Question
The following diagram shows the graphs of the displacement, velocity and acceleration of a moving object as functions of time, t.
Complete the following table by noting which graph A, B or C corresponds to each function.
Write down the value of t when the velocity is greatest.
Answer/Explanation
Markscheme
A2A2 N4
[4 marks]
\(t = 3\) A2 N2
[2 marks]
Question
Let \(f(x) = 3 + \frac{{20}}{{{x^2} – 4}}\) , for \(x \ne \pm 2\) . The graph of f is given below.
The y-intercept is at the point A.
(i) Find the coordinates of A.
(ii) Show that \(f'(x) = 0\) at A.
The second derivative \(f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}\) . Use this to
(i) justify that the graph of f has a local maximum at A;
(ii) explain why the graph of f does not have a point of inflexion.
Describe the behaviour of the graph of \(f\) for large \(|x|\) .
Write down the range of \(f\) .
Answer/Explanation
Markscheme
(i) coordinates of A are \((0{\text{, }} – 2)\) A1A1 N2
(ii) derivative of \({x^2} – 4 = 2x\) (seen anywhere) (A1)
evidence of correct approach (M1)
e.g. quotient rule, chain rule
finding \(f'(x)\) A2
e.g. \(f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)\) , \(\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}\)
substituting \(x = 0\) into \(f'(x)\) (do not accept solving \(f'(x) = 0\) ) M1
at A \(f'(x) = 0\) AG N0
[7 marks]
(i) reference to \(f'(x) = 0\) (seen anywhere) (R1)
reference to \(f”(0)\) is negative (seen anywhere) R1
evidence of substituting \(x = 0\) into \(f”(x)\) M1
finding \(f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}\) \(\left( { = – \frac{5}{2}} \right)\) A1
then the graph must have a local maximum AG
(ii) reference to \(f”(x) = 0\) at point of inflexion (R1)
recognizing that the second derivative is never 0 A1 N2
e.g. \(40(3{x^2} + 4) \ne 0\) , \(3{x^2} + 4 \ne 0\) , \({x^2} \ne – \frac{4}{3}\) , the numerator is always positive
Note: Do not accept the use of the first derivative in part (b).
[6 marks]
correct (informal) statement, including reference to approaching \(y = 3\) A1 N1
e.g. getting closer to the line \(y = 3\) , horizontal asymptote at \(y = 3\)
[1 mark]
correct inequalities, \(y \le – 2\) , \(y > 3\) , FT from (a)(i) and (c) A1A1 N2
[2 marks]
Question
A function f is defined for \( – 4 \le x \le 3\) . The graph of f is given below.
The graph has a local maximum when \(x = 0\) , and local minima when \(x = – 3\) , \(x = 2\) .
Write down the x-intercepts of the graph of the derivative function, \(f’\) .
Write down all values of x for which \(f'(x)\) is positive.
At point D on the graph of f , the x-coordinate is \( – 0.5\). Explain why \(f”(x) < 0\) at D.
Answer/Explanation
Markscheme
x-intercepts at \( – 3\), 0, 2 A2 N2
[2 marks]
\( – 3 < x < 0\) , \(2 < x < 3\) A1A1 N2
[2 marks]
correct reasoning R2
e.g. the graph of f is concave-down (accept convex), the first derivative is decreasing
therefore the second derivative is negative AG
[2 marks]
Question
Consider \(f(x) = \ln ({x^4} + 1)\) .
The second derivative is given by \(f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}\) .
The equation \(f”(x) = 0\) has only three solutions, when \(x = 0\) , \( \pm \sqrt[4]{3}\) \(( \pm 1.316 \ldots )\) .
Find the value of \(f(0)\) .
Find the set of values of \(x\) for which \(f\) is increasing.
(i) Find \(f”(1)\) .
(ii) Hence, show that there is no point of inflexion on the graph of \(f\) at \(x = 0\) .
There is a point of inflexion on the graph of \(f\) at \(x = \sqrt[4]{3}\) \((x = 1.316 \ldots )\) .
Sketch the graph of \(f\) , for \(x \ge 0\) .
