Question
Let y = \(\frac{Inx}{x^{4}}\) for x > 0.
(a) Show that \(\frac{dy}{dx}= \frac{1-4Inx}{x^{5}}\)
Consider the function defined by f (x) \(\frac{Inx}{x^{4}}\) = for x> 0 and its graph y = f (x) .
(b) The graph of f has a horizontal tangent at point P. Find the coordinates of P. [5]
(c) Given that f ” (x) = \(\frac{20Lnx-9}{x^{6}}\) show that P is a local maximum point. [3]
(d) Solve f (x) > 0 for x > 0. [2]
(e) Sketch the graph of f , showing clearly the value of the x-intercept and the approximate position of point P. [3]
Answer/Explanation
Ans
Question
Consider the functions f (x) = -(x – h)2 + 2k and g(x) = e x-2 + k where h , k ∈ R.
(a) Find f ′(x). [1]
The graphs of f and g have a common tangent at x = 3
(b) Show that h + = \(\frac{e+6}{2}\) [3]
(c) Hence, show that k = \(e+ \frac{e^{2}}{4}\) [3]
Answer/Explanation
Ans
(a) f ‘( x) = −2( x − h)
Question
Consider \(f(x) = \frac{1}{3}{x^3} + 2{x^2} – 5x\) . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.
Find \(f'(x)\) .
Find the x-coordinate of M.
Find the x-coordinate of N.
The line L is the tangent to the curve of f at \((3{\text{, }}12)\). Find the equation of L in the form \(y = ax + b\) .
Answer/Explanation
Markscheme
\(f'(x) = {x^2} + 4x – 5\) A1A1A1 N3
[3 marks]
evidence of attempting to solve \(f'(x) = 0\) (M1)
evidence of correct working A1
e.g. \((x + 5)(x – 1)\) , \(\frac{{ – 4 \pm \sqrt {16 + 20} }}{2}\) , sketch
\(x = – 5\), \(x = 1\) (A1)
so \(x = – 5\) A1 N2
[4 marks]
METHOD 1
\(f”(x) = 2x + 4\) (may be seen later) A1
evidence of setting second derivative = 0 (M1)
e.g. \(2x + 4 = 0\)
\(x = – 2\) A1 N2
METHOD 2
evidence of use of symmetry (M1)
e.g. midpoint of max/min, reference to shape of cubic
correct calculation A1
e.g. \(\frac{{ – 5 + 1}}{2}\)
\(x = – 2\) A1 N2
[3 marks]
attempting to find the value of the derivative when \(x = 3\) (M1)
\(f'(3) = 16\) A1
valid approach to finding the equation of a line M1
e.g. \(y – 12 = 16(x – 3)\) , \(12 = 16 \times 3 + b\)
\(y = 16x – 36\) A1 N2
[4 marks]
Question
Let \(f(x) = {{\rm{e}}^x}\cos x\) . Find the gradient of the normal to the curve of f at \(x = \pi \) .
Answer/Explanation
Markscheme
evidence of choosing the product rule (M1)
\(f'(x) = {{\rm{e}}^x} \times ( – \sin x) + \cos x \times {{\rm{e}}^x}\) \(( = {{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x)\) A1A1
substituting \(\pi \) (M1)
e.g. \(f'(\pi ) = {{\rm{e}}^\pi }\cos \pi – {{\rm{e}}^\pi }\sin \pi \) , \({{\rm{e}}^\pi }( – 1 – 0)\) , \( – {{\rm{e}}^\pi }\)
taking negative reciprocal (M1)
e.g. \( – \frac{1}{{f'(\pi )}}\)
gradient is \(\frac{1}{{{{\rm{e}}^\pi }}}\) A1 N3
[6 marks]
Question
Let \(f(x) = \sqrt x \) . Line L is the normal to the graph of f at the point (4, 2) .
In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .
Show that the equation of L is \(y = – 4x + 18\) .
Point A is the x-intercept of L . Find the x-coordinate of A.
Find an expression for the area of R .
The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .
Answer/Explanation
Markscheme
finding derivative (A1)
e.g. \(f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}\)
correct value of derivative or its negative reciprocal (seen anywhere) A1
e.g. \(\frac{1}{{2\sqrt 4 }}\) , \(\frac{1}{4}\)
gradient of normal = \(\frac{1}{{{\text{gradient of tangent}}}}\) (seen anywhere) A1
e.g. \( – \frac{1}{{f'(4)}} = – 4\) , \( – 2\sqrt x \)
substituting into equation of line (for normal) M1
e.g. \(y – 2 = – 4(x – 4)\)
\(y = – 4x + 18\) AG N0
[4 marks]
recognition that \(y = 0\) at A (M1)
e.g. \( – 4x + 18 = 0\)
\(x = \frac{{18}}{4}\) \(\left( { = \frac{9}{2}} \right)\) A1 N2
[2 marks]
splitting into two appropriate parts (areas and/or integrals) (M1)
correct expression for area of R A2 N3
e.g. area of R = \(\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( – 4x + 18){\rm{d}}x} \) , \(\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2\) (triangle)
Note: Award A1 if dx is missing.
