Differentiation: exam review Exam review: second paper One question at a time

The graph of \(y=f(x)\), for \(0\le x\le 6.5\), has a local maximum at A\((2,52)\) and a local minimum at B\((5,25)\). [6]

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

   • A1 for \(x=2\) and \(x=5\)

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

   • M1 for the gradient being negative between the turning points
   • A1 for \(2<x<5\)

3. (c) Sketch the graph of \(f'(x)\) for \(0\le x\le 6.5\), showing where it crosses the \(x\)-axis. [3]

   • A1 for a parabola opening upwards
   • A1 for \(x\)-intercepts at 2 and 5
   • A1 for the graph below the \(x\)-axis between 2 and 5 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 \(x=1\) and \(x=5\). [6]

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

   • A1 for \(x=1\) and \(x=5\) (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 \(1<x<5\)

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 \(x=1\)
   • A1 for a maximum at \(x=5\)
   • R1 for the reason: \(f'\) changes from negative to positive at \(x=1\) and from positive to negative at \(x=5\)

A function is defined by \(f(x)=x^{4} - 9 x^{2}\) for \(-4\le x\le 4\). [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 \(-3\), \(0\) and \(3\)

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

   • A1 for \(x<-3\) or \(x>3\)

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

   • A1 for \(f'(x)=4 x^{3} - 18 x\)
   • M1 for solving \(f'(x)=0\): \(x=0,\ \pm2.12\)
   • A1 for \(x<-2.12\text{ or }0<x<2.12\)

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

1. (a) \(f(x)=7x^4+\frac{8}{x^3}-3x+2\) [2]

   • M1 for a negative power or the rule \(ax^n\to nax^{n-1}\) used on at least two terms
   • A1 for \(f'(x)=28 x^{3} - 3 - \frac{24}{x^{4}}\)

2. (b) \(g(x)=(x+4)(2x-3)\) [2]

   • M1 for expanding: \(g(x)=2 x^{2} + 5 x - 12\)
   • A1 for \(g'(x)=4 x + 5\)

3. (c) \(h(x)=\dfrac{9 x^{4} - 6 x + 10}{x^{3}}\) [2]

   • M1 for splitting the fraction: \(h(x)=9 x - \frac{6}{x^{2}} + \frac{10}{x^{3}}\)
   • A1 for \(h'(x)=9 + \frac{12}{x^{3}} - \frac{30}{x^{4}}\)

The volume \(V\) litres of water in a tank \(t\) hours after a valve is opened is modelled by \(V=1.5t^2-18t+120\), for \(0\le t\le 12\). [6]

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

   • M1 for differentiating a term
   • A1 for \(\dfrac{dV}{dt}=3 t - 18\)

2. (b) Find the rate of change of the volume when \(t=4\). Give a reason for the sign of your answer. [2]

   • A1 for \(\dfrac{dV}{dt}=-6\) when \(t=4\), with units litres per hour
   • R1 for the reason: the rate is negative, so the volume is falling

3. (c) Show that the volume is decreasing for \(0\le t<6\). [2]

   • M1 for \(\dfrac{dV}{dt}<0\): \(3 t - 18<0\)
   • R1 for \(t<6\), so it is decreasing for \(0\le t<6\)

The curve \(C\) has equation \(y=x^{3} - 2 x + 7\). 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 \(\dfrac{dy}{dx}=3 x^{2} - 2\)
   • A1 for \(10\)

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 \(y=11\) at P and \(y-11=10(x-2)\)
   • A1 for \(10x-y=9\)

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

   • M1 for the normal gradient \(-\dfrac{1}{10}\)
   • A1 for \(y=- \frac{1}{10}x+\frac{56}{5}\)

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 \(3 x^{2} - 2=10\): \(x=-2\)

The curve \(C\) has equation \(y=f(x)\), where \(f(x)=x\ln(x+2)\) for \(x>0\). Give all answers to 3 significant figures. [6]

1. (a) Use your calculator to find \(f(3)\) and \(f'(3)\). [2]

   • A1 for \(f(3)=4.83\)
   • A1 for \(f'(3)=2.21\) (from the calculator's derivative at a point)

