Edexcel International A Level Maths exam readiness check

Question 1

Solve $\log_{3}(2x + 1) - \log_{3}(x - 2) = 2$.

State clearly why any solution you reject is invalid.

Accepted Answer

$\log_{3}\dfrac{2x+1}{x-2} = 2 \Rightarrow \dfrac{2x+1}{x-2} = 9$.

$2x + 1 = 9x - 18 \Rightarrow 7x = 19 \Rightarrow x = \dfrac{19}{7}$.

Domain: we need $x - 2 > 0$, i.e. $x > 2$. Since $	frac{19}{7} \approx 2.71 > 2$, the solution is valid.

Question 2

A geometric series has first term $a$ and common ratio $r$, where $|r| < 1$.

The sum to infinity is $50$ and the second term is $12$.

Find the two possible values of $r$, and the corresponding values of $a$.

Accepted Answer

$\dfrac{a}{1-r} = 50 \Rightarrow a = 50(1 - r)$; also $ar = 12$.

$50(1-r)r = 12 \Rightarrow 50r - 50r^{2} = 12 \Rightarrow 50r^{2} - 50r + 12 = 0$.

Dividing by 2: $25r^{2} - 25r + 6 = 0 = (5r - 2)(5r - 3)$.

$r = 0.4 \Rightarrow a = 50(0.6) = 30$.

$r = 0.6 \Rightarrow a = 50(0.4) = 20$.

Question 3

Evaluate $\displaystyle\int_{1}^{e} x\ln x\,dx$, giving your answer in exact form.

Accepted Answer

$u = \ln x$, $\dfrac{dv}{dx} = x \Rightarrow v = \dfrac{x^{2}}{2}$.

$\int x\ln x\,dx = \dfrac{x^{2}}{2}\ln x - \int\dfrac{x^{2}}{2}\cdot\dfrac1x dx = \dfrac{x^{2}}{2}\ln x - \dfrac{x^{2}}{4}$.

At $x = e$: $\dfrac{e^{2}}{2}(1) - \dfrac{e^{2}}{4} = \dfrac{e^{2}}{4}$.

At $x = 1$: $0 - \dfrac14 = -\dfrac14$.

So the integral is $\dfrac{e^{2}}{4} + \dfrac14 = \dfrac{e^{2} + 1}{4}$.

Question 4

A curve is defined by $x = 3\cos t$, $y = 2\sin 2t$, for $0 \le t < 2\pi$.

Find the gradient of the curve at $t = \dfrac{\pi}{6}$, giving your answer in exact form.

Accepted Answer

$\dfrac{dx}{dt} = -3\sin t$, $\dfrac{dy}{dt} = 4\cos 2t$.

$\dfrac{dy}{dx} = \dfrac{4\cos 2t}{-3\sin t}$.

At $t = 	frac{\pi}{6}$: $\cos	frac{\pi}{3} = 	frac12$, $\sin	frac{\pi}{6} = 	frac12$.

$\dfrac{dy}{dx} = \dfrac{4 	imes 	frac12}{-3 	imes 	frac12} = -\dfrac43$.

Question 5

A particle moves in a straight line with velocity $v = 3t^{2} - 12t + 9$ m s⁻¹, $t \ge 0$.

Find the total distance travelled by the particle in the first 4 seconds.

Accepted Answer

$v = 3(t - 1)(t - 3)$, so $v = 0$ at $t = 1$ and $t = 3$.

$s(t) = t^{3} - 6t^{2} + 9t$ (taking $s(0) = 0$).

$s(1) = 4$, $s(3) = 0$, $s(4) = 4$.

Leg distances: $|4 - 0| = 4$, $|0 - 4| = 4$, $|4 - 0| = 4$.

Total distance $= 12$ m (displacement is only 4 m).

Edexcel International A Level Maths — free exam readiness check