Second derivative, stationary points, inflexion and optimisation – IB Maths exam questions with mark scheme

1HardCalculator10 marks

A closed cylindrical tin has radius $r$ cm and height $h$ cm. Its volume is $1000\ \text{cm}^3$.

(a) Show that the total surface area $A\ \text{cm}^2$ of the tin is given by $A=2\pi r^2+\dfrac{2000}{r}$. [3]

(b) Find the value of $r$ for which $A$ is a minimum, giving your answer to 3 significant figures. [4]

(d) Find the minimum surface area, to 3 significant figures. [1]

$\pi r^2h=1000$, so $h=\dfrac{1000}{\pi r^2}$B1
$A=2\pi r^2+2\pi rh$ with $h$ substitutedM1
$A=2\pi r^2+\dfrac{2000}{r}$ (AG, must show the cancellation of $\pi r$)A1
Differentiates: $\dfrac{dA}{dr}=4\pi r-\dfrac{2000}{r^2}$ (attempt, at least one term correct)M1
Both terms correctA1
Sets $\dfrac{dA}{dr}=0$ and rearranges to $r^3=\dfrac{500}{\pi}$M1
$r=5.42$ cmA1
$\dfrac{d^2A}{dr^2}=4\pi+\dfrac{4000}{r^3}$M1
Positive for $r>0$ (so at $r=5.42$), hence minimumR1
$A_{\min}=554\ \text{cm}^2$A1
Answer: (a) shown; (b) $r=5.42$ cm; (c) $A''>0$ so minimum; (d) $554\ \text{cm}^2$
Examiner tip: In a 'show that' part every step must appear; in (c) say why the second derivative is positive, not only that it is.

2CoreNon-calculator8 marks

Let $f(x)=2x^3-3x^2-12x+5$.

(a) Find the coordinates of the two stationary points of the graph of $f$. [4]

(b) Use the second derivative to determine the nature of each stationary point. [2]

$f'(x)=6x^2-6x-12$A1
Sets $f'(x)=0$ and attempts to solve, e.g. $6(x-2)(x+1)=0$M1
$x=-1,\ x=2$A1
$(-1,\,12)$ and $(2,\,-15)$A1
$f''(x)=12x-6$ and substitutes at least one of their $x$-valuesM1
$f''(-1)=-18<0$ so $(-1,12)$ is a local maximum; $f''(2)=18>0$ so $(2,-15)$ is a local minimum (both needed)A1
$12x-6=0\Rightarrow x=\tfrac12$A1
$f''$ changes sign at $x=\tfrac12$ (e.g. $f''(0)=-6<0$, $f''(1)=6>0$), so it is a point of inflexion. Do not accept $f''=0$ aloneR1
Answer: (a) $(-1,12)$, $(2,-15)$ (b) max at $(-1,12)$, min at $(2,-15)$ (c) $x=\tfrac12$
Examiner tip: Setting f''(x)=0 only finds a candidate; the reason mark needs the sign of f'' to change either side.

3CoreNon-calculator6 marks

The diagram shows the graph of $y=f'(x)$, the derivative of a function $f$, for $-2\le x\le 4$. The graph of $f'$ crosses the $x$-axis at $x=-1$ and $x=3$ and has a minimum point at $(1,-4)$.

(a) Write down the $x$-coordinate of the local maximum point on the graph of $f$. Justify your answer. [2]

(b) Write down the $x$-coordinate of the point of inflexion on the graph of $f$. Justify your answer. [2]

(d) State whether $f''(0)$ is positive, negative or zero. [1]

$x=-1$A1
$f'$ changes from positive to negative at $x=-1$ (accept 'graph of $f'$ crosses from above to below the $x$-axis')R1
$x=1$A1
$f'$ has a minimum (turning point) at $x=1$, so $f''$ changes sign there (accept 'gradient of $f'$ changes from negative to positive')R1
$11$, $1A1
Negative (as $f'$ is decreasing at $x=0$)A1
Answer: (a) $x=-1$ (b) $x=1$ (c) $1
Examiner tip: Read the question: this is the graph of f', so the turning point of f' gives an inflexion of f, not a maximum of f.

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4StandardNon-calculator8 marks

Let $f(x)=x^4-4x^3$.

(a) Find the $x$-coordinates of the stationary points of $f$. [3]

(b) Show that the second derivative test fails at one of these stationary points, and determine the nature of this point by another method. [3]

$f'(x)=4x^3-12x^2$A1
$4x^2(x-3)=0$ (sets $f'=0$ and factorises)M1
$x=0,\ x=3$A1
$f''(x)=12x^2-24x$, $f''(0)=0$ (so the test is inconclusive)A1
Valid method: checks sign of $f'$ either side of $x=0$, e.g. $f'(-1)=-16$, $f'(1)=-8$M1
$f'$ negative on both sides so $(0,0)$ is a stationary point of inflexion (horizontal inflexion). Accept 'neither max nor min' only with inflexion statedA1
$12x(x-2)=0\Rightarrow x=2$ (with $f''$ changing sign)M1
$(2,\,-16)$A1
Answer: (a) $x=0,\ 3$ (b) $(0,0)$ is a stationary (horizontal) point of inflexion (c) $(2,-16)$
Examiner tip: f''=0 at a stationary point does NOT mean inflexion: you must test the sign of f' (or f'') either side.

