Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 10: Differentiation worked solutions

Worked solutions to the practice questions in Chapter 10 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[5 marks]
A hot-air balloon is at a height of H(t) metres above the ground, t minutes after it leaves the ground. It is known that
H'(5) = 18, \qquad H'(12) = 0, \qquad H'(16) = -9.
(a) Interpret H'(5) = 18 in the context.
[2]
(b) What does H'(12) = 0 tell you about the height of the balloon at t = 12?
[1]
(c) Interpret H'(16) = -9 in the context.
[1]
(d) Of the three times t = 5, t = 12 and t = 16, state at which the height is changing most rapidly.
[1]
Show solutionHide solution
(a) 5 minutes after leaving the ground, the balloon is rising (height increasing)
A1
at 18 metres per minute
A1
(b) at that instant the height is not changing (the balloon is neither rising nor falling)
A1
(c) at t = 16 the balloon is descending: the height is decreasing at 9 metres per minute
A1
(d) t = 5 (largest size of rate, 18 > 9)
A1

Final answer

(a) Rising at 18 m per minute after 5 minutes. (b) Height is momentarily unchanged. (c) Falling at 9 m per minute. (d) t=5.

Question 2

[6 marks]
For f(x) = x^n, where n \in \mathbb{Z}, the derivative is f'(x) = nx^{n-1}.
Write down f'(x) for each function.
(a) f(x) = x^7 (b) f(x) = x^{10} (c) f(x) = x (d) f(x) = x^2 (e) f(x) = 8
[5]
(f) Explain, using the graph of y = 8, why your answer to (e) is correct.
[1]
Show solutionHide solution
(a) f'(x) = 7x^6
A1
(b) f'(x) = 10x^{9}
A1
(c) f'(x) = 1x^0 = 1
A1
(d) f'(x) = 2x
A1
(e) f(x) = 8 = 8x^0, so f'(x) = 0
A1
(f) the graph of y = 8 is a horizontal line, so its gradient is 0 at every point
R1
Note: in (c) do not accept x^0 left unsimplified.

Final answer

(a)–(e) 7x^6, 10x^9, 1, 2x, 0. (f) A horizontal line has gradient 0.

Question 3

[6 marks]
The power rule f(x) = x^n \Rightarrow f'(x) = nx^{n-1} holds for every rational power n. To use it, write roots as powers first, for example \sqrt{x} = x^{\frac{1}{2}} and \dfrac{1}{x^3} = x^{-3}.
(a) Write each of the following in the form x^n. (i) \sqrt[3]{x} (ii) \dfrac{1}{\sqrt{x}} (iii) x^2\sqrt{x} (iv) \dfrac{1}{\sqrt[3]{x^4}}
[2]
(b) Hence differentiate each of the four functions in part (a), for x > 0.
[4]
Show solutionHide solution
(a) (i) x^{\frac{1}{3}} (ii) x^{-\frac{1}{2}} (iii) x^{\frac{5}{2}} (iv) x^{-\frac{4}{3}}
A2
Note: award A1 for two or three correct.
(b) (i) \dfrac{1}{3}x^{-\frac{2}{3}}
A1
(ii) -\dfrac{1}{2}x^{-\frac{3}{2}}
A1
(iii) \dfrac{5}{2}x^{\frac{3}{2}}
A1
(iv) -\dfrac{4}{3}x^{-\frac{7}{3}}
A1
Note: accept equivalent forms, e.g. \dfrac{1}{3\sqrt[3]{x^2}} for (i). FT from part (a).

Final answer

(a) x^{1/3}, x^{-1/2}, x^{5/2}, x^{-4/3}. (b) \frac13x^{-2/3}, -\frac12x^{-3/2}, \frac52x^{3/2}, -\frac43x^{-7/3}.

Question 4

[5 marks]
Let f(x) = \dfrac{(x^2 + 2)(x^2 - 3)}{x^2}, x \ne 0.
(a) Show that f(x) = x^2 - 1 - 6x^{-2}.
[2]
(b) Hence show that f'(x) = \dfrac{2(x^4 + 6)}{x^3}.
[3]
Show solutionHide solution
(a) (x^2 + 2)(x^2 - 3) = x^4 - 3x^2 + 2x^2 - 6 = x^4 - x^2 - 6 M1A1
f(x) = \dfrac{x^4}{x^2} - \dfrac{x^2}{x^2} - \dfrac{6}{x^2} = x^2 - 1 - 6x^{-2}
AG
(b) f'(x) = 2x + 12x^{-3} A1A1
= \dfrac{2x^4}{x^3} + \dfrac{12}{x^3} = \dfrac{2x^4 + 12}{x^3}
M1
= \dfrac{2(x^4 + 6)}{x^3}
AG
Note: in (b) the M1 is for writing both terms over the common denominator x^3. Working backwards from the given answer earns no marks.

Final answer

(a) x^2-1-6x^{-2}. (b) 2(x^4+6)/x^3.