Answer/Explanation
Markscheme
substitute \(0\) into \(f\) (M1)
eg \(\ln (0 + 1)\) , \(\ln 1\)
\(f(0) = 0\) A1 N2
[2 marks]
\(f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}\) (seen anywhere) A1A1
Note: Award A1 for \(\frac{1}{{{x^4} + 1}}\) and A1 for \(4{x^3}\) .
recognizing \(f\) increasing where \(f'(x) > 0\) (seen anywhere) R1
eg \(f'(x) > 0\) , diagram of signs
attempt to solve \(f'(x) > 0\) (M1)
eg \(4{x^3} = 0\) , \({x^3} > 0\)
\(f\) increasing for \(x > 0\) (accept \(x \ge 0\) ) A1 N1
[5 marks]
(i) substituting \(x = 1\) into \(f”\) (A1)
eg \(\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}\) , \(\frac{{4 \times 2}}{4}\)
\(f”(1) = 2\) A1 N2
(ii) valid interpretation of point of inflexion (seen anywhere) R1
eg no change of sign in \(f”(x)\) , no change in concavity,
\(f’\) increasing both sides of zero
attempt to find \(f”(x)\) for \(x < 0\) (M1)
eg \(f”( – 1)\) , \(\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}\) , diagram of signs
correct working leading to positive value A1
eg \(f”( – 1) = 2\) , discussing signs of numerator and denominator
there is no point of inflexion at \(x = 0\) AG N0
[5 marks]
A1A1A1 N3
Notes: Award A1 for shape concave up left of POI and concave down right of POI.
Only if this A1 is awarded, then award the following:
A1 for curve through (\(0\), \(0\)) , A1 for increasing throughout.
Sketch need not be drawn to scale. Only essential features need to be clear.
[3 marks]
Question
The following diagram shows part of the graph of \(y = f(x)\).
The graph has a local maximum at \(A\), where \(x = – 2\), and a local minimum at \(B\), where \(x = 6\).
On the following axes, sketch the graph of \(y = f'(x)\).
Write down the following in order from least to greatest: \(f(0),{\text{ }}f'(6),{\text{ }}f”( – 2)\).
Answer/Explanation
Markscheme
A1A1A1A1 N4
Note: Award A1 for x-intercept in circle at \(-2\), A1 for x-intercept in circle at \(6\).
Award A1 for approximately correct shape.
Only if this A1 is awarded, award A1 for a negative y-intercept.
[4 marks]
\(f”( – 2),{\text{ }}f'(6),{\text{ }}f(0)\) A2 N2
[2 marks]
Question
A function \(f\) has its derivative given by \(f'(x) = 3{x^2} – 2kx – 9\), where \(k\) is a constant.
Find \(f”(x)\).
The graph of \(f\) has a point of inflexion when \(x = 1\).
Show that \(k = 3\).
Find \(f'( – 2)\).
Find the equation of the tangent to the curve of \(f\) at \(( – 2,{\text{ }}1)\), giving your answer in the form \(y = ax + b\).
Given that \(f'( – 1) = 0\), explain why the graph of \(f\) has a local maximum when \(x = – 1\).
Answer/Explanation
Markscheme
\(f”(x) = 6x – 2k\) A1A1 N2
[2 marks]
substituting \(x = 1\) into \(f”\) (M1)
eg\(\;\;\;f”(1),{\text{ }}6(1) – 2k\)
recognizing \(f”(x) = 0\;\;\;\)(seen anywhere) M1
correct equation A1
eg\(\;\;\;6 – 2k = 0\)
\(k = 3\) AG N0
[3 marks]
correct substitution into \(f'(x)\) (A1)
eg\(\;\;\;3{( – 2)^2} – 6( – 2) – 9\)
\(f'( – 2) = 15\) A1 N2
[2 marks]
recognizing gradient value (may be seen in equation) M1
eg\(\;\;\;a = 15,{\text{ }}y = 15x + b\)
attempt to substitute \(( – 2,{\text{ }}1)\) into equation of a straight line M1
eg\(\;\;\;1 = 15( – 2) + b,{\text{ }}(y – 1) = m(x + 2),{\text{ }}(y + 2) = 15(x – 1)\)
correct working (A1)
eg\(\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1\)
\(y = 15x + 31\) A1 N2
[4 marks]
METHOD 1 (\({{\text{2}}^{{\text{nd}}}}\) derivative)
recognizing \(f” < 0\;\;\;\)(seen anywhere) R1
substituting \(x = – 1\) into \(f”\) (M1)
eg\(\;\;\;f”( – 1),{\text{ }}6( – 1) – 6\)
\(f”( – 1) = – 12\) A1
therefore the graph of \(f\) has a local maximum when \(x = – 1\) AG N0
METHOD 2 (\({{\text{1}}^{{\text{st}}}}\) derivative)
recognizing change of sign of \(f'(x)\;\;\;\)(seen anywhere) R1
eg\(\;\;\;\)sign chart\(\;\;\;\)
correct value of \(f’\) for \( – 1 < x < 3\) A1
eg\(\;\;\;f'(0) = – 9\)
correct value of \(f’\) for \(x\) value to the left of \( – 1\) A1
eg\(\;\;\;f'( – 2) = 15\)
therefore the graph of \(f\) has a local maximum when \(x = – 1\) AG N0
[3 marks]
Total [14 marks]
Question
Let \(y = f(x)\), for \( – 0.5 \le \) x \( \le \) \(6.5\). The following diagram shows the graph of \(f’\), the derivative of \(f\).