2. (b) Find the equation of the tangent to \(C\) at the point where \(x=3\), in the form \(y=mx+c\). [2]

   • M1 for \(y-4.83=2.21(x-3)\)
   • A1 for \(y=2.21x-1.80\)

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 1.5. [2]

   • M1 for graphing \(y=f'(x)\) with \(y=1.5\) and finding where they meet
   • A1 for \(x=1.13\)

The curve \(C\) has equation \(y=f(x)\), where \(f(x)=x^{4} + \frac{48}{x}\) and \(x>0\). [14]

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

   • M1 for writing \(\dfrac{48}{x}\) as \(48x^{-1}\) and differentiating a term
   • A1 for \(f'(x)=4 x^{3} - \frac{48}{x^{2}}\)

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

   • A1 for \(f(2)=40\) and \(f'(2)=20\)
   • M1 for \(y-40=20(x-2)\)
   • R1 for \(y=20x\), 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 \(-\dfrac{1}{20}\)
   • M1 for \(y-40=-\dfrac{1}{20}(x-2)\)
   • A1 for \(x+20y=802\)

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 \(y=- \frac{1}{20}x+\frac{401}{10}\)
   • A1 for \(x=1.29\)
   • A1 for \(y=40.0\)

5. (e) The tangent to \(C\) at the point R is parallel to the line \(y=-44x+5\). Find the coordinates of R. [3]

   • M1 for \(f'(x)=-44\): \(4 x^{3} - \frac{48}{x^{2}}=-44\)
   • A1 for \(x=1\)
   • A1 for \(y=49\)

The function \(f\) is given by \(f(x)=x^3+ax^2+bx+9\), where \(a\) and \(b\) are constants. The graph of \(y=f(x)\) has a horizontal tangent at \(x=-2\) and at \(x=4\). [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 \(f'(-2)=0\): \(- 4 a + b + 12=0\)
   • M1 for \(f'(4)=0\): \(8 a + b + 48=0\)
   • M1 for solving the pair of equations (calculator solver)
   • A1 for \(a=-3\) and \(b=-24\)

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

   • A1 for \(9\)

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 \(f(x)=x^{3} - 3 x^{2} - 24 x + 9\)
   • A1 for the local maximum \((-2,37)\)
   • A1 for the local minimum \((4,-71)\)

5. (e) The domain of \(f\) is \(-3\le x\le 8\). Find the greatest value of \(f(x)\) on this domain. [2]

   • M1 for comparing the local values with \(f(-3)=27\) and \(f(8)=137\)
   • A1 for \(137\)

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

   • M1 for comparing the local minimum \(-71\) with \(f(-3)=27\)
   • A1 for the least value \(-71\)
   • A1 for \(-71\le f(x)\le 137\)

An open fish tank with a square base of side \(x\) dm and height \(h\) dm is made from 432 dm² of glass. The tank has no lid. [14]

1. (a) Show that \(h=\dfrac{432-x^2}{4x}\). [2]

   • M1 for the surface area \(x^2+4xh=432\)
   • A1 for rearranging to \(h=\dfrac{432-x^2}{4x}\)

2. (b) Hence show that the volume \(V\) of the tank is \(V=108x-\dfrac{x^3}{4}\). [2]

   • M1 for \(V=x^2h\)
   • A1 for \(V=\dfrac{x^2(432-x^2)}{4x}=108x-\dfrac{x^3}{4}\)

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

   • M1 for differentiating a term
   • A1 for \(\dfrac{dV}{dx}=108 - \frac{3 x^{2}}{4}\)

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

   • M1 for graphing \(V=108x-\dfrac{x^3}{4}\) with a window showing the peak
   • M1 for using the calculator's maximum tool
   • A1 for \(x=12\)

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

   • A1 for \(V=864\)
   • M1 for substituting \(x=12\) into \(h\)
   • A1 for \(h=6\)

6. (f) Sketch the graph of \(V\) against \(x\) for \(0<x<20.8\), showing the maximum point. [2]

   • A1 for the shape: a maximum, then falling
   • A1 for the maximum at \((12,864)\) and the \(x\)-intercept at \(20.8\)