Graduation practice paper 2

We study the sign of $f'(x)$; the factor $(x-2)^2$ is never negative, so it does not make $f'$ change sign:

$f'$ changes sign from $-$ to $+$ only at $x=4$, so the function has a local minimum at $x=4$. Option A treats every root of $f'$ as a turning point, and option B is the local maximum point.

Revise: derivatives and graphs of functions.

Divide the numerator by the denominator:

$$\frac{2x^2+3x-1}{x+2}=2x-1+\frac{1}{x+2}.$$

Since $\dfrac1{x+2}\to0$ as $x\to\pm\infty$, the oblique asymptote is $y=2x-1$. Option B just takes the first two coefficients of the numerator, option C forgets the coefficient $2$ of $x^2$, and option D is the vertical asymptote.

Revise: derivatives and graphs of functions.

$y=x^3+x-5$ has $y'=3x^2+1>0$ for every $x$, so it is increasing on $\mathbb{R}$. Function A has $y'=4x(x^2+1)<0$ for $x<0$; function B is not defined at $x=3$; function D has $y'=3x^2-3<0$ on $(-1, 1)$.

Revise: derivatives and graphs of functions.

We use $\int a^x\,dx=\dfrac{a^x}{\ln a}+C$ and $\int\dfrac1x\,dx=\ln|x|+C$ (for $x>0$, $|x|=x$). Option A is the derivative of $3^x$, option C wrongly uses the rule for powers of $x$, and option D differentiates $\dfrac1x$.

Revise: antiderivatives and integrals.

We split at $x=2$, where $x-2$ changes sign:

$$\int_0^3|x-2|\,dx=\int_0^2(2-x)\,dx+\int_2^3(x-2)\,dx=2+\frac12=\frac52.$$

Option A ignores the modulus, option B subtracts the two parts instead of adding them, and option C only computes the part on $[0, 2]$.

Revise: antiderivatives and integrals.

We have $u_4=u_1q^3$, so $q^3=\dfrac{-54}{2}=-27$ and $q=-3$. Option B loses the minus sign, option C forgets to take the cube root, and option D wrongly uses the arithmetic sequence formula $u_4=u_1+3d$.

Revise: trigonometry, exponentials, logarithms and sequences.

A plane perpendicular to $d$ has the direction vector $\vec u=(2, -1, 3)$ of $d$ as a normal vector:

$$2(x-1)-(y+1)+3(z-2)=0\iff 2x-y+3z-9=0.$$

Option A has the wrong sign of the constant, option B has the wrong sign of the $y$ coefficient, and option D is the plane through the point $(1, 0, -3)$ of $d$ instead of through $A$.

Revise: coordinates in space.

The radius is $IA$:

$$R^2=IA^2=(0-2)^2+(1+1)^2+(2-1)^2=9.$$

So the sphere is $(x-2)^2+(y+1)^2+(z-1)^2=9$. Option A writes $R$ instead of $R^2$; options C and D change the signs of the centre's coordinates.

Revise: coordinates in space.

$ABC'D'$ is a rectangle, so $BC'\parallel AD'$ and the required angle equals the angle between $AB'$ and $AD'$. The three sides of triangle $AB'D'$ are diagonals of three faces of the cube, so the triangle is equilateral and the angle is $60^\circ$. Option A confuses it with the angle between a face diagonal and an edge; option B comes from the faces $(ABB'A')$ and $(BCC'B')$ being perpendicular.

Revise: geometry in space.

The angle between $SC$ and the base is $\angle SCA=45^\circ$, so triangle $SAC$ is a right-angled isosceles triangle at $A$ and $SA=AC=\sqrt{3^2+4^2}=5$. Hence

$$V=\frac13\cdot AB\cdot AD\cdot SA=\frac13\cdot12\cdot5=20.$$

Option B forgets the factor $\dfrac13$, option C takes $SA=AD=4$ by mistake, and option D uses the area of triangle $ABC$ as the base area.

Revise: geometry in space.

The class with the highest frequency is $[60, 70)$ with $n_m=15$; the class before it has $n_{m-1}=9$ and the class after it has $n_{m+1}=10$:

$$M_o=60+\frac{15-9}{(15-9)+(15-10)}\cdot10=60+\frac{60}{11}\approx65.45.$$

Option A is the midpoint of the modal class, option B swaps the two differences in the numerator ($15-10$ instead of $15-9$), and option C is the highest frequency.

Revise: statistics and probability.

"Exactly one scores" consists of two mutually exclusive cases: An scores and Binh misses, or An misses and Binh scores. The throws are independent, so

$$P=0.7\cdot0.4+0.3\cdot0.6=0.28+0.18=0.46.$$

Option A is the probability that both score, option C the probability that at least one scores, and option D the probability of the complementary event.

Revise: statistics and probability.

a) True: $\lim\limits_{x\to1^+}\dfrac{x+3}{x-1}=+\infty$.

b) False: $y'=\dfrac{1\cdot(-1)-3\cdot1}{(x-1)^2}=\dfrac{-4}{(x-1)^2}<0$, so the function is decreasing on each interval of its domain.

c) True: the function is decreasing on $[2, 5]$, so its maximum value is $y(2)=\dfrac51=5$.

d) True: write $y=1+\dfrac{4}{x-1}$. For integer $x$, $y$ is an integer exactly when $x-1$ is a divisor of $4$, that is $x-1\in\{\pm1, \pm2, \pm4\}$: there are $6$ values of $x$, so $6$ points.