Question 5

[5 marks]
A curve contains P(2,\ 5), and f'(2) = 3.
(a) State the tangent gradient at P.
[1]
(b) Find the tangent at P in the form y=mx+c.
[2]
(c) Find the normal at P in the form ax + by + d = 0, where a,\ b,\ d \in \mathbb{Z}.
[2]
Show solutionHide solution
(a) 3
A1
(b) y - 5 = 3(x - 2)
M1
y = 3x - 1
A1
(c) gradient of normal -\dfrac{1}{3}, so y - 5 = -\dfrac{1}{3}(x - 2)
M1
3y - 15 = -x + 2 \Rightarrow x + 3y - 17 = 0
A1
Note: accept any integer multiple, e.g. -x - 3y + 17 = 0.

Final answer

(a) 3. (b) y=3x-1. (c) x+3y-17=0.

Question 6

[6 marks]
(a) Let f(x) = 2x^2 - 6x + 1. Find f'(x) and hence find the interval on which f is increasing and the interval on which f is decreasing.
[3]
(b) Let y = 7 + 5x - x^2. Find \dfrac{dy}{dx} and hence find the interval on which y is increasing and the interval on which y is decreasing.
[3]
Show solutionHide solution
(a) f'(x) = 4x - 6
A1
4x - 6 > 0 \Leftrightarrow x > \dfrac{3}{2}
M1
increasing for x > \dfrac{3}{2}, decreasing for x < \dfrac{3}{2}
A1
(b) \dfrac{dy}{dx} = 5 - 2x
A1
5 - 2x > 0 \Leftrightarrow x < \dfrac{5}{2}
M1
increasing for x < \dfrac{5}{2}, decreasing for x > \dfrac{5}{2}
A1
Note: accept 1.5 and 2.5.

Final answer

(a) 4x-6; increasing x>3/2, decreasing x<3/2. (b) 5-2x; increasing x<5/2, decreasing x>5/2.

Question 7

[8 marks]
(a) For a function f, f'(4) = 0 and f''(4) = -3. State the nature of the stationary point at x = 4, giving a reason.
[2]
(b) For a function g, g'(-1) = 0 and g''(-1) = 5. State the nature of the stationary point at x = -1.
[1]
(c) For both curves y = x^3 and y = x^4, the first and second derivatives are 0 at x = 0.
(i) By considering the sign of \dfrac{dy}{dx} on each side of x = 0, state the nature of the stationary point at the origin for each curve.
[2]
(ii) Hence state what can be concluded about the nature of a stationary point at x = a when f''(a) = 0.
[1]
(d) For a function p, p''(x) < 0 for x < 2, p''(2) = 0, p''(x) > 0 for x > 2, and p'(2) = 3. Describe the point on the graph of p where x = 2.
[2]
Show solutionHide solution
(a) local maximum
A1
because f'(4) = 0 and f''(4) = -3 < 0
R1
(b) local minimum (since g''(-1) = 5 > 0)
A1
(c) (i) y = x^3: \dfrac{dy}{dx} = 3x^2 > 0 on both sides of 0, so the origin is a point of inflexion with zero gradient
A1
y = x^4: \dfrac{dy}{dx} = 4x^3 changes from negative to positive at 0, so the origin is a local minimum point
A1
(ii) no conclusion can be drawn (it could be a maximum, a minimum or a point of inflexion); the sign of f'(x) on each side of x = a must be checked
R1
(d) a point of inflexion with non-zero gradient (gradient 3)
A1
because p''(2) = 0 and p'' changes sign there: the graph changes from concave-down to concave-up
R1

Final answer

(a) Maximum, f\prime\prime(4)<0. (b) Minimum. (c) x^3: stationary inflexion; x^4: minimum. Second derivative zero gives no conclusion. (d) Inflexion with gradient 3, concave down then up.

Question 8

[9 marks]
A function f is defined for all x \in \mathbb{R}, and its derivative exists everywhere. It is given that
f(2) = 5, \quad f'(2) = 4, \quad f(4) = 3, \quad f'(4) = -2.
(a) Let g(x) = f\left(x^2\right). Find g'(2).
[2]
(b) Let h(x) = x\,f(2x). Find h'(1).
[3]
(c) Let k(x) = e^{f(x)}. Find the exact value of k'(4).
[2]
(d) Find the tangent's equation at the point (2,g(2)).
[2]
Show solutionHide solution
(a) g'(x) = 2x\,f'\left(x^2\right) (M1)
g'(2) = 4f'(4) = -8
A1
(b) h'(x) = f(2x) + x \times 2f'(2x) M1A1
h'(1) = f(2) + 2f'(2) = 5 + 8 = 13
A1
(c) k'(x) = f'(x)e^{f(x)} (M1)
k'(4) = -2e^3
A1
(d) g(2) = f(4) = 3
A1
y - 3 = -8(x - 2), i.e. y = -8x + 19
A1

