Graduation practice paper 4

Differentiate and write over a common denominator:

$$y'=\dfrac{2x}{x^2+1}-1=\dfrac{2x-x^2-1}{x^2+1}=-\dfrac{(x-1)^2}{x^2+1}\le0\quad\text{for all } x.$$

$y'=0$ only at $x=1$ and $y'$ does not change sign there, so the function is decreasing on $\mathbb{R}$ and has no turning points. Option A assumes every root of $y'=0$ is a turning point.

Revise: Revision: derivatives and graphs of functions.

The domain is $(-\infty, -2)\cup(2, +\infty)$. For the limits at infinity, remember that $\sqrt{x^2}=|x|$:

$$\lim_{x\to+\infty}y=\lim_{x\to+\infty}\dfrac{1+\frac1x}{\sqrt{1-\frac{4}{x^2}}}=1,\qquad \lim_{x\to-\infty}y=\lim_{x\to-\infty}\dfrac{1+\frac1x}{-\sqrt{1-\frac{4}{x^2}}}=-1.$$

Also $\displaystyle\lim_{x\to2^+}y=+\infty$ and $\displaystyle\lim_{x\to-2^-}y=-\infty$ (the numerator tends to $3$ and to $-1$, not $0$). So there are $2$ horizontal asymptotes $y=\pm1$ and $2$ vertical asymptotes $x=\pm2$, $4$ in total. Option C misses $y=-1$ by writing $\sqrt{x^2}=x$.

Revise: Revision: derivatives and graphs of functions.

On $(-2, 3)$ we have $f'(x)<0$, equal to $0$ only at the single point $x=1$, so $f$ is decreasing on $(-2, 3)$. The derivative does not change sign at $x=1$, so it is not a turning point: the function has only two turning points, $x=-2$ and $x=3$. Since $f$ is decreasing on $[-2, 3]$, $f(1)>f(3)$. Options A and B treat $x=1$ as a turning point just because $f'(1)=0$.

Revise: Revision: derivatives and graphs of functions.

The antiderivatives are $F(x)=x^2-\cos x+C$. Use the condition $F(0)=3$, remembering that $\cos0=1$:

$$0-1+C=3\iff C=4.$$

So $F(x)=x^2-\cos x+4$. Option A forgets that $\cos0=1$; option B takes the antiderivative of $\sin x$ to be $\cos x$; option D moves the $-1$ across with the wrong sign, giving $C=3-1=2$.

Revise: Revision: antiderivatives and integrals.

Expand the square, then use the standard results:

$$\int_0^{\ln2}\big(e^{2x}+2e^x+1\big)dx=\Big[\dfrac12e^{2x}+2e^x+x\Big]_0^{\ln2}=\big(2+4+\ln2\big)-\Big(\dfrac12+2\Big)=\dfrac72+\ln2.$$

Option A forgets the factor $\dfrac12$ in the antiderivative of $e^{2x}$; option B takes $\dfrac{(e^x+1)^3}{3}$ as the antiderivative; option C misses the term $2e^x$.

Revise: Revision: antiderivatives and integrals.

The line through $M$ perpendicular to $(P)$ has direction vector $\vec n=(1, 2, -2)$ and consists of the points $H(3+t, 3+2t, -1-2t)$. Substitute into the equation of $(P)$:

$$(3+t)+2(3+2t)-2(-1-2t)-2=0\iff 9+9t=0\iff t=-1.$$

So $H(2, 1, 1)$. Option A takes $t=1$ by mistake; option B is the reflection of $M$ in $(P)$ (which corresponds to $t=-2$).

Revise: Revision: coordinates in space.

With $\vec u=(1, 0, -1)$ and $\vec n=(1, 1, 0)$, the angle $\varphi$ between a line and a plane is found with the sine:

$$\sin\varphi=\dfrac{|\vec u\cdot\vec n|}{|\vec u|\cdot|\vec n|}=\dfrac{1}{\sqrt2\cdot\sqrt2}=\dfrac12\ \Rightarrow\ \varphi=30^\circ.$$

Option A uses the cosine formula, as for the angle between two lines.

Revise: Revision: coordinates in space.

Let $O$ be the centre of the square. Since $BD\perp AO$ and $BD\perp SA$, $BD\perp SO$; the plane angle of the dihedral angle is $\angle SOA$:

$$AO=\dfrac{a\sqrt2}{2},\qquad \tan\angle SOA=\dfrac{SA}{AO}=\dfrac{a\sqrt6/2}{a\sqrt2/2}=\sqrt3\ \Rightarrow\ \angle SOA=60^\circ.$$

Option A uses $\dfrac{AO}{SA}$ by mistake; option C uses $\angle SBA$ (the dihedral angle along $BC$); option D uses $AC$ instead of $AO$.

