Graduation practice paper 1

$f'(x)$ changes sign from $+$ to $-$ at $x=2$, so the function has a local maximum at $x=2$ and the local maximum value is $f(2)=1$. Option A is the $x$-coordinate of the maximum point, option C is the local minimum value and option D is the $x$-coordinate of the minimum point.

Revise: derivatives and graphs of functions.

We find the limit as $x\to\pm\infty$ by dividing the numerator and the denominator by $x$:

$$\lim_{x\to\pm\infty}\frac{4x+1}{2x-6}=\lim_{x\to\pm\infty}\frac{4+\frac1x}{2-\frac6x}=\frac42=2.$$

So the horizontal asymptote is $y=2$. Option A uses only the coefficient in the numerator, option B is the vertical asymptote and option D is the ratio of the constants $\dfrac{1}{-6}$.

Revise: derivatives and graphs of functions.

We study the sign of the derivative:

$$y'=3x^2-6x=3x(x-2),\qquad y'<0\iff 0<x<2.$$

So the function is decreasing on $(0, 2)$. Options B and C are the intervals where it is increasing ($y'>0$); option D comes from solving $y'=0$ wrongly as $x=0$, $x=-2$.

Revise: derivatives and graphs of functions.

We use $\int\cos x\,dx=\sin x+C$ and $\int x^2\,dx=\dfrac{x^3}{3}+C$:

$$\int\left(2\cos x-3x^2\right)dx=2\sin x-3\cdot\frac{x^3}{3}+C=2\sin x-x^3+C.$$

Option A has the wrong sign for the antiderivative of $\cos x$, option B differentiates $3x^2$ instead of integrating it, and option D forgets to divide by $3$.

Revise: antiderivatives and integrals.

The definite integral is linear, so

$$\int_1^3\left[2f(x)-3g(x)\right]dx=2\cdot4-3\cdot(-1)=8+3=11.$$

Option A drops the minus sign of $-1$ ($8-3$), option C forgets the factor $2$ and option D forgets the factor $3$.

Revise: antiderivatives and integrals.

Each coordinate of the midpoint is the mean of the coordinates of the endpoints:

$$I\left(\frac{2+4}{2}, \frac{-1+3}{2}, \frac{3+(-1)}{2}\right)=I(3, 1, 1).$$

Option A is half the difference, option B is the sum (forgetting to divide by $2$) and option C is the vector $\overrightarrow{AB}$.

Revise: coordinates in space.

The three points lie on the three axes, so we use the intercept form of the equation of a plane and multiply both sides by $6$:

$$\frac x2+\frac{y}{-3}+\frac z6=1\iff 3x-2y+z-6=0.$$

Option B drops the minus sign of $-3$, option C uses the intercepts as coefficients, and option D has the wrong sign of the constant (substituting $M$ gives $6+6\neq0$).

Revise: coordinates in space.

The sphere $(x-a)^2+(y-b)^2+(z-c)^2=R^2$ has centre $I(a, b, c)$ and radius $R$. Here $a=1$, $b=-2$, $c=3$ and $R^2=16$, so $R=4$. Option A changes the signs of the centre's coordinates and option B forgets to take the square root.

Revise: coordinates in space.

$AC$ is the orthogonal projection of $SC$ onto $(ABCD)$, so the required angle is $\angle SCA$. The diagonal of the square is $AC=a\sqrt2$, so

$$\tan\angle SCA=\frac{SA}{AC}=\frac{a\sqrt6}{a\sqrt2}=\sqrt3\ \Rightarrow\ \angle SCA=60^\circ.$$

Option A comes from the upside-down ratio $\dfrac{AC}{SA}$; option D wrongly takes $SC$ to be perpendicular to the base.

Revise: geometry in space.

We use the midpoint of each class, $10$, $30$, $50$, $70$, $90$:

$$\bar x=\frac{5\cdot10+12\cdot30+15\cdot50+6\cdot70+2\cdot90}{40}=\frac{1760}{40}=44.$$

Option A uses the lower class boundaries, option B is the midpoint of the middle class and option D is the mean of the frequencies.

Revise: statistics and probability.

By the definition of conditional probability:

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.3}{0.5}=0.6.$$

Option A is $P(B\mid A)=\dfrac{0.3}{0.6}$ (dividing by $P(A)$ by mistake), option B multiplies $P(A\cap B)$ by $P(B)$ instead of dividing, and option C is $P(A\cap B)$.

Revise: statistics and probability.

Write $27=3^3$; the base $3$ is greater than $1$, so the inequality keeps its direction:

$$3^{2x-1}\le3^3\iff 2x-1\le3\iff x\le2.$$

Option A loses the equals sign, option C reverses the inequality (that only happens when the base is less than $1$), and option D compares $2x-1$ with $27$ instead of with the exponent $3$.

Revise: trigonometry, exponentials, logarithms and sequences.

a) True: differentiate term by term, $(x^3)'=3x^2$, $(-6x^2)'=-12x$, $(9x)'=9$.

b) False: $f'(x)=3(x-1)(x-3)<0$ for $1<x<3$, so the function is decreasing on $(1, 3)$.

c) True: $f'$ changes sign from $+$ to $-$ at $x=1$, so the local maximum value is $f(1)=1-6+9+1=5$.

d) False: compare $f(0)=1$, $f(1)=5$, $f(3)=1$, $f(4)=64-96+36+1=5$. The maximum value is $5$, but it is reached at two points, $x=1$ and $x=4$.

