Rom

\(f(x) = a x^3 + b x^2 + c x + d\\ f(-1) = -a + b - c + d = 15\\ f(0) = d = 0\\ f(1) = a + b + c + d = -5\\ f(2) = 8a+4b+2c+d=12\\ \text{I like to use matrices, there are other ways to solve the following}\\ \begin{pmatrix} -1&1 &-1 \\ 1 &1 &1 \\ 8 &4 &2 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}= \begin{pmatrix}15\\-5\\12\end{pmatrix}\)

\(\text{Gaussian reducing this we get}\\ \begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}2\\5\\-12\end{pmatrix}\)

\(f(x) = 2x^3 + 5x^2 -12x = \\ x(2x^2 + 5x - 12) = \\ x(2x-3 )(x+4 )\\ \text{x intercepts at }x \in \left\{0,~\dfrac 3 2,~-4\right\}\)

I'd say 1 and 4 are the yes answers

2 and 3 should be no

\(h(y) = \dfrac{1+y}{2-y}\\ x = \dfrac{1+y}{2-y}\\ (2-y)x = 1+y\\ 2x-1=(1+x)y\\ y = \dfrac{2x-1}{x+1}\\ \text{now substitute }x \to y\\ h^{-1}(y) = \dfrac{2y-1}{y+1}\\ h^{-1}(5) = \dfrac{2(5)-1}{5+1}=\dfrac{9}{6}=\dfrac 3 2\)

\(\sum \limits_{k=2}^{100}\dfrac{1}{\log_k(100!)} = \\ \sum \limits_{k=2}^{100}\dfrac{\log(k)}{\log(100!)} = \\ \sum \limits_{k=2}^{100}\dfrac{\log(k)}{\sum \limits_{j=2}^{100}\log(j)} =\\ \dfrac{\sum \limits_{k=2}^{100} \log(k)}{\sum \limits_{j=2}^{100} \log(j)} = \\ 1\)

\(\text{relatively prime means no factors in common}\\ \text{12 and 25 are relatively prime}\\ \text{12 and 27 have 3 as a common factor}\\ \text{14 and 21 have 7 as a common factor}\\ \text{15 and 32 are relatively prime} \)

quick rule of thumb

16% of the population is a z-score of -1, i.e. 1 sigma less than the mean.

\(\dfrac{x - \mu}{\sigma} = -1 \Rightarrow x = \mu - \sigma\)

So we would have scores less than 73-4 = 69 being 16%

This doesn't match any of the choices so we note that it could be the top 16% in which case the z-score is 1 and we have

scores greater than x = 73+4 = 77 being 16%

this matches the 3rd choice.

Again Lagrange multipliers make short work of this.  This problem requires the use of 2 of them.

\(s = 2+3\sqrt{2} \text{ is the largest value of }s \text{ satisfying the constraints}\)

\(\text{This one I'd do using Lagrange multipliers}\\ f(a,b,c,d) = a \cdot \sqrt{b} \cdot \sqrt[3]{c} \cdot \sqrt[4]{d}\\ g(a,b,c,d) = 36a+4b+4c+3d-25\\ \text{Solve}\\ \nabla(f-\lambda g) = 0\)

\(y = \pm \sqrt{6-(x-3)^2}+ 3\\ \dfrac{y}{x} = \dfrac{ \pm \sqrt{6-(x-3)^2}+ 3}{x}\\ \text{The maximum will be when the numerator has a positive sign}\\ \dfrac{y}{x} = \dfrac{ \sqrt{6-(x-3)^2}+ 3}{x}\\ \dfrac{d}{dx} \dfrac{y}{x} = -\dfrac{3 (x-1)}{x^2 \sqrt{-x^2+6 x-3}}\\ \text{and this clearly equals 0 at }x=1\)

\(\text{The second derivative evaluated at 1 is less than zero so this is a maximum}\)

\(\dfrac{y}{x} = \sqrt{6-(1-3)^2}+3 = 3+\sqrt{2}\)

\(\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}\\ n! = n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot 3 \cdot 2\)