IB Maths AA exam readiness check

Question 1

In an arithmetic sequence, the 4th term is $17$ and the 9th term is $42$.

(a) Find the common difference and the first term.

(b) Find the sum of the first 20 terms.

Accepted Answer

(a) $u_1 + 3d = 17$ and $u_1 + 8d = 42$. Subtracting: $5d = 25 \Rightarrow d = 5$, so $u_1 = 17 - 15 = 2$.

(b) $S_{20} = 	frac{20}{2}(2(2) + 19(5)) = 10(4 + 95) = 990$.

Question 2

The quadratic $f(x) = kx^{2} + (k + 2)x + 3$ has two distinct real roots.

Find the set of possible values of $k$, $k 
eq 0$.

Accepted Answer

$b^{2} - 4ac = (k+2)^{2} - 4(k)(3) = k^{2} + 4k + 4 - 12k = k^{2} - 8k + 4 > 0$.

Critical values: $k = \dfrac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3}$.

Since the coefficient of $k^{2}$ is positive, the expression is positive outside the roots:

$k < 4 - 2\sqrt{3}$ or $k > 4 + 2\sqrt{3}$, with $k 
eq 0$.

Question 3

The curve $C$ has equation $y = \dfrac{4}{x} + x^{2}$, $x > 0$.

Find the equation of the normal to $C$ at the point where $x = 2$, in the form $ax + by + c = 0$ with integer coefficients.

Accepted Answer

$y = 4x^{-1} + x^{2} \Rightarrow \dfrac{dy}{dx} = -4x^{-2} + 2x$.

At $x = 2$: gradient $= -1 + 4 = 3$, and $y = 2 + 4 = 6$.

Normal gradient $= -\dfrac13$, through $(2, 6)$:

$y - 6 = -	frac13(x - 2) \Rightarrow 3y - 18 = -x + 2 \Rightarrow x + 3y - 20 = 0$.

Question 4

The masses of apples are normally distributed with mean $180$ g and standard deviation $22$ g.

(a) Find the probability that a randomly chosen apple has mass greater than $200$ g.

(b) 15% of apples are classified as small. Find the maximum mass of a small apple, to the nearest gram.

Accepted Answer

(a) $z = \dfrac{200 - 180}{22} = 0.909$, so $P(X > 200) = 0.182$ (3 s.f.).

(b) $P(X < x) = 0.15 \Rightarrow z = -1.0364$.

$x = 180 + (-1.0364)(22) = 157.2$ g, so the maximum mass of a small apple is $	extbf{157 g}$.

Question 5

The curve $y = 6x - x^{2}$ meets the line $y = 2x$ at two points.

Find the exact area of the region enclosed between the curve and the line.

Accepted Answer

$6x - x^{2} = 2x \Rightarrow 4x - x^{2} = 0 \Rightarrow x(4 - x) = 0$, so $x = 0, 4$.

Area $= \displaystyle\int_{0}^{4}(4x - x^{2})dx = \left[2x^{2} - \dfrac{x^{3}}{3}\right]_{0}^{4} = 32 - \dfrac{64}{3} = \dfrac{32}{3}$.

IB Maths AA SL — free exam readiness check