Final answer

(a) -8. (b) 13. (c) -2e^3. (d) y=-8x+19.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[5 marks]
The depth of water in a rainwater tank is h metres, t minutes after a pump is switched on. It is given that h'(10)=0.35 and h'(30)=-0.1.
(a) Write down the units of h'(t).
[1]
(b) Interpret the statement h'(10)=0.35.
[1]
(c) Interpret the statement h'(30)=-0.1.
[1]
(d) The depth at t=10 is 2.4 m. Use the value of h'(10) to estimate the depth at t=12.
[2]
Show solutionHide solution
(a) Metres per minute (m min^{-1}).
[1]
(b) Ten minutes after the pump is switched on, the depth is increasing at an instantaneous rate of 0.35 m per minute.
[1]
(c) Thirty minutes after the pump is switched on, the depth is decreasing at 0.1 m per minute (the negative sign means it is falling).
[1]
(d) Over a short time the curve is close to its tangent, so the depth rises by about 0.35\times 2 metres.
[1]
Estimated depth =2.4+0.7=3.1 m.
[1]

Final answer

(a) m per minute. (b) Rising at 0.35 m per minute after 10 minutes. (c) Falling at 0.1 m per minute after 30 minutes. (d) 3.1 m.

Question 10

[5 marks]
Let y=\dfrac{3x^{2}-4}{\sqrt{x}}, for x>0.
(a) Write y in the form ax^{p}+bx^{q}, where a,b,p,q are constants.
[1]
(b) Hence find \dfrac{\mathrm{d}y}{\mathrm{d}x}.
[2]
(c) Calculate the gradient at x=4.
[2]
Show solutionHide solution
(a) y=\dfrac{3x^{2}}{x^{1/2}}-\dfrac{4}{x^{1/2}}=3x^{3/2}-4x^{-1/2}.
[1]
(b) \dfrac{\mathrm{d}y}{\mathrm{d}x}=3\cdot\dfrac32 x^{1/2}-4\cdot\left(-\dfrac12\right)x^{-3/2}
[1]
=\dfrac92 x^{1/2}+2x^{-3/2}.
[1]
(c) At x=4: \sqrt{4}=2 and 4^{-3/2}=\dfrac18.
[1]
Gradient =\dfrac92(2)+2\cdot\dfrac18=9+\dfrac14=\dfrac{37}{4}.
[1]

Final answer

(a) 3x^{3/2}-4x^{-1/2}. (b) \frac92\sqrt x+2x^{-3/2}. (c) 37/4.

Question 11

[6 marks]
Let f(x)=e^{\sin 2x}, for 0\le x\le\pi.
(a) Find f'(x).
[2]
(b) Find the coordinates of each contact point of a horizontal tangent.
[2]
(c) Find the equation of the tangent at x=0.
[2]
Show solutionHide solution
(a) Outer function \mathrm{e}^{u}, inner u=\sin 2x with \dfrac{\mathrm{d}u}{\mathrm{d}x}=2\cos 2x.
[1]
f'(x)=2\cos 2x\;\mathrm{e}^{\sin 2x}.
[1]
(b) f'(x)=0 and \mathrm{e}^{\sin 2x}>0 for all x, so \cos 2x=0.
[1]
2x=\dfrac{\pi}{2},\dfrac{3\pi}{2}, so x=\dfrac{\pi}{4} or x=\dfrac{3\pi}{4}.
[1]
(c) f(0)=\mathrm{e}^{0}=1 and f'(0)=2\cos 0\cdot\mathrm{e}^{0}=2.
[1]
Tangent: y-1=2(x-0), so y=2x+1.
[1]

Final answer

(a) 2e^{\sin2x}\cos2x. (b) (\pi/4,e) and (3\pi/4,e^{-1}); the tangents are y=e and y=e^{-1}. (c) y=2x+1.

Question 12

[5 marks]
Let f(x)=\dfrac{x}{x^{2}+4}.
(a) Show that f'(x)=\dfrac{4-x^{2}}{(x^{2}+4)^{2}}.
[3]
(b) Hence find the coordinates of the stationary points of f.
[2]
Show solutionHide solution
(a) u=x, v=x^{2}+4, so u'=1, v'=2x.
[1]
f'(x)=\dfrac{(x^{2}+4)(1)-x(2x)}{(x^{2}+4)^{2}}.
[1]
Numerator: x^{2}+4-2x^{2}=4-x^{2}.
[1]
f'(x)=\dfrac{4-x^{2}}{(x^{2}+4)^{2}}.
AG
(b) f'(x)=0\Rightarrow 4-x^{2}=0\Rightarrow x=\pm2.
[1]
f(2)=\dfrac{2}{8}=\dfrac14 and f(-2)=-\dfrac14, so the points are \left(2,\tfrac14\right) and \left(-2,-\tfrac14\right).
[1]

Final answer

(a) As shown. (b) (-2,-1/4), (2,1/4).