5StandardNon-calculator10 marks

A ferry makes a journey of 120 km at a constant speed of $v$ km h$^{-1}$, where $v>0$. The running cost of the ferry is $\left(50+0.08v^2\right)$ dollars per hour.

(a) Show that the total cost, $C$ dollars, of the journey is $C=\dfrac{6000}{v}+9.6v$. [2]

(b) Find the speed that minimises the cost of the journey. [3]

(d) Find the minimum cost. [1]

(e) The ferry cannot travel faster than 20 km h$^{-1}$. Find the minimum cost of the journey in this case, justifying your answer. [2]

Time taken $=\dfrac{120}{v}$ hoursB1
$C=\dfrac{120}{v}\left(50+0.08v^2\right)$ leading to $\dfrac{6000}{v}+9.6v$ (AG)M1
$\dfrac{dC}{dv}=-\dfrac{6000}{v^2}+9.6$A1
Sets $\dfrac{dC}{dv}=0$, e.g. $v^2=625$M1
$v=25$ km h$^{-1}$ (reject $-25$)A1
$\dfrac{d^2C}{dv^2}=\dfrac{12000}{v^3}$A1
$>0$ at $v=25$ (or for all $v>0$), hence minimumR1
$\$480$A1
$\dfrac{dC}{dv}<0$ for $0R1
$\$492$A1
Answer: (b) $25$ km h$^{-1}$ (c) $C''=0.768>0$ (d) $\$480$ (e) $\$492$ at $v=20$
Examiner tip: In (e) the stationary point is outside the allowed domain, so the minimum is at the end point; you must say why C is decreasing up to 20.

6HardCalculator7 marks

The point $P(x,\ \ln x)$, where $x>0$, lies on the curve $y=\ln x$. The point $A$ has coordinates $(2,\,0)$.

(a) Write down an expression for $AP^2$ in terms of $x$. [1]

(b) Hence find the coordinates of the point $P$ on the curve that is closest to $A$. [3]

(d) Verify that, at this point $P$, the line $AP$ is perpendicular to the tangent to the curve. [2]

$AP^2=(x-2)^2+(\ln x)^2$A1
Attempt to differentiate $AP^2$ (or $AP$): $2(x-2)+\dfrac{2\ln x}{x}$, or uses GDC to find the minimum of the graph of $AP^2$M1
$x=1.68957\ldots$ (from $\dfrac{d}{dx}=0$ or GDC)A1
$P(1.69,\ 0.524)$A1
$0.609$A1
Gradient of tangent $=\dfrac1x=0.592$ and gradient of $AP=\dfrac{0.524}{1.69-2}=-1.69$ (both attempted)M1
Product $=-1$ (allow $-0.999$ to $-1.00$ from rounded values), so perpendicularA1
Answer: (a) $(x-2)^2+(\ln x)^2$ (b) $(1.69,\ 0.524)$ (c) $0.609$ (d) gradients $0.592$ and $-1.69$, product $-1$
Examiner tip: Minimise AP^2 rather than AP: the same x gives both minima and the algebra is much easier; remember to square-root at the end.

7StandardNon-calculator8 marks

Let $f(x)=x\,\mathrm{e}^{-x}$, for $x\in\mathbb{R}$.

(a) Show that $f'(x)=(1-x)\mathrm{e}^{-x}$. [2]

(b) Find the exact coordinates of the maximum point on the graph of $f$. [2]

(d) Hence find the exact coordinates of the point of inflexion, and explain why the graph of $f$ is concave down for $x<2$. [2]

Uses product rule: $1\cdot\mathrm{e}^{-x}+x\cdot(-\mathrm{e}^{-x})$M1
$=(1-x)\mathrm{e}^{-x}$ (AG, factorisation seen)A1
$\mathrm{e}^{-x}\ne0$ so $1-x=0$, $x=1$M1
$\left(1,\ \mathrm{e}^{-1}\right)$ (accept $\left(1,\tfrac{1}{\mathrm{e}}\right)$)A1
Product rule on $(1-x)\mathrm{e}^{-x}$: $-\mathrm{e}^{-x}-(1-x)\mathrm{e}^{-x}$M1
$=(x-2)\mathrm{e}^{-x}$ (AG)A1
$\left(2,\ 2\mathrm{e}^{-2}\right)$A1
$\mathrm{e}^{-x}>0$ and $x-2<0$ for $x<2$, so $f''(x)<0$ (concave down); sign changes at $x=2$R1
Answer: (b) $(1,\mathrm{e}^{-1})$ (d) $(2,2\mathrm{e}^{-2})$; $f''<0$ for $x<2$
Examiner tip: When the answer is given (AG), every line of the product rule must be shown; and always state that e^(-x) is never zero/always positive.

