Graduation practice paper 3

The function has a turning point where $f'(x)$ changes sign, that is, at the roots of odd multiplicity:

$$x=1\ (\text{multiplicity } 1),\qquad x=-2\ (\text{multiplicity } 3),\qquad x=0\ (\text{multiplicity } 2\text{, no sign change}).$$

So the function has $2$ turning points. Option C also counts the double root $x=0$; option D adds the exponents $2+1+3$.

Revise: Revision: derivatives and graphs of functions.

Factorise the numerator and denominator, then cancel (for $x\neq1$):

$$y=\dfrac{(2x-1)(x-1)}{(x-1)(x+1)}=\dfrac{2x-1}{x+1}.$$

As $x\to1$, $y\to\dfrac12$ (finite), so $x=1$ is not a vertical asymptote. The graph has the vertical asymptote $x=-1$ and the horizontal asymptote $y=2$, two lines in total. Option C forgets to check the limit at $x=1$ and only looks at the roots of the denominator.

Revise: Revision: derivatives and graphs of functions.

Rewrite the equation as $f(x)=k$ and count the intersections with the line $y=k$:

$$2f(x)+5=0\iff f(x)=-\dfrac52.$$

Since $-4<-\dfrac52<2$, the line $y=-\dfrac52$ meets the graph exactly once on each of $(-\infty, -1)$, $(-1, 3)$ and $(3, +\infty)$, so the equation has $3$ solutions. Option A makes a sign error and solves $f(x)=\dfrac52>2$.

Revise: Revision: derivatives and graphs of functions.

Use the standard results $\displaystyle\int e^{ax}\,dx=\dfrac1a e^{ax}+C$ and $\displaystyle\int\dfrac1x\,dx=\ln|x|+C$; for $x>0$, $|x|=x$:

$$\int\Big(e^{2x}-\dfrac1x\Big)dx=\dfrac12e^{2x}-\ln x+C.$$

Option A differentiates $e^{2x}$ instead of integrating it; option B differentiates $\dfrac1x$; option D has the wrong sign.

Revise: Revision: antiderivatives and integrals.

Split the interval of integration, then use linearity:

$$\int_2^5 f(x)\,dx=7-3=4,\qquad \int_2^5\big[2f(x)-1\big]dx=2\cdot4-(5-2)=5.$$

Option A forgets the term $-1$; option C takes $\displaystyle\int_2^5 1\,dx=1$; option D uses $\displaystyle\int_0^5$ instead of $\displaystyle\int_2^5$: $2\cdot7-3$.

Revise: Revision: antiderivatives and integrals.

The perpendicular bisector plane passes through the midpoint $I$ of $AB$ and has normal vector $\overrightarrow{AB}$:

$$I(2, 1, 1),\qquad \overrightarrow{AB}=(2, -2, 4)=2(1, -1, 2).$$

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

Option A passes through $A$ and option B through $B$ instead of the midpoint; option C wrongly uses the coordinates of $I$ as the normal vector.

Revise: Revision: coordinates in space.

Find the centre and radius of the sphere, then compare the distance from the centre to the plane with the radius:

$$I(2, -1, 3),\qquad R=\sqrt{2^2+1^2+3^2+2}=4.$$

$$d\big(I,(P)\big)=\dfrac{|2\cdot2-(-1)+2\cdot3+1|}{\sqrt{4+1+4}}=\dfrac{12}{3}=4=R.$$

So $(P)$ is tangent to $(S)$. Option C comes from forgetting the $+2$ in $R$ (giving $R=\sqrt{14}<4$) or forgetting to divide by $3$; option A comes from taking the centre with the wrong signs, $I(-2, 1, -3)$.

Revise: Revision: coordinates in space.

We find the projection of $C$ onto $(SAB)$. Since $BC\perp AB$ and $BC\perp SA$, $BC\perp(SAB)$, so the projection of $SC$ onto $(SAB)$ is $SB$ and the required angle is $\angle CSB$:

$$SB=\sqrt{SA^2+AB^2}=a\sqrt3,\qquad \tan\angle CSB=\dfrac{BC}{SB}=\dfrac{a}{a\sqrt3}=\dfrac{1}{\sqrt3}.$$

So the angle is $30^\circ$. Option A is the angle between $SC$ and the base ($\tan=\dfrac{SA}{AC}=1$); option B is $\angle SCB$, the other acute angle of the right-angled triangle $SBC$.

Revise: Revision: geometry in space.