Revise: Revision: geometry in space.

The right-angled isosceles triangle has hypotenuse $2a$, so $AB=AC=a\sqrt2$. The projection of $A'B$ onto the base is $AB$, so $\angle A'BA=60^\circ$:

$$AA'=AB\tan60^\circ=a\sqrt6,\qquad V=\dfrac12\big(a\sqrt2\big)^2\cdot a\sqrt6=a^3\sqrt6.$$

Option B uses the formula for the volume of a pyramid; option C forgets the factor $\dfrac12$ in the area of the triangle; option D takes $AB=a$.

Revise: Revision: geometry in space.

The cumulative frequencies are $6$, $16$, $30$, $37$, $40$; $\dfrac n2=20$ lies in the class $[60, 70)$, and the cumulative frequency before it is $16$:

$$M_e=60+\dfrac{20-16}{14}\cdot10=60+\dfrac{20}{7}\approx62.86.$$

Option A is the midpoint of the class; option B divides by the frequency $10$ of the previous class; option D is the lower boundary of the class.

Revise: Revision: statistics and probability.

By the law of total probability, split by the box chosen:

$$P(W)=\dfrac12\cdot\dfrac35+\dfrac12\cdot\dfrac26=\dfrac{3}{10}+\dfrac16=\dfrac{7}{15}.$$

Option A merges the two boxes into one ($5$ white balls out of $11$); option C multiplies $\dfrac35\cdot\dfrac13$; option D forgets to multiply by $\dfrac12$.

Revise: Revision: statistics and probability.

Reduce to a basic trigonometric inequality:

$$h(t)\ge4\iff\sin\dfrac{\pi t}{6}\ge\dfrac12\iff\dfrac{\pi}{6}+2k\pi\le\dfrac{\pi t}{6}\le\dfrac{5\pi}{6}+2k\pi\iff1+12k\le t\le5+12k.$$

For $0\le t\le24$ this gives $t\in[1, 5]\cup[13, 17]$, $8$ hours in total. Option A only considers one $12$-hour period; option D solves $\sin\dfrac{\pi t}{6}\ge-\dfrac12$ by mistake (that is, $h\ge2$).

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

a) True: $f'(x)=2xe^x+(x^2-3)e^x=(x^2+2x-3)e^x=(x-1)(x+3)e^x$.

b) False: $f'(x)$ changes from $-$ to $+$ at $x=1$, so $x=1$ is a local minimum; the local maximum is at $x=-3$.

c) True: compare $f(0)=-3$, $f(1)=-2e\approx-5.44$, $f(2)=e^2$; the least is $-2e$.

d) False: as $x\to-\infty$, $f(x)\to0$, and $f(x)>0$ for $x<-\sqrt3$; $f$ increases on $(-\infty, -3)$ up to the local maximum $f(-3)=6e^{-3}\approx0.2987<0.3$, then decreases on $(-3, 1)$ and increases on $(1, +\infty)$ from $-2e$ to $+\infty$. So the line $y=0.3$ meets the graph at only one point (on $(1, +\infty)$).

Revise: Revision: derivatives and graphs of functions.

a) True: $\displaystyle\int_0^8 2t\,dt=\Big[t^2\Big]_0^8=64$.

b) True: for $0\le t\le12$ the distance between them is $16t-t^2$, greatest when the velocities are equal ($2t=16$, $t=8$), and equal to $128-64=64$ m; for $t>12$ car B is faster, so the gap shrinks.

c) False: solving $t^2=16t$ gives $t=16$, but the formula $v_B=2t$ only holds for $t\le12$. For $t>12$ car B has travelled $144+24(t-12)$, so $144+24(t-12)=16t\iff t=18$.

d) True: $16\cdot18=288$ m and $144+24\cdot6=288$ m.

Revise: Revision: antiderivatives and integrals.

a) True: the sphere has centre $I(1, -2, 3)$ and radius $5$.

b) False: $IA=\sqrt{3^2+4^2+0^2}=5$, so $A$ lies on the sphere and is still covered.

c) True: the distance from $I$ to the floor is $3$, so the disc has radius $\sqrt{5^2-3^2}=4$ and centre $(1, -2, 0)$.

d) True: on the $x$-axis $y=0$ and $z=0$: $(x-1)^2+4+9\le25\iff(x-1)^2\le12$, so $x$ runs over an interval of length $2\sqrt{12}=4\sqrt3\approx6.93$ m.