Question 13

[6 marks]
The tangent to the graph of y=e^{2x} at the point P makes an angle of 45^{\circ} with the positive x-axis.
(a) Find the exact coordinates of P.
[4]
(b) Find the equation of the normal at P.
[2]
Show solutionHide solution
(a) The gradient of the tangent is \tan45^{\circ}=1.
[1]
\dfrac{\mathrm{d}y}{\mathrm{d}x}=2\mathrm{e}^{2x}=1\Rightarrow\mathrm{e}^{2x}=\dfrac12.
[1]
2x=\ln\dfrac12=-\ln2, so x=-\dfrac12\ln2.
[1]
y=\mathrm{e}^{2x}=\dfrac12, so \mathrm{P}\left(-\dfrac12\ln2,\ \dfrac12\right).
[1]
(b) Normal gradient =-\dfrac11=-1.
[1]
y-\dfrac12=-\left(x+\dfrac12\ln2\right), so y=-x+\dfrac12-\dfrac12\ln2.
[1]

Final answer

(a) (-\frac12\ln2,\frac12). (b) y=-x+\frac12-\frac12\ln2.

Question 14

[6 marks]
Let f(x)=\dfrac{x^{3}}{3}-x^{2}+kx, with constant k.
(a) Find f'(x).
[1]
(b) Find the values of k for which the derivative is positive at every real x.
[3]
(c) Given that k=-3, state the decreasing interval of f.
[2]
Show solutionHide solution
(a) f'(x)=x^{2}-2x+k.
[1]
(b) x^{2}-2x+k>0 for all x means the quadratic has no real roots (it opens upwards), so its discriminant is negative.
[1]
\Delta=(-2)^{2}-4(1)(k)=4-4k.
[1]
4-4k<0\Rightarrow k>1.
[1]
(c) With k=-3: f'(x)=x^{2}-2x-3=(x-3)(x+1), which is negative for -1<x<3.
[1]
So f is decreasing for -1<x<3.
[1]

Final answer

(a) x^2-2x+k. (b) k>1. (c) -1<x<3.

Question 15

[8 marks]
Let f(x)=(x-3)e^{x}, for x\in\mathbb{R}.
(a) Find f'(x) and f''(x), simplifying each as far as possible.
[3]
(b) State the intervals of each concavity for f.
[2]
(c) Find the exact coordinates of the point of inflexion and justify it.
[2]
(d) Find the exact gradient of the graph at the point of inflexion.
[1]
Show solutionHide solution
(a) Product rule: f'(x)=\mathrm{e}^{x}+(x-3)\mathrm{e}^{x}.
[1]
f'(x)=(x-2)\mathrm{e}^{x}.
[1]
f''(x)=\mathrm{e}^{x}+(x-2)\mathrm{e}^{x}=(x-1)\mathrm{e}^{x}.
[1]
(b) \mathrm{e}^{x}>0, so f''(x)=0 only at x=1.
[1]
f''<0 for x<1 (concave down) and f''>0 for x>1 (concave up).
[1]
(c) f'' changes sign at x=1 (from negative to positive), so it is a point of inflexion.
[1]
f(1)=(1-3)\mathrm{e}=-2\mathrm{e}: the point is (1,-2\mathrm{e}).
[1]
(d) f'(1)=(1-2)\mathrm{e}=-\mathrm{e}.
[1]

Final answer

(a) (x-2)e^x, (x-1)e^x. (b) Down x<1, up x>1. (c) (1,-2e), f\prime\prime changes sign. (d) -e.

Question 16

[6 marks]
Let f(x)=x^{2}e^{-0.3x}, for x\ge0.
(a) Show that f'(x)=x(2-0.3x)e^{-0.3x}.
[2]
(b) Use your GDC to find the greatest gradient of the graph of f, and the value of x at which it occurs.
[3]
(c) Explain why (b) locates an inflexion of f.
[1]
Show solutionHide solution
(a) Product rule: f'(x)=2x\mathrm{e}^{-0.3x}+x^{2}(-0.3)\mathrm{e}^{-0.3x}.
[1]
f'(x)=x(2-0.3x)\mathrm{e}^{-0.3x}.
[1]
(b) The gradient is f'(x). Graph y=x(2-0.3x)\mathrm{e}^{-0.3x} and use the Maximum function.
[1]
x=1.95 (unrounded 1.9526).
[1]
Greatest gradient f'(1.9526)=1.54.
[1]
(c) f' has a maximum there, so f''=0 and f'' changes sign (from positive to negative): a point of inflexion.
[1]

Final answer

(a) As shown. (b) Greatest gradient 1.54 at x=1.95. (c) f\prime has a maximum, so f\prime\prime changes from positive to negative.

Question 17

[6 marks]
Two positive numbers x and y satisfy x+y=18. Let P=x^{2}y.
(a) Show that P=18x^{2}-x^{3}.
[1]
(b) Find \dfrac{\mathrm{d}P}{\mathrm{d}x} and hence find the value of x for which P is stationary, for 0<x<18.
[3]
(c) Justify the maximum of P at this x.
[1]
(d) Find the maximum value of P.
[1]
Show solutionHide solution
(a) y=18-x, so P=x^{2}(18-x)=18x^{2}-x^{3}.
[1]
(b) \dfrac{\mathrm{d}P}{\mathrm{d}x}=36x-3x^{2}.
[1]
36x-3x^{2}=0\Rightarrow3x(12-x)=0.
[1]
x=12 (the root x=0 is not in the domain).
[1]
(c) \dfrac{\mathrm{d}^{2}P}{\mathrm{d}x^{2}}=36-6x, which is -36<0 at x=12, so it is a maximum.
[1]
(d) P=18(144)-1728=864.
[1]