The modal class is $[20, 30)$, with $n_{m-1}=12$, $n_m=15$, $n_{m+1}=6$ and class width $h=10$:

$$M_o=20+\dfrac{15-12}{(15-12)+(15-6)}\cdot10=20+\dfrac{3}{12}\cdot10=22.5.$$

Option A is the midpoint of the class; option C puts $15-6$ in the numerator by mistake; option D is the mean.

Revise: Revision: statistics and probability.

Use the midpoints $5$, $7$, $9$:

$$\bar x=\dfrac{3\cdot5+5\cdot7+2\cdot9}{10}=6.8,\qquad s^2=\dfrac{3\cdot25+5\cdot49+2\cdot81}{10}-6.8^2=48.2-46.24=1.96.$$

So $s=\sqrt{1.96}=1.4$. Option A is the variance (no square root); option B is the range $10-4$; option D is the class width.

Revise: Revision: statistics and probability.

By the definition of conditional probability, divide by the probability of the known event $B$:

$$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.12}{0.4}=0.3.$$

Option A divides by $P(A)$, which gives $P(B\mid A)$; option B is the product $P(A)\cdot P(B)$, which is only correct for independent events; option C is $P(A\cap B)$.

Revise: Revision: statistics and probability.

We need $x>1$. The base $\dfrac12$ is less than $1$, so the inequality reverses when we remove the logarithm:

$$\log_{\frac12}(x-1)\ge-2\iff x-1\le\Big(\dfrac12\Big)^{-2}=4\iff x\le5.$$

Together with the condition, $S=(1, 5]$. Option A forgets to reverse the inequality; option C forgets the condition $x>1$; option D takes $\Big(\dfrac12\Big)^{-2}=\dfrac14$.

Revise: Revision: trigonometry, exponentials, logarithms and sequences.

a) True: $f'(x)=\dfrac{(2x-3)(x-1)-(x^2-3x+6)}{(x-1)^2}=\dfrac{x^2-2x-3}{(x-1)^2}$.

b) False: $f'(x)=\dfrac{(x+1)(x-3)}{(x-1)^2}<0$ on $(-1, 1)$ and on $(1, 3)$, but the function is not defined at $x=1$, so it is not decreasing on the whole of $(-1, 3)$ (for example $f(0)=-6<f(2)=4$).

c) True: polynomial division gives $f(x)=x-2+\dfrac{4}{x-1}$, and $\dfrac{4}{x-1}\to0$ as $x\to\pm\infty$.

d) True: the turning points are $(-1, -5)$ and $(3, 3)$, at distance $\sqrt{4^2+8^2}=\sqrt{80}=4\sqrt5$.

Revise: Revision: derivatives and graphs of functions.

a) True: $v(2)=16+12-4=24$.

b) False: $v'(t)=6-2t$ changes sign at $t=3$, so the rate increases on $[0, 3]$ and decreases on $[3, 8]$ (the greatest rate is $v(3)=25$).

c) True: the amount after $t$ minutes is $W(t)=200+\displaystyle\int_0^t v(s)\,ds=200+16t+3t^2-\dfrac{t^3}{3}$, so $W(8)=200+128+192-\dfrac{512}{3}=\dfrac{1048}{3}\approx349.3$ litres.

d) False: $W(4)=200+64+48-\dfrac{64}{3}\approx290.7<300$; since $v(t)>0$, $W$ is increasing, so the tank only reaches $300$ litres after $t=4$.

Revise: Revision: antiderivatives and integrals.

a) True: $|\vec u|=\sqrt{4+1+4}=3$, which equals the speed, so each second the drone moves by exactly the vector $\vec u$.

b) True: substituting $t=2$ gives $(5, 0, 7)$.

c) False: the ground has normal vector $\vec k=(0, 0, 1)$, so $\sin\varphi=\dfrac{|\vec u\cdot\vec k|}{|\vec u|\cdot|\vec k|}=\dfrac23$ and $\varphi\approx41.8^\circ<45^\circ$.

d) True: with $P(t)=(1+2t, 2-t, 3+2t)$, $\overrightarrow{KP}=(2t-6, -t, 2t-3)$ and $KP^2=9t^2-36t+45=9(t-2)^2+9\ge9$, least when $t=2$, so the least distance is $3$ m.

Revise: Revision: coordinates in space.