Revise: Revision: coordinates in space.

Let $R$ be "the email is spam" and $D$ "the email is flagged".

a) False: $0.276=0.3\cdot0.92$ is only the flagged spam. By the law of total probability, $P(D)=0.3\cdot0.92+0.7\cdot0.04=0.276+0.028=0.304$.

b) True: $P(R\mid D)=\dfrac{0.276}{0.304}\approx0.908>0.9$.

c) False: $P(R\mid\overline D)=\dfrac{0.3\cdot0.08}{1-0.304}=\dfrac{0.024}{0.696}\approx0.034<0.05$.

d) True: the probability of being flagged twice is $0.3\cdot0.92^2=0.25392$ for spam and $0.7\cdot0.04^2=0.00112$ for ordinary email, so the required probability is $\dfrac{0.25392}{0.25392+0.00112}\approx0.9956>0.99$.

Revise: Revision: statistics and probability.

Let the radius of the semicircle be $r$ and the height of the rectangle $h$. From the perimeter:

$$2r+2h+\pi r=6\iff h=\dfrac{6-(2+\pi)r}{2}.$$

$$S(r)=2rh+\dfrac{\pi r^2}{2}=6r-\Big(2+\dfrac{\pi}{2}\Big)r^2,\qquad S'(r)=6-(4+\pi)r=0\iff r=\dfrac{6}{4+\pi}.$$

$S$ is a quadratic with negative $r^2$ coefficient, so it is greatest there (and then $h=r>0$): $S_{\max}=3r=\dfrac{18}{4+\pi}\approx2.52$ m$^2$.

Revise: Revision: optimisation and real-life problems.

Let $O$ be the centre of the base and $M$ the midpoint of $CD$. The line $AO$ meets $(SCD)$ at $C$ and $AC=2OC$, so $d\big(A,(SCD)\big)=2\,d\big(O,(SCD)\big)$. In the triangle $SOM$, right-angled at $O$, the distance from $O$ is the altitude:

$$OM=2,\qquad d\big(O,(SCD)\big)=\dfrac{SO\cdot OM}{\sqrt{SO^2+OM^2}}=\dfrac{3\cdot2}{\sqrt{13}}=\dfrac{6}{\sqrt{13}}.$$

So $d\big(A,(SCD)\big)=\dfrac{12}{\sqrt{13}}\approx3.33$.

Revise: Revision: geometry in space.

The square of the distance is a quadratic in $t$:

$$\overrightarrow{QP}=(3t-20, -t, 6-t),\qquad PQ^2=(3t-20)^2+t^2+(6-t)^2=11t^2-132t+436=11(t-6)^2+40.$$

$PQ^2$ is least, equal to $40$, when $t=6$, so the least distance is $\sqrt{40}=2\sqrt{10}\approx6.32$ m.

Revise: Revision: coordinates in space.

The event "the total is at least $9$" has $10$ outcomes:

$$(3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6).$$

Of these, $7$ contain a $6$ (all except $(4, 5)$, $(5, 4)$, $(5, 5)$). So the required probability is $\dfrac{7}{10}=0.7$.

Revise: Revision: statistics and probability.

Expand the square so that the standard results apply:

$$V=\pi\int_0^4\big(1+\sqrt x\big)^2dx=\pi\int_0^4\big(1+2\sqrt x+x\big)dx=\pi\Big[x+\dfrac43x^{\frac32}+\dfrac{x^2}{2}\Big]_0^4=\pi\Big(4+\dfrac{32}{3}+8\Big)=\dfrac{68\pi}{3}.$$

So $V\approx71.2\ \text{dm}^3$.

Revise: Revision: antiderivatives and integrals.

After $n$ months the balance is the sum of a geometric sequence with common ratio $q=1.005$:

$$S_n=5\cdot1.005\cdot\dfrac{1.005^n-1}{0.005}=1005\big(1.005^n-1\big).$$

$$S_n\ge100\iff1.005^n\ge1+\dfrac{100}{1005}\iff n\ge\dfrac{\ln\big(1+\frac{100}{1005}\big)}{\ln1.005}\approx19.02.$$

So at least $20$ months are needed (check: $S_{19}\approx99.9$ million, $S_{20}\approx105.4$ million).

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