The graph of \(f’\) has a local maximum when \(x = 2\), a local minimum when \(x = 4\), and it crosses the \(x\)-axis at the point \((5,{\text{ }}0)\).
Explain why the graph of \(f\) has a local minimum when \(x = 5\).
Find the set of values of \(x\) for which the graph of \(f\) is concave down.
The following diagram shows the shaded regions \(A\), \(B\) and \(C\).
The regions are enclosed by the graph of \(f’\), the \(x\)-axis, the \(y\)-axis, and the line \(x = 6\).
The area of region \(A\) is 12, the area of region \(B\) is 6.75 and the area of region \(C\) is 6.75.
Given that \(f(0) = 14\), find \(f(6)\).
The following diagram shows the shaded regions \(A\), \(B\) and \(C\).
The regions are enclosed by the graph of \(f’\), the x-axis, the y-axis, and the line \(x = 6\).
The area of region \(A\) is 12, the area of region \(B\) is 6.75 and the area of region \(C\) is 6.75.
Let \(g(x) = {\left( {f(x)} \right)^2}\). Given that \(f'(6) = 16\), find the equation of the tangent to the graph of \(g\) at the point where \(x = 6\).
Answer/Explanation
Markscheme
METHOD 1
\(f'(5) = 0\) (A1)
valid reasoning including reference to the graph of \(f’\) R1
eg\(\;\;\;f’\) changes sign from negative to positive at \(x = 5\), labelled sign chart for \(f’\)
so \(f\) has a local minimum at \(x = 5\) AG N0
Note: It must be clear that any description is referring to the graph of \(f’\), simply giving the conditions for a minimum without relating them to \(f’\) does not gain the R1.
METHOD 2
\(f'(5) = 0\) A1
valid reasoning referring to second derivative R1
eg\(\;\;\;f”(5) > 0\)
so \(f\) has a local minimum at \(x = 5\) AG N0
[2 marks]
attempt to find relevant interval (M1)
eg\(\;\;\;f’\) is decreasing, gradient of \(f’\) is negative, \(f” < 0\)
\(2 < x < 4\;\;\;\)(accept “between 2 and 4”) A1 N2
Notes: If no other working shown, award M1A0 for incorrect inequalities such as \(2 \le \) \(x\) \( \le \) 4, or “from 2 to 4”
[2 marks]
METHOD 1 (one integral)
correct application of Fundamental Theorem of Calculus (A1)
eg\(\;\;\;\int_0^6 {f'(x){\text{d}}x = } f(6) – f(0),{\text{ }}f(6) = 14 + \int_0^6 {f'(x){\text{d}}x} \)
attempt to link definite integral with areas (M1)
eg\(\;\;\;\int_0^6 {f'(x){\text{d}}x = – 12 – 6.75 + 6.75,{\text{ }}\int_0^6 {f'(x){\text{d}}x = {\text{Area }}A + {\text{Area }}B + {\text{ Area }}C} } \)
correct value for \(\int_0^6 {f'(x){\text{d}}x} \) (A1)
eg\(\;\;\;\int_0^6 {f'(x){\text{d}}x} = – 12\)
correct working A1
eg\(\;\;\;f(6) – 14 = – 12,{\text{ }}f(6) = – 12 + f(0)\)
\(f(6) = 2\) A1 N3
METHOD 2 (more than one integral)
correct application of Fundamental Theorem of Calculus (A1)
eg\(\;\;\;\int_0^2 {f'(x){\text{d}}x} = f(2) – f(0),{\text{ }}f(2) = 14 + \int_0^2 {f'(x)} \)
attempt to link definite integrals with areas (M1)
eg\(\;\;\;\int_0^2 {f'(x){\text{d}}x} = 12,{\text{ }}\int_2^5 {f'(x){\text{d}}x = – 6.75} ,{\text{ }}\int_0^6 {f'(x)} = 0\)
correct values for integrals (A1)
eg\(\;\;\;\int_0^2 {f'(x){\text{d}}x} = – 12,{\text{ }}\int_5^2 {f'(x)} {\text{d}}x = 6.75,{\text{ }}f(6) – f(2) = 0\)
one correct intermediate value A1
eg\(\;\;\;f(2) = 2,{\text{ }}f(5) = – 4.75\)
\(f(6) = 2\) A1 N3
[5 marks]
correct calculation of \(g(6)\) (seen anywhere) A1
eg\(\;\;\;{2^2},{\text{ }}g(6) = 4\)
choosing chain rule or product rule (M1)
eg\(\;\;\;g’\left( {f(x)} \right)f'(x),{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}f(x)f'(x) + f'(x)f(x)\)
correct derivative (A1)
eg\(\;\;\;g'(x) = 2f(x)f'(x),{\text{ }}f(x)f'(x) + f'(x)f(x)\)
correct calculation of \(g'(6)\) (seen anywhere) A1
eg\(\;\;\;2(2)(16),{\text{ }}g'(6) = 64\)
attempt to substitute their values of \(g'(6)\) and \(g(6)\) (in any order) into equation of a line (M1)
eg\(\;\;\;{2^2} = (2 \times 2 \times 16)6 + b,{\text{ }}y – 6 = 64(x – 4)\)
correct equation in any form A1 N2
eg\(\;\;\;y – 4 = 64(x – 6),{\text{ }}y = 64x – 380\)
[6 marks]
[Total 15 marks]
Question
The following diagram shows the graph of \(f’\), the derivative of \(f\).
