-C = 2e-1$ . So it's 2.a)

to #3:

$\int_1^2\left(\frac a{x^2}+3\right)dx=\left. -\frac ax+3x\right|_1^2 = \left(-\frac12 a + 6\right) - \left(-a+3\right) = \frac12 a + 3$

According to your question you have to solve for a:

$\frac12 a + 3 = 4~\iff~\boxed{a = 2}$ Therefore it's answer 3.b)

3. Multi

Thanks for everything earboth

So far I can tell the answer for:

5 c stathionary point= 1
is that right!!

sorry I am not that very good in math

4. Originally Posted by soso love
So far I can tell the answer for:

5 c stathionary point= 1
is that right!! ..... Sorry, but no.
$f(x)=\sqrt{x^2-2x}~,~x\leq 0~\vee~x\geq2$

$f'(x)=\frac12 \cdot (x^2-2x)^{-\frac12} \cdot (2x-2) = \frac{2x-2}{2\sqrt{x^2-2x}}$

$f'(x) = 0~\implies~x=1$ ..... But f is not define for x = 1 therefore f don't have any stationary points. So it's answer 5.d)

5. Hlp

May you answers the rest of the questions 6 to 10

even if you dont write the whole mathmatical asnwer or rule you used to solve the question it is okey because I want to try my best

thanks very much

6. Originally Posted by soso love
May you answers the rest of the questions 6 to 10

even if you dont write the whole mathmatical asnwer or rule you used to solve the question it is okey because I want to try my best

thanks very much
#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 $\frac{d^2 y}{dx^2}$ .....

#9 Evaluate $\int_2^3 f(x) \, dx$.

#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.