Revise: derivatives and graphs of functions.

a) True: $v(1)=12-3=9$ (m/s).

b) False: $v'(t)=12-6t<0$ for $t>2$, so the velocity increases only during the first $2$ seconds and then decreases.

c) True: $h(t)$ is the antiderivative of $v(t)$ with $h(0)=0$: $h(t)=6t^2-t^3+C$ and $h(0)=0$ gives $C=0$.

d) True: $v(t)=3t(4-t)\ge0$ on $[0, 4]$, so the height increases and is greatest at $t=4$: $h(4)=\displaystyle\int_0^4\left(12t-3t^2\right)dt=96-64=32$ (m).

Revise: antiderivatives and integrals.

a) True: $\overrightarrow{AB}=(3-1, 2-0, 0-2)=(2, 2, -2)$.

b) True: $I(2, 1, 1)$ and $2\cdot2+1-1-4=0$.

c) False: the perpendicular bisector plane passes through $I(2, 1, 1)$ with normal vector $\vec n=(1, 1, -1)$, so its equation is $(x-2)+(y-1)-(z-1)=0$, that is $x+y-z-2=0$. The plane $x+y-z=0$ does not pass through $I$.

d) True: with $\vec u=(1, 1, -1)$ and $\vec n_P=(2, 1, -1)$,

$$\sin\varphi=\frac{|2+1+1|}{\sqrt3\cdot\sqrt6}=\frac{4}{3\sqrt2}\approx0.943\ \Rightarrow\ \varphi\approx70.5^\circ>60^\circ.$$

Revise: coordinates in space.

Let $A$ be the event "the product was made on line I" and $L$ the event "the product is defective".

a) True: we are given $P(L\mid A)=0.03$.

b) False: by the law of total probability, $P(L)=0.6\cdot0.03+0.4\cdot0.05=0.018+0.02=0.038$. The number $0.04$ is the plain mean of the two rates, which ignores how much each line produces.

c) False: $P(A\cap L)=P(A)\cdot P(L\mid A)=0.6\cdot0.03=0.018$; the number $0.03$ is the conditional probability.

d) True: by Bayes' theorem, $P(\overline A\mid L)=\dfrac{0.02}{0.038}=\dfrac{10}{19}\approx0.526>0.5$.

Revise: statistics and probability.

Let the printed width be $x$ cm ($x>0$); the printed height is $\dfrac{384}{x}$ cm. The area of the poster is

$$S(x)=(x+4)\left(\frac{384}{x}+6\right)=408+6x+\frac{1536}{x}.$$

$S'(x)=6-\dfrac{1536}{x^2}=0\iff x^2=256\iff x=16$, and $S'$ changes sign from $-$ to $+$ at $x=16$. Then the printed area is $16\times24$ cm and the poster is $20\times30$ cm, so the smallest area is $600$ cm².

Revise: optimisation and real-life problems.

We set up coordinates with $A$ at the origin and the axes along $AB$, $AD$, $AA'$: $B(3, 0, 0)$, $D(0, 4, 0)$, $A'(0, 0, 12)$. A normal vector $\vec n=(a, b, c)$ of $(A'BD)$ is perpendicular to $\overrightarrow{BD}=(-3, 4, 0)$ and $\overrightarrow{BA'}=(-3, 0, 12)$:

$$-3a+4b=0,\qquad -3a+12c=0\ \Rightarrow\ \text{choose } a=4,\ b=3,\ c=1.$$

So $(A'BD)\colon 4(x-3)+3y+z=0$, that is $4x+3y+z-12=0$, and

$$d(A, (A'BD))=\frac{|4\cdot0+3\cdot0+0-12|}{\sqrt{4^2+3^2+1^2}}=\frac{12}{\sqrt{26}}\approx2.353.$$

To 2 decimal places the distance is $2.35$. Note: because $AB$, $AD$, $AA'$ are pairwise perpendicular, the result can be checked with $\dfrac1{h^2}=\dfrac1{AB^2}+\dfrac1{AD^2}+\dfrac1{AA'^2}$, but this formula is not in the textbook.

Revise: geometry in space.

The line has direction vector $\overrightarrow{AB}=(30, 40, -15)$ and the ground has normal vector $\vec k=(0, 0, 1)$. Then

$$\sin\varphi=\frac{|-15|}{\sqrt{30^2+40^2+15^2}}=\frac{15}{\sqrt{2725}}\approx0.287\ \Rightarrow\ \varphi\approx16.7^\circ.$$

To the nearest degree, the angle between the zip line and the ground is about $17^\circ$.

Revise: coordinates in space.

"Exactly two sensors detect it" is the union of three mutually exclusive events (one sensor misses, the other two detect). The sensors are independent, so we multiply the probabilities in each case and add:

$$P=0.9\cdot0.8\cdot0.5+0.9\cdot0.2\cdot0.5+0.1\cdot0.8\cdot0.5=0.36+0.09+0.04=0.49.$$

Revise: statistics and probability.

The intersections: $x^2-2x=x\iff x^2-3x=0\iff x=0$ or $x=3$. On $[0, 3]$ the line lies above the parabola (test $x=1$: $1>-1$), so

$$S=\int_0^3\left(x-x^2+2x\right)dx=\int_0^3\left(3x-x^2\right)dx=\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3=\frac{27}{2}-9=4.5.$$

Revise: antiderivatives and integrals.

From $T(5)=60$ we get $70e^{-5k}=35$, so $e^{-5k}=\dfrac12$ and $k=\dfrac{\ln2}{5}$. Solve $T(t)=30$:

$$70e^{-kt}=5\iff e^{-kt}=\frac1{14}\iff t=\frac{\ln14}{k}=\frac{5\ln14}{\ln2}\approx19.04.$$

So the tea reaches $30^\circ$C after about $19$ minutes.

Revise: trigonometry, exponentials, logarithms and sequences.