IB Mathematics Practice Exam — Free

By the binomial theorem, the general term of $(a + b)^6$ is $\dbinom{6}{k} a^{6-k} b^{k}$. With $a = x^2$ and $b = \dfrac{2}{x} = 2x^{-1}$,

The constant term is the one with no $x$, so set the exponent to zero: $12-3k = 0$ gives $k = 4$. Substituting,

(Here $\dbinom{6}{4} = \dfrac{6 \cdot 5}{2} = 15$, the same as $\dbinom{6}{2}$.)

The tempting wrong answer is 15, which is just the binomial coefficient and forgets that the 2 in $\dfrac{2}{x}$ is also raised to the fourth power. The answer 60 uses $2^2$ instead of $2^4$ (confusing the power of $b$ with the power of $a$), and 160 is $\dbinom{6}{3}2^3$, the coefficient of $x^{3}$ rather than the constant term.

The constant term is 240 (option D).

Apply the transformations one at a time, in the order given, since the order matters when stretches and translations are mixed.

Translation 3 units right replaces $x$ by $x - 3$: $y = (x - 3)^2$.

Vertical stretch with scale factor 2 multiplies every $y$-value by 2: $y = 2(x - 3)^2$.

Translation 5 units down subtracts 5: $y = 2(x - 3)^2 - 5$.

(a) Expanding, $g(x) = 2\left(x^2 - 6x + 9\right) - 5 = 2x^2 - 12x + 18 - 5 = 2x^2 - 12x + 13$. The vertex is $(3, -5)$, which is where the original vertex $(0, 0)$ ends up, a useful confirmation that the transformations were applied in the right order.

(b) The $x$-intercepts solve $g(x) = 0$. The vertex form is easier than the quadratic formula here:

So $x = 3 \pm \dfrac{\sqrt{10}}{2}$, approximately $1.42$ and $4.58$. (Both are real because the vertex is below the $x$-axis and the parabola opens upward; the quadratic formula on $2x^2 - 12x + 13 = 0$ gives the same $x = \dfrac{12 \pm \sqrt{40}}{4}$.)

(a) $g(x) = 2x^2 - 12x + 13$; (b) $x = 3 - \dfrac{\sqrt{10}}{2}$ and $x = 3 + \dfrac{\sqrt{10}}{2}$

(a) Standardize so that a normal table or calculator can be used: $Z = \dfrac{M - 180}{15}$.

So about 9.12% of oranges are lighter than 160 g.

(b) Let $m$ be the minimum premium mass. "Heaviest 8%" means $P(M > m) = 0.08$, so $P(M < m) = 0.92$. The $z$-value with $P(Z < z) = 0.92$ is $z = 1.405$ (from the inverse normal). Convert back to grams:

This is about 1.4 standard deviations above the mean, as expected for the top 8%.

(c) Each orange is premium with probability $p = 0.08$ (by the definition in part (b), so no new calculation is needed), independently of the others, so the number of premium oranges $X$ among six is binomial: $X \sim \mathrm{B}(6,\ 0.08)$.

(Here $0.92^4 = 0.71639$ and $15\times 0.0064 = 0.096$, so the product is $0.0688$ to 3 significant figures.)

(a) $P(M < 160) = 0.0912$; (b) minimum premium mass $= 201$ g; (c) $P(X = 2) = 0.0688$.