Let $L$ be the event "the product is defective" and $D_1$, $D_2$ the events "made by line I, line II".

a) True: by the law of total probability, $P(L)=0.6\cdot0.03+0.4\cdot0.05=0.018+0.02=0.038$.

b) False: by Bayes' theorem, $P(D_1\mid L)=\dfrac{0.018}{0.038}=\dfrac{9}{19}\approx0.474<0.5$. Line I makes more products but has a lower defect rate.

c) False: $P(D_2\mid\overline L)=\dfrac{0.4\cdot0.95}{1-0.038}=\dfrac{0.38}{0.962}\approx0.395<0.4$.

d) True: using the complement, $1-0.962^2\approx0.0746<0.08$.

Revise: Revision: statistics and probability.

Let the base radius be $r$; the height is $h=\dfrac{1000}{\pi r^2}$. Taking the curved surface material as $1$ unit per $\text{cm}^2$, the cost is proportional to

$$C(r)=2\cdot2\pi r^2+2\pi rh=4\pi r^2+\dfrac{2000}{r},\qquad r>0.$$

$$C'(r)=8\pi r-\dfrac{2000}{r^2}=0\iff r^3=\dfrac{250}{\pi}\iff r=\sqrt[3]{\dfrac{250}{\pi}}\approx4.30.$$

$C'(r)$ changes from negative to positive there, so the cost is least when $r\approx4.3$ cm.

Revise: Revision: optimisation and real-life problems.

Construct the common perpendicular. Let $O$ be the centre of the square. Since $BD\perp AC$ and $BD\perp SA$, $BD\perp(SAC)$. In $(SAC)$ draw $OH\perp SC$ at $H$; then $OH\perp BD$, so $OH$ is the common perpendicular and $d(BD, SC)=OH$.

$$AC=2\sqrt2,\qquad SC=\sqrt{SA^2+AC^2}=2\sqrt3,\qquad OC=\sqrt2.$$

The right-angled triangles $CHO$ and $CAS$ share the angle at $C$, so they are similar:

$$OH=\dfrac{OC\cdot SA}{SC}=\dfrac{\sqrt2\cdot2}{2\sqrt3}=\dfrac{\sqrt6}{3}\approx0.816.$$

So the distance is about $0.82$.

Revise: Revision: geometry in space.

The angle $\varphi$ between the line $AB$ and the plane $(Oxy)$, whose normal vector is $\vec k=(0, 0, 1)$:

$$\overrightarrow{AB}=(40, 30, -15),\qquad AB=\sqrt{1600+900+225}=5\sqrt{109}\approx52.2.$$

$$\sin\varphi=\dfrac{|-15|}{5\sqrt{109}}\approx0.2873\ \Rightarrow\ \varphi\approx16.7^\circ.$$

Revise: Revision: coordinates in space.

Let $A$ be "both balls are red" and $B$ "at least one ball is red"; since $A\subset B$, $A\cap B=A$:

$$P(A)=\dfrac{\binom52}{\binom92}=\dfrac{10}{36},\qquad P(B)=1-\dfrac{\binom42}{\binom92}=1-\dfrac{6}{36}=\dfrac{30}{36}.$$

$$P(A\mid B)=\dfrac{10}{30}=\dfrac13\approx0.33.$$

Revise: Revision: statistics and probability.

The $x$-coordinates of the intersections satisfy $\dfrac{x^2}{2}=x+4\iff x^2-2x-8=0\iff x=-2$ or $x=4$. On $[-2, 4]$ the line lies above the parabola, so the area is

$$S=\int_{-2}^{4}\Big(x+4-\dfrac{x^2}{2}\Big)dx=\Big[\dfrac{x^2}{2}+4x-\dfrac{x^3}{6}\Big]_{-2}^{4}=\dfrac{40}{3}-\Big(-\dfrac{14}{3}\Big)=18\ (\text{m}^2).$$

The cost is $18\cdot250\,000=4\,500\,000$ đồng, that is $4.5$ million đồng.

Revise: Revision: antiderivatives and integrals.

Use $T(10)=60$ to find $k$:

$$25+70e^{-10k}=60\iff e^{-10k}=\dfrac12\iff k=\dfrac{\ln2}{10}.$$

Solve $T(t)=35$:

$$70e^{-kt}=10\iff e^{-kt}=\dfrac17\iff t=\dfrac{\ln7}{k}=\dfrac{10\ln7}{\ln2}\approx28.07.$$

So after about $28.1$ minutes.

Revise: Revision: trigonometry, exponentials, logarithms and sequences.