Final answer

(a) As shown. (b) 36x-3x^2; x=12. (c) P\prime\prime(12)=-36<0. (d) 864.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 18

[5 marks]
Let f(x)=x^{3}+kx^{2}+27x, where k>0. The graph of f has exactly one stationary point.
(a) Find the value of k.
[3]
(b) Show that the stationary point is a point of inflexion.
[2]
Show solutionHide solution
(a) f'(x)=3x^{2}+2kx+27.
[1]
Exactly one stationary point means f'(x)=0 has one repeated root, so the discriminant is zero: (2k)^{2}-4(3)(27)=0.
[1]
4k^{2}=324, so k=9 (as k>0).
[1]
(b) With k=9: f'(x)=3x^{2}+18x+27=3(x+3)^{2}\ge0.
[1]
f' does not change sign at x=-3, so it is not a maximum or minimum: a (stationary) point of inflexion.
[1]

Final answer

(a) k=9. (b) f\prime\prime=6x+18 changes sign at x=-3.

Question 19

[6 marks]
Let y=\dfrac{x}{\sqrt{x^{2}+9}}.
(a) Show that \dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{9}{(x^{2}+9)^{3/2}}.
[5]
(b) Hence explain why y is an increasing function of x for all real x.
[1]
Show solutionHide solution
(a) Quotient rule with u=x, v=(x^{2}+9)^{1/2}: u'=1.
[1]
Chain rule: v'=\dfrac12(x^{2}+9)^{-1/2}(2x)=\dfrac{x}{\sqrt{x^{2}+9}}.
[1]
\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\sqrt{x^{2}+9}-\dfrac{x^{2}}{\sqrt{x^{2}+9}}}{x^{2}+9}.
[1]
Combine the numerator over \sqrt{x^{2}+9}: \dfrac{(x^{2}+9)-x^{2}}{\sqrt{x^{2}+9}}=\dfrac{9}{\sqrt{x^{2}+9}}.
[2]
\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{9}{\sqrt{x^{2}+9}\,(x^{2}+9)}=\dfrac{9}{(x^{2}+9)^{3/2}}.
AG
(b) (x^{2}+9)^{3/2}>0 for all x, so \dfrac{\mathrm{d}y}{\mathrm{d}x}>0 for all x.
[1]

Final answer

(a) 9/(x^2+9)^{3/2}. (b) The derivative is positive for every x.

Question 20

[6 marks]
Let y=\sqrt{x}\,\ln x, for x>0.
(a) Find \dfrac{\mathrm{d}y}{\mathrm{d}x}.
[2]
(b) Find the exact coordinates of the stationary point of the curve.
[2]
(c) Find the equation of the tangent at the intercept on the x-axis.
[2]
Show solutionHide solution
(a) Product rule: \dfrac{\mathrm{d}y}{\mathrm{d}x}=x^{1/2}\cdot\dfrac1x+\dfrac12x^{-1/2}\ln x.
[1]
\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2+\ln x}{2\sqrt{x}}.
[1]
(b) \sqrt{x}>0, so 2+\ln x=0, giving \ln x=-2 and x=\mathrm{e}^{-2}.
[1]
y=\mathrm{e}^{-1}\cdot(-2)=-\dfrac{2}{\mathrm{e}}; the point is \left(\mathrm{e}^{-2},\,-\dfrac{2}{\mathrm{e}}\right).
[1]
(c) y=0 with x>0 gives \ln x=0, so the curve meets the x-axis at (1,0), where the gradient is \dfrac{2+0}{2}=1.
[1]
Tangent: y-0=1(x-1), so y=x-1.
[1]

Final answer

(a) (2+\ln x)/(2\sqrt x). (b) (e^{-2},-2/e). (c) y=x-1.

Question 21

[5 marks]
Let f(x)=x^{3}+px^{2}+qx, with constants p,q. The function f is decreasing exactly when -1<x<3.
Determine both constants.
Show solutionHide solution
f'(x)=3x^{2}+2px+q.
[1]
f is decreasing exactly when f'(x)<0, so the quadratic f'(x) (opening upwards) is negative exactly between its roots: its roots are x=-1 and x=3.
[1]
f'(x)=3(x+1)(x-3)=3x^{2}-6x-9.
[1]
Comparing coefficients: 2p=-6, so p=-3.
[1]
q=-9.
[1]

Final answer

p=-3, q=-9.