[3 marks]
correct expression for the volume from \(x = 0\) to \(x = 4\) (A1)
e.g. \(V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x\) , \({\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x\) , \(\int_0^4 {\pi x{\rm{d}}x} \)
\(V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4\) A1
\(V = \pi \left( {\frac{1}{2} \times 16 – \frac{1}{2} \times 0} \right)\) (A1)
\(V = 8\pi \) A1
finding the volume from \(x = 4\) to \(x = 4.5\)
EITHER
recognizing a cone (M1)
e.g. \(V = \frac{1}{3}\pi {r^2}h\)
\(V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}\) (A1)
\( = \frac{{2\pi }}{3}\) A1
total volume is \(8\pi + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\) A1 N4
OR
\(V = \pi \int_4^{4.5} {{{( – 4x + 18)}^2}{\rm{d}}x} \) (M1)
\( = \int_4^{4.5} {\pi (16{x^2} – 144x + 324){\rm{d}}x} \)
\( = \pi \left[ {\frac{{16}}{3}{x^3} – 72{x^2} + 324x} \right]_4^{4.5}\) A1
\( = \frac{{2\pi }}{3}\) A1
total volume is \(8\pi + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\) A1 N4
[8 marks]
Question
Let \(f(x) = k{x^4}\) . The point \({\text{P}}(1{\text{, }}k)\) lies on the curve of f . At P, the normal to the curve is parallel to \(y = – \frac{1}{8}x\) . Find the value of k.
Answer/Explanation
Markscheme
gradient of tangent \(= 8\) (seen anywhere) (A1)
\(f'(x) = 4k{x^3}\) (seen anywhere) A1
recognizing the gradient of the tangent is the derivative (M1)
setting the derivative equal to 8 (A1)
e.g. \(4k{x^3} = 8\) , \(k{x^3} = 2\)
substituting \(x = 1\) (seen anywhere) (M1)
\(k = 2\) A1 N4
[6 marks]
Question
The following diagram shows part of the graph of the function \(f(x) = 2{x^2}\) .
The line T is the tangent to the graph of f at \(x = 1\) .
Show that the equation of T is \(y = 4x – 2\) .
Find the x-intercept of T .
The shaded region R is enclosed by the graph of f , the line T , and the x-axis.
(i) Write down an expression for the area of R .
(ii) Find the area of R .
Answer/Explanation
Markscheme
\(f(1) = 2\) (A1)
\(f'(x) = 4x\) A1
evidence of finding the gradient of f at \(x = 1\) M1
e.g. substituting \(x = 1\) into \(f'(x)\)
finding gradient of f at \(x = 1\) A1
e.g. \(f'(1) = 4\)
evidence of finding equation of the line M1
e.g. \(y – 2 = 4(x – 1)\) , \(2 = 4(1) + b\)
\(y = 4x – 2\) AG N0
[5 marks]
appropriate approach (M1)
e.g. \(4x – 2 = 0\)
\(x = \frac{1}{2}\) A1 N2
[2 marks]
(i) bottom limit \(x = 0\) (seen anywhere) (A1)
approach involving subtraction of integrals/areas (M1)
e.g. \(\int {f(x) – {\text{area of triangle}}} \) , \(\int {f – \int l } \)
correct expression A2 N4
e.g. \(\int_0^1 {2{x^2}{\rm{d}}x – } \int_{0.5}^1 {(4x – 2){\rm{d}}x} \) , \(\int_0^1 {f(x){\rm{d}}x – \frac{1}{2}} \) , \(\int_0^{0.5} {2{x^2}{\rm{d}}x} + \int_{0.5}^1 {(f(x) – (4x – 2)){\rm{d}}x} \)
(ii) METHOD 1 (using only integrals)
correct integration (A1)(A1)(A1)
\(\int {2{x^2}{\rm{d}}x} = \frac{{2{x^3}}}{3}\) , \(\int {(4x – 2){\rm{d}}x = } 2{x^2} – 2x\)
substitution of limits (M1)
e.g. \(\frac{1}{{12}} + \frac{2}{3} – 2 + 2 – \left( {\frac{1}{{12}} – \frac{1}{2} + 1} \right)\)
area = \(\frac{1}{6}\) A1 N4
METHOD 2 (using integral and triangle)
area of triangle = \(\frac{1}{2}\) (A1)
correct integration (A1)
\(\int {2{x^2}{\rm{d}}x = } \frac{{2{x^3}}}{3}\)
substitution of limits (M1)
e.g. \(\frac{2}{3}{(1)^3} – \frac{2}{3}{(0)^3}\) , \(\frac{2}{3} – 0\)
correct simplification (A1)
e.g. \(\frac{2}{3} – \frac{1}{2}\)
area = \(\frac{1}{6}\) A1 N4
[9 marks]
Question
Let \(f(x) = \frac{1}{4}{x^2} + 2\) . The line L is the tangent to the curve of f at (4, 6) .
