Differentiation: exam review Exam review Worked solutions

The graph of \(y=f(x)\), for \(0\le x\le 5.5\), has a local maximum at A\((1,11)\) and a local minimum at B\((4,-16)\). [6]

1. (a) Write down the values of \(x\) for which \(f'(x)=0\). [1]

   • A1 for \(x=1\) and \(x=4\)

2. (b) Write down the interval on which \(f\) is decreasing. [2]

   • M1 for the gradient being negative between the turning points
   • A1 for \(1<x<4\)

3. (c) Sketch the graph of \(f'(x)\) for \(0\le x\le 5.5\), showing where it crosses the \(x\)-axis. [3]

   • A1 for a parabola opening upwards
   • A1 for \(x\)-intercepts at 1 and 4
   • A1 for the graph below the \(x\)-axis between 1 and 4 and above it outside

The diagram shows the graph of \(y=f'(x)\), the gradient function of a function \(f\). The graph crosses the \(x\)-axis at \(x=-2\) and \(x=3\). [6]

1. (a) Write down the values of \(x\) at which \(f\) has a stationary point. [1]

   • A1 for \(x=-2\) and \(x=3\) (where \(f'(x)=0\))

2. (b) Write down the interval on which \(f\) is increasing. [2]

   • M1 for \(f'(x)>0\) (graph above the \(x\)-axis)
   • A1 for \(-2<x<3\)

3. (c) For each stationary point, state whether it is a local maximum or a local minimum. Give a reason for each answer. [3]

   • A1 for a minimum at \(x=-2\)
   • A1 for a maximum at \(x=3\)
   • R1 for the reason: \(f'\) changes from negative to positive at \(x=-2\) and from positive to negative at \(x=3\)

A function is defined by \(f(x)=x^{4} - 4 x^{2}\) for \(-3\le x\le 3\). [6]

1. (a) Use your calculator to sketch the graph of \(y=f(x)\), showing the \(x\)-intercepts. [2]

   • A1 for a W shape (two minima, a local maximum at \(x=0\))
   • A1 for \(x\)-intercepts at \(-2\), \(0\) and \(2\)

2. (b) Write down the values of \(x\) for which \(f(x)>0\). [1]

   • A1 for \(x<-2\) or \(x>2\)

3. (c) Find the values of \(x\) for which \(f'(x)<0\), that is, where \(f\) is decreasing. [3]

   • A1 for \(f'(x)=4 x^{3} - 8 x\)
   • M1 for solving \(f'(x)=0\): \(x=0,\ \pm1.41\)
   • A1 for \(x<-1.41\text{ or }0<x<1.41\)

Differentiate each of the following with respect to \(x\). [6]

1. (a) \(f(x)=4x^3-\frac{6}{x^2}+5x-1\) [2]

   • M1 for a negative power or the rule \(ax^n\to nax^{n-1}\) used on at least two terms
   • A1 for \(f'(x)=12 x^{2} + 5 + \frac{12}{x^{3}}\)

2. (b) \(g(x)=(3x+2)(x-5)\) [2]

   • M1 for expanding: \(g(x)=3 x^{2} - 13 x - 10\)
   • A1 for \(g'(x)=6 x - 13\)

3. (c) \(h(x)=\dfrac{6 x^{3} - 4 x + 9}{x^{2}}\) [2]

   • M1 for splitting the fraction: \(h(x)=6 x - \frac{4}{x} + \frac{9}{x^{2}}\)
   • A1 for \(h'(x)=6 + \frac{4}{x^{2}} - \frac{18}{x^{3}}\)

The temperature \(T\) °C of a laboratory sample \(t\) minutes after it is removed from a heater is modelled by \(T=0.25t^2-5t+80\), for \(0\le t\le 20\). [6]

1. (a) Find \(\dfrac{dT}{dt}\). [2]

   • M1 for differentiating a term
   • A1 for \(\dfrac{dT}{dt}=\frac{t}{2} - 5\)

2. (b) Find the rate of change of the temperature when \(t=6\). Give a reason for the sign of your answer. [2]

   • A1 for \(\dfrac{dT}{dt}=-2\) when \(t=6\), with units °C per minute
   • R1 for the reason: the rate is negative, so the temperature is falling

3. (c) Show that the temperature is decreasing for \(0\le t<10\). [2]

   • M1 for \(\dfrac{dT}{dt}<0\): \(\frac{t}{2} - 5<0\)
   • R1 for \(t<10\), so it is decreasing for \(0\le t<10\)

The curve \(C\) has equation \(y=x^{3} - 4 x + 6\). The point P on \(C\) has \(x\)-coordinate \(2\). [7]

1. (a) Find the gradient of the tangent to \(C\) at P. [2]

   • M1 for \(\dfrac{dy}{dx}=3 x^{2} - 4\)
   • A1 for \(8\)

2. (b) Find the equation of the tangent to \(C\) at P, in the form \(ax+by=c\), where \(a, b, c\in\mathbb{Z}\). [2]

   • M1 for \(y=6\) at P and \(y-6=8(x-2)\)
   • A1 for \(8x-y=10\)

3. (c) Find the equation of the normal to \(C\) at P, in the form \(y=mx+c\). [2]

   • M1 for the normal gradient \(-\dfrac{1}{8}\)
   • A1 for \(y=- \frac{1}{8}x+\frac{25}{4}\)

4. (d) There is one other point on \(C\) at which the gradient is the same as at P. Find its \(x\)-coordinate. [1]

   • A1 for solving \(3 x^{2} - 4=8\): \(x=-2\)

The curve \(C\) has equation \(y=f(x)\), where \(f(x)=x\sqrt{x+5}\) for \(x>0\). Give all answers to 3 significant figures. [6]

