Question 8 of 12

Question 8

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) $f_1(x)=x^3$, $f_2(x)=x^4+x$, both $R\to R$. (b) A 13-card hand dealt from a standard 52-card deck (4 suits of 13 cards each).

Find. (a) Invertibility of each function, and $f^{-1}$ where it exists. (b) $P(\text{hand contains all 4 suits})$.

Approach. A function $R\to R$ is invertible iff it is a bijection; check monotonicity (via the derivative) to settle both injectivity and surjectivity for odd/even-degree polynomials. Use inclusion–exclusion over "missing suit $i$" for (b).

1. a1) $f_1(x)=x^3$. $f_1'(x)=3x^2\ge0$ everywhere, and $=0$ only at the single point $x=0$, so $f_1$ is strictly increasing on all of $R$ (a single stationary point does not break strict monotonicity) — hence injective. As an odd-degree polynomial, $f_1(x)\to\pm\infty$ as $x\to\pm\infty$, so it is also surjective onto $R$. $\boxed{f_1 \text{ is invertible}}$, with $\boxed{f_1^{-1}(y)=\sqrt[3]{y}}$ (the real cube root, defined for all $y\in R$).
2. a2) $f_2(x)=x^4+x$. $f_2'(x)=4x^3+1$, which is zero at $x=-\left(\tfrac14\right)^{1/3}\approx-0.63$ and changes sign there (negative for $x<-0.63$, positive after) — so $f_2$ decreases then increases, i.e. it is NOT monotonic on $R$. Concretely $f_2$ has a strict local minimum at that point, so it takes some output values twice (once on each side of the minimum). $\boxed{f_2 \text{ is NOT invertible}}$ (not injective — e.g. as an even-degree-leading polynomial with a genuine local min below its behaviour elsewhere, some horizontal line crosses the graph twice).
3. b) Count hands containing all four suits, by inclusion–exclusion. Let $A_i$ = "hand has no card of suit $i$" ($i=1,\dots,4$). Want $\big|\overline{A_1\cup A_2\cup A_3\cup A_4}\big|$: $$\Big|\bigcap_i \overline{A_i}\Big| = \sum_{i=0}^{4}(-1)^i\binom{4}{i}\binom{52-13i}{13}$$ Evaluating term by term ($i=0,1,2,3,4$; the $i=4$ term is $\binom{4}{4}\binom{0}{13}=0$): $$\binom{52}{13} - 4\binom{39}{13} + 6\binom{26}{13} - 4\binom{13}{13} = 635{,}013{,}559{,}600 - \dots = 602{,}586{,}261{,}420$$
4. b) Form the probability. $$P=\frac{602{,}586{,}261{,}420}{635{,}013{,}559{,}600}=\frac{30{,}129{,}313{,}071}{31{,}750{,}677{,}980}$$ $\boxed{P\approx0.9489}$