Let \(g(x) = \frac{{90}}{{3x + 4}}\) , for \(2 \le x \le 12\) . The following diagram shows the graph of g .
Find the equation of L .
Find the area of the region enclosed by the curve of g , the x-axis, and the lines \(x = 2\) and \(x = 12\) . Give your answer in the form \(a\ln b\) , where \(a,b \in \mathbb{Z}\) .
The graph of g is reflected in the x-axis to give the graph of h . The area of the region enclosed by the lines L , \(x = 2\) , \(x = 12\) and the x-axis is 120 \(120{\text{ c}}{{\text{m}}^2}\) .
Find the area enclosed by the lines L , \(x = 2\) , \(x = 12\) and the graph of h .
Answer/Explanation
Markscheme
finding \(f'(x) = \frac{1}{2}x\) A1
attempt to find \(f'(4)\) (M1)
correct value \(f'(4) = 2\) A1
correct equation in any form A1 N2
e.g. \(y – 6 = 2(x – 4)\) , \(y = 2x – 2\)
[4 marks]
\({\rm{area}} = \int_2^{12} {\frac{{90}}{{3x + 4}}} {\rm{d}}x\)
correct integral A1A1
e.g. \(30\ln (3x + 4)\)
substituting limits and subtracting (M1)
e.g. \(30\ln (3 \times 12 + 4) – 30\ln (3 \times 2 + 4)\) , \(30\ln 40 – 30\ln 10\)
correct working (A1)
e.g. \(30(\ln 40 – \ln 10)\)
correct application of \(\ln b – \ln a\) (A1)
e.g. \(30\ln \frac{{40}}{{10}}\)
\({\rm{area}} = 30\ln 4\) A1 N4
[6 marks]
valid approach (M1)
e.g. sketch, area h = area g , 120 + their answer from (b)
\({\rm{area}} = 120 + 30\ln 4\) A2 N3
[3 marks]
Question
Let \(f(x) = {{\rm{e}}^{6x}}\) .
Write down \(f'(x)\) .
The tangent to the graph of f at the point \({\text{P}}(0{\text{, }}b)\) has gradient m .
(i) Show that \(m = 6\) .
(ii) Find b .
Hence, write down the equation of this tangent.
Answer/Explanation
Markscheme
\(f'(x) = 6{{\rm{e}}^{6x}}\) A1 N1
[1 mark]
(i) evidence of valid approach (M1)
e.g. \(f'(0)\) , \(6{{\rm{e}}^{6 \times 0}}\)
correct manipulation A1
e.g. \(6{{\rm{e}}^0}\) , \(6 \times 1\)
\(m = 6\) AG N0
(ii) evidence of finding \(f(0)\) (M1)
e.g. \(y = {{\rm{e}}^{6(0)}}\)
\(b = 1\) A1 N2
[4 marks]
\(y = 6x + 1\) A1 N1
[1 mark]
Question
Let \(f(x) = \sin x + \frac{1}{2}{x^2} – 2x\) , for \(0 \le x \le \pi \) .
Let \(g\) be a quadratic function such that \(g(0) = 5\) . The line \(x = 2\) is the axis of symmetry of the graph of \(g\) .
The function \(g\) can be expressed in the form \(g(x) = a{(x – h)^2} + 3\) .
Find \(f'(x)\) .
Find \(g(4)\) .
(i) Write down the value of \(h\) .
(ii) Find the value of \(a\) .
Find the value of \(x\) for which the tangent to the graph of \(f\) is parallel to the tangent to the graph of \(g\) .
Answer/Explanation
Markscheme
\(f'(x) = \cos x + x – 2\) A1A1A1 N3
Note: Award A1 for each term.
[3 marks]
recognizing \(g(0) = 5\) gives the point (\(0\), \(5\)) (R1)
recognize symmetry (M1)
eg vertex, sketch
\(g(4) = 5\) A1 N3
[3 marks]
(i) \(h = 2\) A1 N1
(ii) substituting into \(g(x) = a{(x – 2)^2} + 3\) (not the vertex) (M1)
eg \(5 = a{(0 – 2)^2} + 3\) , \(5 = a{(4 – 2)^2} + 3\)
working towards solution (A1)
eg \(5 = 4a + 3\) , \(4a = 2\)
\(a = \frac{1}{2}\) A1 N2
[4 marks]
\(g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5\)
correct derivative of \(g\) A1A1
eg \(2 \times \frac{1}{2}(x – 2)\) , \(x – 2\)
evidence of equating both derivatives (M1)
eg \(f’ = g’\)
correct equation (A1)
eg \(\cos x + x – 2 = x – 2\)
working towards a solution (A1)
eg \(\cos x = 0\) , combining like terms
\(x = \frac{\pi }{2}\) A1 N0
Note: Do not award final A1 if additional values are given.