Question 22

[9 marks]
A straight line passes through the point P(3,4) and meets the positive x-axis at A(a,0) and the positive y-axis at B(0,b), where a>3. Let S be the area of triangle OAB, where O is the origin.
(a) Show that b=\dfrac{4a}{a-3}.
[2]
(b) Hence show that S=\dfrac{2a^{2}}{a-3}.
[1]
(c) Find the value of a for which S is a minimum, justifying that it is a minimum.
[4]
(d) Find the minimum area, and show that P is the midpoint of AB for this line.
[2]
Show solutionHide solution
(a) The line has equation \dfrac{x}{a}+\dfrac{y}{b}=1, and \mathrm{P} lies on it: \dfrac3a+\dfrac4b=1.
[1]
\dfrac4b=\dfrac{a-3}{a}, so b=\dfrac{4a}{a-3}.
[1]
(b) Area of the right-angled triangle: S=\dfrac12ab.
[1]
S=\dfrac12\cdot a\cdot\dfrac{4a}{a-3}=\dfrac{2a^{2}}{a-3}.
AG
(c) \dfrac{\mathrm{d}S}{\mathrm{d}a}=\dfrac{4a(a-3)-2a^{2}}{(a-3)^{2}}.
[2]
\dfrac{\mathrm{d}S}{\mathrm{d}a}=\dfrac{2a(a-6)}{(a-3)^{2}}=0\Rightarrow a=6 (as a>3).
[1]
For 3<a<6 the derivative is negative and for a>6 it is positive, so S has a minimum at a=6.
[1]
(d) S(6)=\dfrac{2(36)}{3}=24.
[1]
Then b=\dfrac{4(6)}{3}=8, so \mathrm{A}(6,0) and \mathrm{B}(0,8) have midpoint (3,4)=\mathrm{P}.
[1]

Final answer

(a)–(b) As shown. (c) a=6, derivative changes from negative to positive. (d) S=24, b=8, midpoint (3,4).

Question 23

[8 marks]
Let f(x)=x^{3}-3kx, with constant k.
(a) Find f'(x), and explain why the graph of f has two stationary points only if k>0.
[2]
(b) For k>0, find the coordinates of the two stationary points in terms of k, and state which is the local maximum.
[3]
(c) The vertical distance between the local maximum point and the local minimum point is 32. Find the value of k.
[3]
Show solutionHide solution
(a) f'(x)=3x^{2}-3k.
[1]
f'(x)=0\Rightarrow x^{2}=k, which has two distinct real solutions only if k>0 (for k<0 there are none, and k=0 gives only x=0).
[1]
(b) x=\sqrt{k} or x=-\sqrt{k}.
[1]
f(\sqrt k)=k\sqrt k-3k\sqrt k=-2k\sqrt k and f(-\sqrt k)=2k\sqrt k: the points are \left(\sqrt k,-2k\sqrt k\right) and \left(-\sqrt k,2k\sqrt k\right).
[1]
f''(x)=6x is negative at x=-\sqrt k, so the local maximum is \left(-\sqrt k,\,2k\sqrt k\right).
[1]
(c) Vertical distance =2k\sqrt k-(-2k\sqrt k)=4k^{3/2}.
[1]
4k^{3/2}=32\Rightarrow k^{3/2}=8.
[1]
k=8^{2/3}=4.
[1]

Final answer

(a) 3x^2-3k; two roots need k>0. (b) (-\sqrt k,2k\sqrt k) maximum, (\sqrt k,-2k\sqrt k) minimum. (c) k=4.

Question 24

[13 marks]
Let f(x)=(x^{2}-3x+1)e^{-0.4x}, for x\ge0.
(a) Show that f'(x)=e^{-0.4x}\left(-0.4x^{2}+3.2x-3.4\right).
[3]
(b) Use your GDC to find the x-coordinates of the stationary points of the graph of f.
[2]
(c) Find the coordinates of both local extrema and justify each classification.
[3]
(d) Find the equation of the tangent at x=0.
[2]
(e) Given that f(x)\to0 as x\to\infty, find the values of k for which exactly three real roots satisfy f(x)=k. Justify your answer.
[3]
Show solutionHide solution
(a) Product rule: u=x^{2}-3x+1, v=\mathrm{e}^{-0.4x}, so u'=2x-3 and v'=-0.4\mathrm{e}^{-0.4x}.
[1]
f'(x)=(2x-3)\mathrm{e}^{-0.4x}-0.4(x^{2}-3x+1)\mathrm{e}^{-0.4x}.
[1]
f'(x)=\mathrm{e}^{-0.4x}\left(2x-3-0.4x^{2}+1.2x-0.4\right)=\mathrm{e}^{-0.4x}\left(-0.4x^{2}+3.2x-3.4\right).
[1]
(b) Solve -0.4x^{2}+3.2x-3.4=0 (as \mathrm{e}^{-0.4x}>0) on the GDC, or graph f' and find its zeros.
[1]
x=1.26 and x=6.74 (exactly 4\mp\sqrt{7.5}).
[1]
(c) f(1.2614)=-0.720 and f(6.7386)=1.77.
[2]
f'(0)=-3.4<0 and f' changes from - to + at x=1.26: local minimum (1.26,-0.720); then from + to - at x=6.74: local maximum (6.74,1.77).
[1]
(d) f(0)=1 and f'(0)=-3.4.
[1]
y=-3.4x+1.
[1]
(e) The graph starts at (0,1), falls to the minimum value -0.720, rises to the maximum value 1.77, then falls towards 0 without reaching it for large x. It crosses a horizontal line y=k three times exactly when the line cuts the initial falling part, the rising part and the final falling part.
[1]
The initial falling part covers heights from 1 down to -0.720, and the final falling part covers heights below 1.77 and above 0. Both are crossed only when 0<k\le1 (at k=1 one solution is x=0).
[1]
So 0<k\le1 (for 1<k<1.77 there are only two solutions and for k\le0 at most two).
[1]