The graph of \(f’\) has a local minimum at A, a local maximum at B and passes through \((4,{\text{ }} – 2)\).
The point \({\text{P}}(4,{\text{ }}3)\) lies on the graph of the function, \(f\).
Write down the gradient of the curve of \(f\) at P.
Find the equation of the normal to the curve of \(f\) at P.
Determine the concavity of the graph of \(f\) when \(4 < x < 5\) and justify your answer.
Answer/Explanation
Markscheme
\( – 2\) A1 N1
[1 mark]
gradient of normal \( = \frac{1}{2}\) (A1)
attempt to substitute their normal gradient and coordinates of P (in any order) (M1)
eg\(\,\,\,\,\,\)\(y – 4 = \frac{1}{2}(x – 3),{\text{ }}3 = \frac{1}{2}(4) + b,{\text{ }}b = 1\)
\(y – 3 = \frac{1}{2}(x – 4),{\text{ }}y = \frac{1}{2}x + 1,{\text{ }}x – 2y + 2 = 0\) A1 N3
[3 marks]
correct answer and valid reasoning A2 N2
answer: eg graph of \(f\) is concave up, concavity is positive (between \(4 < x < 5\))
reason: eg slope of \(f’\) is positive, \(f’\) is increasing, \(f’’ > 0\),
sign chart (must clearly be for \(f’’\) and show A and B)
Note: The reason given must refer to a specific function/graph. Referring to “the graph” or “it” is not sufficient.
[2 marks]
Question
A function f (x) has derivative f ′(x) = 3x2 + 18x. The graph of f has an x-intercept at x = −1.
Find f (x).
The graph of f has a point of inflexion at x = p. Find p.
Find the values of x for which the graph of f is concave-down.
Answer/Explanation
Markscheme
evidence of integration (M1)
eg \(\int {f’\left( x \right)} \)
correct integration (accept absence of C) (A1)(A1)
eg \({x^3} + \frac{{18}}{2}{x^2} + C,\,\,{x^3} + 9{x^2}\)
attempt to substitute x = −1 into their f = 0 (must have C) M1
eg \({\left( { – 1} \right)^3} + 9{\left( { – 1} \right)^2} + C = 0,\,\, – 1 + 9 + C = 0\)
Note: Award M0 if they substitute into original or differentiated function.
correct working (A1)
eg \(8 + C = 0,\,\,\,C = – 8\)
\(f\left( x \right) = {x^3} + 9{x^2} – 8\) A1 N5
[6 marks]
METHOD 1 (using 2nd derivative)
recognizing that f” = 0 (seen anywhere) M1
correct expression for f” (A1)
eg 6x + 18, 6p + 18
correct working (A1)
6p + 18 = 0
p = −3 A1 N3
METHOD 1 (using 1st derivative)
recognizing the vertex of f′ is needed (M2)
eg \( – \frac{b}{{2a}}\) (must be clear this is for f′)
correct substitution (A1)
eg \(\frac{{ – 18}}{{2 \times 3}}\)
p = −3 A1 N3
[4 marks]
valid attempt to use f” (x) to determine concavity (M1)
eg f” (x) < 0, f” (−2), f” (−4), 6x + 18 ≤ 0 
correct working (A1)
eg 6x + 18 < 0, f” (−2) = 6, f” (−4) = −6 
f concave down for x < −3 (do not accept x ≤ −3) A1 N2
[3 marks]