8HardNon-calculatorHL only9 marks

A curve has equation $x^2-2xy+y^3=4$.

(a) Find an expression for $\dfrac{dy}{dx}$ in terms of $x$ and $y$. [3]

(b) Show that the only point on the curve at which the tangent is horizontal is $(2,\,2)$. [3]

(c) Show that $\dfrac{d^2y}{dx^2}=-\dfrac14$ at the point $(2,\,2)$, and hence determine the nature of this stationary point. [3]

Attempt at implicit differentiation, including the product rule on $2xy$ and $3y^2\dfrac{dy}{dx}$ for $y^3$M1
$2x-2y-2x\dfrac{dy}{dx}+3y^2\dfrac{dy}{dx}=0$A1
$\dfrac{dy}{dx}=\dfrac{2y-2x}{3y^2-2x}$ (accept equivalent, e.g. $\dfrac{2(x-y)}{2x-3y^2}$)A1
Sets numerator $=0$: $y=x$, and substitutes into the curveM1
$x^3-x^2-4=0$, i.e. $(x-2)(x^2+x+2)=0$A1
$x^2+x+2=0$ has no real roots (discriminant $-7<0$), so $x=2$ only, giving $(2,2)$; denominator $3(4)-4=8\ne0$ there (AG)R1
Differentiates $\dfrac{dy}{dx}$ again (quotient rule, or implicit differentiation of the line in (a)), with $\dfrac{dy}{dx}$ terms presentM1
Substitutes $x=y=2$, $\dfrac{dy}{dx}=0$: e.g. $\dfrac{(-2)(8)-0}{8^2}=-\dfrac14$ (AG)A1
$\dfrac{d^2y}{dx^2}<0$ so $(2,2)$ is a local maximum (of $y$)R1
Answer: (a) $\dfrac{2y-2x}{3y^2-2x}$ (b) shown (c) $\dfrac{d^2y}{dx^2}=-\dfrac14<0$, so $(2,2)$ is a local maximum
Examiner tip: When finding d²y/dx² at a stationary point, substitute dy/dx = 0 before simplifying; and in (b) you must show the quadratic factor has no real roots to claim (2,2) is the only point.

9StretchNon-calculator10 marks

Let $f(x)=x^3-3kx^2+4$, where $k$ is a positive constant.

(a) Show that the graph of $f$ has stationary points at $x=0$ and $x=2k$. [3]

(b) Determine the nature of each stationary point. [3]

(d) Find the set of values of $k$ for which the equation $f(x)=0$ has three distinct real solutions. [2]

$f'(x)=3x^2-6kx$A1
Sets $f'(x)=0$: $3x(x-2k)=0$M1
$x=0$ or $x=2k$ (AG)A1
$f''(x)=6x-6k$A1
$f''(0)=-6k<0$ (since $k>0$) so local maximum at $x=0$R1
$f''(2k)=6k>0$ so local minimum at $x=2k$R1
$f(2k)=8k^3-12k^3+4=4-4k^3=0$M1
$k=1$A1
Recognises local max $f(0)=4>0$ and requires local min $4-4k^3<0$ (accept sketch reasoning)M1
$k>1$A1
Answer: (b) max at $(0,4)$, min at $(2k,4-4k^3)$ (c) $k=1$ (d) $k>1$
Examiner tip: The reason marks need the sign of k used explicitly: -6k is only negative because k>0 is given.

10HardNon-calculator8 marks

A water trough is made by bending a sheet of metal 30 cm wide. The cross-section is a trapezium with a horizontal base of 10 cm and two sloping sides of 10 cm, each making an angle $\theta$ with the horizontal, where $0<\theta<\dfrac{\pi}{2}$, as shown in the diagram.

(a) Show that the area of the cross-section is $A=100\sin\theta\,(1+\cos\theta)$. [3]

(b) Find the value of $\theta$ that gives the maximum area. [4]

Height $=10\sin\theta$B1
Top width $=10+20\cos\theta$ (or each overhang $10\cos\theta$)B1
$A=\tfrac12(10+10+20\cos\theta)(10\sin\theta)$ leading to $100\sin\theta(1+\cos\theta)$ (AG)M1
$\dfrac{dA}{d\theta}=100\cos\theta+100\cos^2\theta-100\sin^2\theta$ (accept $100\cos\theta+100\cos2\theta$)A1
Sets $=0$ and uses $\sin^2\theta=1-\cos^2\theta$ to obtain a quadratic in $\cos\theta$M1
$(2\cos\theta-1)(\cos\theta+1)=0$, $\cos\theta=\tfrac12$ (reject $\cos\theta=-1$)A1
$\theta=\dfrac{\pi}{3}$ (accept $60^\circ$)A1
$A=75\sqrt3\ \text{cm}^2$A1
Answer: (b) $\theta=\dfrac{\pi}{3}$ (c) $75\sqrt3\ \text{cm}^2$
Examiner tip: Differentiate sinθcosθ with the product rule (or rewrite as ½sin2θ first); many lose the A1 by writing its derivative as -sin²θ.

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