Final answer

(a) As shown. (b) 1.26,6.74. (c) (1.26,-0.720) minimum; (6.74,1.77) maximum. (d) y=-3.4x+1. (e) 0<k\le1.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Why a dish focuses

G1 Worked example

[6 marks]
The lower arc of a circular mirror is y=5-\sqrt{25-x^2}, -5<x<5. A downward vertical ray hitting a tangent of gradient m\ne0 is reflected along a line of gradient (m^2-1)/(2m). (a) Find dy/dx. [2] (b) Rays hit at P(3,1) and Q(4,2). Find where their reflected lines meet the y-axis. [3] (c) Determine whether these rays share one focus.
[1]
Show solutionHide solution
(a) By the chain rule, y\prime=-\frac12(25-x^2)^{-1/2}(-2x)=x/\sqrt{25-x^2}. M1A1 (b) At P, m=3/4 and the reflected gradient is -7/24; at Q, m=4/3 and the reflected gradient is 7/24. M1 The lines y-1=-7(x-3)/24 and y-2=7(x-4)/24 meet the axis at (0,15/8) and (0,5/6). A1A1 (c) Different crossings mean there is no shared focus.
R1

Final answer

(0,15/8) and (0,5/6); no shared focus.

G1 Investigation

[12 marks]
Rays travel vertically downwards onto the mirror y=\dfrac{x^2}{4}, each along a line x=p with p\ne0. Use the reflection rule from the Key result: a ray meeting the mirror where the tangent has gradient m, m\ne0, leaves along a line with gradient \dfrac{m^2-1}{2m}.
(a) A ray travels down the line x=2. Show that its reflected ray is horizontal, and write down the point where it crosses the y-axis.
[2]
(b) Find the point where the reflected ray crosses the y-axis for the ray along x=6 and for the ray along x=-1. State a conjecture.
[3]
(c) Prove your conjecture for the ray along x=p, where p\ne0.
[3]
(d) Find the point F through which every reflected ray passes for the curve y=kx^2, where k>0.
[2]
(e) A dish has a cross-section of the form y=kx^2. It is 60 cm across and 9 cm deep. Find how far above the bottom of the dish the point F lies.
[2]
Show solutionHide solution
(a) \dfrac{dy}{dx}=\dfrac x2=1 at x=2; reflected gradient \dfrac{1-1}{2}=0
M1
the ray reflects along y=1, crossing the y-axis at (0,1)
A1
(b) x=6: point (6,9), m=3, gradient \dfrac{9-1}{6}=\dfrac43, crosses at y=9-\dfrac43(6)=1
A1
x=-1: point \left(-1,\frac14\right), m=-\frac12, gradient \dfrac{\frac14-1}{-1}=\dfrac34, crosses at y=\dfrac14+\dfrac34=1
A1
conjecture: every reflected ray passes through (0,1)
A1
(c) the point is \left(p,\frac{p^2}4\right) with m=\dfrac p2
M1
reflected gradient \dfrac{\frac{p^2}4-1}{p}=\dfrac{p^2-4}{4p}
A1
at x=0: y=\dfrac{p^2}4-p\cdot\dfrac{p^2-4}{4p}=\dfrac{p^2}{4}-\dfrac{p^2}{4}+1=1, for every p\ne0
A1
(d) at (p,kp^2), m=2kp, reflected gradient \dfrac{4k^2p^2-1}{4kp}; at x=0, y=kp^2-\dfrac{4k^2p^2-1}{4k}
M1
y=\dfrac1{4k}, so F=\left(0,\dfrac{1}{4k}\right)
A1
(e) the rim is at (30,9): 9=k(30)^2, k=\dfrac1{100}
M1
F is \dfrac{1}{4k}=25 cm above the bottom
A1

Final answer

  • (a) (0,1)
  • (b) (0,1) both times; every reflected ray passes through (0,1)
  • (c) the crossing is at y=1 for every p
  • (d) F\left(0,\frac1{4k}\right)
  • (e) 25 cm