Revise: derivatives and graphs of functions.

a) True: $v(t)$ is an antiderivative of $a(t)$, $v(t)=2t-0.05t^2+C$, and $v(0)=0$ gives $C=0$.

b) False: $v(20)=40-0.05\cdot400=20$ (m/s). The value $40$ keeps only $2t$ and forgets the term $-0.05t^2$.

c) False: $\displaystyle\int_0^{10}\left(2t-0.05t^2\right)dt=\left[t^2-\frac{0.05t^3}{3}\right]_0^{10}=100-\frac{50}{3}\approx83.3$ (m).

d) True: in the first $20$ seconds the train travels $\displaystyle\int_0^{20}\left(2t-0.05t^2\right)dt=400-\frac{400}{3}=\frac{800}{3}\approx266.7$ (m); in the next $40$ seconds it travels $20\cdot40=800$ (m). In total about $1066.7$ m $>1000$ m.

Revise: antiderivatives and integrals.

a) True: $\overrightarrow{AB}=(4-0, 6-0, 0-12)=(4, 6, -12)$.

b) True: $AB=\sqrt{16+36+144}=\sqrt{196}=14$ (m).

c) False: the ground has normal vector $\vec k=(0, 0, 1)$, so $\sin\varphi=\dfrac{12}{14}\approx0.857$ and $\varphi\approx59.0^\circ<60^\circ$.

d) True: the plane $(OAB)$ contains the vectors $\overrightarrow{OA}=(0, 0, 12)$ and $\overrightarrow{OB}=(4, 6, 0)$. A normal vector $\vec n=(a, b, c)$ is perpendicular to both:

$$12c=0,\qquad 4a+6b=0\ \Rightarrow\ \text{choose } c=0,\ a=3,\ b=-2.$$

The plane passes through $O$ with normal vector $\vec n=(3, -2, 0)$, so its equation is $3x-2y=0$.

Revise: coordinates in space.

Let $D$ be the event "the marble moved is red" and $E$ the event "the marble taken from box II is red".

a) True: $P(D)=\dfrac46=\dfrac23$.

b) False: after the move, box II has $9$ marbles, $4$ of them red, so $P(E\mid D)=\dfrac49$. The value $\dfrac12=\dfrac48$ forgets that box II now has one more marble.

c) True: by the law of total probability, $P(E)=\dfrac23\cdot\dfrac49+\dfrac13\cdot\dfrac39=\dfrac{8}{27}+\dfrac{3}{27}=\dfrac{11}{27}$.

d) False: by Bayes' theorem, $P(D\mid E)=\dfrac{8/27}{11/27}=\dfrac{8}{11}\approx0.727<0.75$.

Revise: statistics and probability.

Let $x$ be the number of price increases ($0\le x\le50$). Then the price is $400+20x$, $50-x$ rooms are taken, and the daily profit (thousand dong) is

$$L(x)=(400+20x-40)(50-x)=(360+20x)(50-x).$$

$L'(x)=20(50-x)-(360+20x)=640-40x=0\iff x=16$. Compare $L(0)=18\,000$, $L(16)=680\cdot34=23\,120$, $L(50)=0$: the profit is greatest for $x=16$, that is at a price of $400+320=720$ thousand dong.

Revise: optimisation and real-life problems.

Let $O$ be the centre of the base and $M$ the midpoint of $CD$. Since $AC$ meets $(SCD)$ at $C$ and $AC=2OC$, we have $d(A, (SCD))=2\,d(O, (SCD))$. Draw $OH\perp SM$ at $H$. Since $CD\perp OM$ and $CD\perp SO$, we have $CD\perp(SOM)$, so $CD\perp OH$; hence $OH\perp(SCD)$. In the right-angled triangle $SOM$ with $OM=3$ and $SO=4$:

$$OH=\frac{SO\cdot OM}{\sqrt{SO^2+OM^2}}=\frac{4\cdot3}{5}=2.4.$$

So $d(A, (SCD))=2\cdot2.4=4.8$.

Revise: geometry in space.

We have $\overrightarrow{MN}=(40-5t, 10-3t, 10-t)$, so

$$f(t)=MN^2=(40-5t)^2+(10-3t)^2+(10-t)^2=35t^2-480t+1800.$$

$f'(t)=70t-480=0\iff t=\dfrac{48}{7}$, and $f'$ changes sign from $-$ to $+$ there. The minimum value is $f\left(\dfrac{48}{7}\right)=\dfrac{1080}{7}$, so the smallest distance is $\sqrt{\dfrac{1080}{7}}\approx12.42$, which rounds to $12.4$ m.

Revise: coordinates in space.

Let $K$ be the event "Lan knows the answer" and $C$ the event "Lan answers correctly". Then $P(C\mid K)=1$ and $P(C\mid\overline K)=0.25$. By the law of total probability and Bayes' theorem:

$$P(C)=0.6\cdot1+0.4\cdot0.25=0.7,\qquad P(K\mid C)=\frac{0.6}{0.7}=\frac67\approx0.857.$$

To 2 decimal places this is $0.86$.

Revise: statistics and probability.

The volume of the solid of revolution is

$$V=\pi\int_0^2\left(4-x^2\right)^2dx=\pi\int_0^2\left(16-8x^2+x^4\right)dx=\pi\left(32-\frac{64}{3}+\frac{32}{5}\right)=\frac{256\pi}{15}\approx53.62.$$

To 1 decimal place this is $53.6$ cm³.

Revise: antiderivatives and integrals.

Dividing $N_0a^5=2400$ by $N_0a^2=300$ gives $a^3=8$, so $a=2$ and $N_0=\dfrac{300}{4}=75$. Solve $N(t)=10\,000$:

$$75\cdot2^t=10\,000\iff 2^t=\frac{400}{3}\iff t=\log_2\frac{400}{3}\approx7.06.$$

So the culture contains $10\,000$ bacteria after about $7.1$ hours.

Revise: trigonometry, exponentials, logarithms and sequences.