1. (a) Use your calculator to find \(f(4)\) and \(f'(4)\). [2]

   • A1 for \(f(4)=12\)
   • A1 for \(f'(4)=3.67\) (from the calculator's derivative at a point)

2. (b) Find the equation of the tangent to \(C\) at the point where \(x=4\), in the form \(y=mx+c\). [2]

   • M1 for \(y-12=3.67(x-4)\)
   • A1 for \(y=3.67x-2.67\)

3. (c) Graph \(y=f'(x)\) and a suitable line on your calculator. Hence find the value of \(x\), for \(0<x<10\), at which the gradient of \(C\) is 4. [2]

   • M1 for graphing \(y=f'(x)\) with \(y=4\) and finding where they meet
   • A1 for \(x=5.17\)

The curve \(C\) has equation \(y=f(x)\), where \(f(x)=x^{3} + \frac{16}{x}\) and \(x>0\). [14]

1. (a) Find \(f'(x)\). [2]

   • M1 for writing \(\dfrac{16}{x}\) as \(16x^{-1}\) and differentiating a term
   • A1 for \(f'(x)=3 x^{2} - \frac{16}{x^{2}}\)

2. (b) Show that the tangent to \(C\) at the point where \(x=2\) passes through the origin. [3]

   • A1 for \(f(2)=16\) and \(f'(2)=8\)
   • M1 for \(y-16=8(x-2)\)
   • R1 for \(y=8x\), which gives \(y=0\) when \(x=0\)

3. (c) Find the equation of the normal to \(C\) at the point where \(x=2\). Give your answer in the form \(ax+by=c\), where \(a, b, c\in\mathbb{Z}\). [3]

   • A1 for the normal gradient \(-\dfrac{1}{8}\)
   • M1 for \(y-16=-\dfrac{1}{8}(x-2)\)
   • A1 for \(x+8y=130\)

4. (d) The normal meets \(C\) again at the point Q. Use your calculator to find the coordinates of Q. [3]

   • M1 for graphing \(y=f(x)\) and \(y=- \frac{1}{8}x+\frac{65}{4}\)
   • A1 for \(x=1.08\)
   • A1 for \(y=16.1\)

5. (e) The tangent to \(C\) at the point R is parallel to the line \(y=-13x+5\). Find the coordinates of R. [3]

   • M1 for \(f'(x)=-13\): \(3 x^{2} - \frac{16}{x^{2}}=-13\)
   • A1 for \(x=1\)
   • A1 for \(y=17\)

The function \(f\) is given by \(f(x)=x^3+ax^2+bx+7\), where \(a\) and \(b\) are constants. The graph of \(y=f(x)\) has a horizontal tangent at \(x=-1\) and at \(x=3\). [14]

1. (a) Find \(f'(x)\) in terms of \(a\) and \(b\). [1]

   • A1 for \(f'(x)=3x^2+2ax+b\)

2. (b) Hence write down two equations in \(a\) and \(b\), and find the value of \(a\) and the value of \(b\). [4]

   • M1 for \(f'(-1)=0\): \(- 2 a + b + 3=0\)
   • M1 for \(f'(3)=0\): \(6 a + b + 27=0\)
   • M1 for solving the pair of equations (calculator solver)
   • A1 for \(a=-3\) and \(b=-9\)

3. (c) Write down the \(y\)-intercept of the graph. [1]

   • A1 for \(7\)

4. (d) Use your calculator to find the coordinates of the local maximum point and of the local minimum point on the graph of \(f\). [3]

   • M1 for graphing \(f(x)=x^{3} - 3 x^{2} - 9 x + 7\)
   • A1 for the local maximum \((-1,12)\)
   • A1 for the local minimum \((3,-20)\)

5. (e) The domain of \(f\) is \(-2\le x\le 6\). Find the greatest value of \(f(x)\) on this domain. [2]

   • M1 for comparing the local values with \(f(-2)=5\) and \(f(6)=61\)
   • A1 for \(61\)

6. (f) Find the range of \(f\) on this domain. [3]

   • M1 for comparing the local minimum \(-20\) with \(f(-2)=5\)
   • A1 for the least value \(-20\)
   • A1 for \(-20\le f(x)\le 61\)

An open box with a square base of side \(x\) cm and height \(h\) cm is made from 1200 cm² of card. The box has no lid. [14]

1. (a) Show that \(h=\dfrac{1200-x^2}{4x}\). [2]

   • M1 for the surface area \(x^2+4xh=1200\)
   • A1 for rearranging to \(h=\dfrac{1200-x^2}{4x}\)

2. (b) Hence show that the volume \(V\) of the box is \(V=300x-\dfrac{x^3}{4}\). [2]

   • M1 for \(V=x^2h\)
   • A1 for \(V=\dfrac{x^2(1200-x^2)}{4x}=300x-\dfrac{x^3}{4}\)

3. (c) Find \(\dfrac{dV}{dx}\). [2]

   • M1 for differentiating a term
   • A1 for \(\dfrac{dV}{dx}=300 - \frac{3 x^{2}}{4}\)

4. (d) Use your calculator to find the value of \(x\) for which \(V\) is a maximum. [3]

   • M1 for graphing \(V=300x-\dfrac{x^3}{4}\) with a window showing the peak
   • M1 for using the calculator's maximum tool
   • A1 for \(x=20\)

5. (e) Find the maximum volume and the height of the box when its volume is a maximum. [3]

   • A1 for \(V=4000\)
   • M1 for substituting \(x=20\) into \(h\)
   • A1 for \(h=10\)

6. (f) Sketch the graph of \(V\) against \(x\) for \(0<x<34.6\), showing the maximum point. [2]

   • A1 for the shape: a maximum, then falling
   • A1 for the maximum at \((20,4000)\) and the \(x\)-intercept at \(34.6\)