G2 Light takes the quickest path

G2 Worked example

[9 marks]
A lifeguard at L is 20 m from a straight shoreline. A swimmer at S is 30 m from the shoreline, and the point of the shoreline nearest to S is 60 m along the shore from the point nearest to L. The lifeguard runs at 6 m s^{-1} to a point P on the shoreline, x m along the shore, and then swims at 1.5 m s^{-1} straight to S.
(a) Show that the time taken, T seconds, is given by T(x)=\dfrac{\sqrt{400+x^2}}{6}+\dfrac{\sqrt{900+(60-x)^2}}{1.5}.
[2]
(b) Find the value of x that minimises T, and the least time.
[2]
(c) Find the time taken if the lifeguard heads straight for S, and the time taken if the lifeguard runs to the point of the shoreline nearest to S.
[2]
(d) Let \theta_1 and \theta_2 be the angles that LP and PS make with the line through P perpendicular to the shoreline. By considering T'(x), show that at the minimum \dfrac{\sin\theta_1}{\sin\theta_2}=4.
[3]
Show solutionHide solution
(a) LP=\sqrt{20^2+x^2} and PS=\sqrt{30^2+(60-x)^2}, by Pythagoras
M1
time = distance \div speed for each stage: T(x)=\dfrac{\sqrt{400+x^2}}{6}+\dfrac{\sqrt{900+(60-x)^2}}{1.5}
A1AG
(b) by GDC, x=52.8
A1
least time T=30.0 s (29.98\ldots)
A1
(c) straight for S: the path crosses the shore where \dfrac{x}{20}=\dfrac{60-x}{30}, x=24; T(24)=36.4 s
A1
run to the nearest point: T(60)=\dfrac{\sqrt{4000}}{6}+\dfrac{30}{1.5}=30.5 s
A1
(d) T'(x)=\dfrac{x}{6\sqrt{400+x^2}}-\dfrac{60-x}{1.5\sqrt{900+(60-x)^2}}
M1
\sin\theta_1=\dfrac{x}{\sqrt{400+x^2}} and \sin\theta_2=\dfrac{60-x}{\sqrt{900+(60-x)^2}}, so T'(x)=\dfrac{\sin\theta_1}{6}-\dfrac{\sin\theta_2}{1.5}
M1
at the minimum T'(x)=0: \dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{6}{1.5}=4
A1AG
Check: at x=52.79, \sin\theta_1=0.935 and \sin\theta_2=0.234.

Final answer

  • (b) x=52.8, T=30.0 s
  • (c) 36.4 s; 30.5 s
  • (d) T'(x)=\frac{\sin\theta_1}{6}-\frac{\sin\theta_2}{1.5}=0

G2 Investigation

[11 marks]
The shoreline is the x-axis. A lifeguard starts at L(0,30) on the beach and must reach a swimmer at S(60,-20) in the sea, entering the water at P(x,0). The lifeguard runs k times as fast as they swim, where k\ge1. In units where the swimming speed is 1, the time taken is T(x)=\dfrac{\sqrt{900+x^2}}{k}+\sqrt{400+(60-x)^2}. Let x^* be the value of x that minimises T, and let \theta_1 and \theta_2 be the angles that LP and PS make with the line through P perpendicular to the shoreline. You may use the result that at the minimum \sin\theta_1=k\sin\theta_2.
(a) For k=1, explain why the quickest route is the straight line LS, and find x^*.
[2]
(b) Use your GDC to find x^* for k=2, k=4 and k=8. State a conjecture about x^* as k increases.
[3]
(c) Show that 60-x^*=20\tan\theta_2, and hence show that 0<60-x^*<\dfrac{20}{\sqrt{k^2-1}} for k>1. Hence prove your conjecture.
[4]
(d) Find the least value of k for which (c) guarantees that the lifeguard enters the water within 1 m of the point nearest to the swimmer.
[2]
Show solutionHide solution
(a) for k=1, T is the total length LP+PS, which is least when L, P and S are collinear
R1
LS crosses the shore where \dfrac{x}{30}=\dfrac{60-x}{20}, so x^*=36
A1
(b) k=2: x^*=50.5; k=4: x^*=55.5; k=8: x^*=57.8
A2
Note: award A1 for two correct values.
conjecture: x^* increases towards 60, so the lifeguard enters the water ever closer to the point opposite the swimmer
A1
(c) in the right-angled triangle with hypotenuse PS, the vertical side is 20 and the horizontal side is 60-x^*, so \tan\theta_2=\dfrac{60-x^*}{20}
A1
\sin\theta_2=\dfrac{\sin\theta_1}{k}<\dfrac1k, since \sin\theta_1<1
M1
\theta_2 is acute and \tan^2\theta_2=\dfrac{\sin^2\theta_2}{1-\sin^2\theta_2} increases with \sin^2\theta_2<\dfrac{1}{k^2}, so \tan^2\theta_2<\dfrac{1}{k^2-1}, giving 0<\tan\theta_2<\dfrac{1}{\sqrt{k^2-1}}
A1
0<60-x^*<\dfrac{20}{\sqrt{k^2-1}}\to0 as k\to\infty, so x^*\to60
R1
(d) \dfrac{20}{\sqrt{k^2-1}}\le1\iff k^2\ge401
M1
k\ge\sqrt{401}=20.0 (20.02\ldots)
A1

Final answer

  • (a) x^*=36
  • (b) 50.5, 55.5, 57.8; x^*\to60
  • (c) bound \frac{20}{\sqrt{k^2-1}}\to0
  • (d) k=\sqrt{401}\approx20.0

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 12.

JobTI-84 Plus CE keystrokes
Gradient at a pointMATH, 8:nDeriv(expression, X, value)
IntersectionsGraph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL

Chapter 10 teaches differentiation from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book

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