[6 marks]
Question
Consider the functions \(f(x)\) , \(g(x)\) and \(h(x)\) . The following table gives some values associated with these functions.
The following diagram shows parts of the graphs of \(h\) and \(h”\) .
There is a point of inflexion on the graph of \(h\) at P, when \(x = 3\) .
Given that \(h(x) = f(x) \times g(x)\) ,
Write down the value of \(g(3)\) , of \(f'(3)\) , and of \(h”(2)\) .
Explain why P is a point of inflexion.
find the \(y\)-coordinate of P.
find the equation of the normal to the graph of \(h\) at P.
Answer/Explanation
Markscheme
\(g(3) = – 18\) , \(f'(3) = 1\) , \(h”(2) = – 6\) A1A1A1 N3
[3 marks]
\(h”(3) = 0\) (A1)
valid reasoning R1
eg \({h”}\) changes sign at \(x = 3\) , change in concavity of \(h\) at \(x = 3\)
so P is a point of inflexion AG N0
[2 marks]
writing \(h(3)\) as a product of \(f(3)\) and \(g(3)\) A1
eg \(f(3) \times g(3)\) , \(3 \times ( – 18)\)
\(h(3) = – 54\) A1 N1
[2 marks]
recognizing need to find derivative of \(h\) (R1)
eg \({h’}\) , \(h'(3)\)
attempt to use the product rule (do not accept \(h’ = f’ \times g’\) ) (M1)
eg \(h’ = fg’ + gf’\) , \(h'(3) = f(3) \times g'(3) + g(3) \times f'(3)\)
correct substitution (A1)
eg \(h'(3) = 3( – 3) + ( – 18) \times 1\)
\(h'(3) = – 27\) A1
attempt to find the gradient of the normal (M1)
eg \( – \frac{1}{m}\) , \( – \frac{1}{{27}}x\)
attempt to substitute their coordinates and their normal gradient into the equation of a line (M1)
eg \( – 54 = \frac{1}{{27}}(3) + b\) , \(0 = \frac{1}{{27}}(3) + b\) , \(y + 54 = 27(x – 3)\) , \(y – 54 = \frac{1}{{27}}(x + 3)\)
correct equation in any form A1 N4
eg \(y + 54 = \frac{1}{{27}}(x – 3)\) , \(y = \frac{1}{{27}}x – 54\frac{1}{9}\)
[7 marks]
Question
Let \(f(x) = {{\text{e}}^{2x}}\). The line \(L\) is the tangent to the curve of \(f\) at \((1,{\text{ }}{{\text{e}}^2})\).
Find the equation of \(L\) in the form \(y = ax + b\).
Answer/Explanation
Markscheme
recognising need to differentiate (seen anywhere) R1
eg \(f’,{\text{ }}2{{\text{e}}^{2x}}\)
attempt to find the gradient when \(x = 1\) (M1)
eg \(f'(1)\)
\(f'(1) = 2{{\text{e}}^2}\) (A1)
attempt to substitute coordinates (in any order) into equation of a straight line (M1)
eg \(y – {{\text{e}}^2} = 2{{\text{e}}^2}(x – 1),{\text{ }}{{\text{e}}^2} = 2{{\text{e}}^2}(1) + b\)
correct working (A1)
eg \(y – {{\text{e}}^2} = 2{{\text{e}}^2}x – 2{{\text{e}}^2},{\text{ }}b = – {{\text{e}}^2}\)
\(y = 2{{\text{e}}^2}x – {{\text{e}}^2}\) A1 N3
[6 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
Let \(f(x) = \sqrt {4x + 5} \), for \(x \geqslant – 1.25\).
Consider another function \(g\). Let R be a point on the graph of \(g\). The \(x\)-coordinate of R is 1. The equation of the tangent to the graph at R is \(y = 3x + 6\).
Find \(f'(1)\).
Write down \(g'(1)\).
Find \(g(1)\).
Let \(h(x) = f(x) \times g(x)\). Find the equation of the tangent to the graph of \(h\) at the point where \(x = 1\).
Answer/Explanation
Markscheme
choosing chain rule (M1)
eg\(\,\,\,\,\,\)\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}u = 4x + 5,{\text{ }}u’ = 4\)
correct derivative of \(f\) A2
eg\(\,\,\,\,\,\)\(\frac{1}{2}{(4x + 5)^{ – \frac{1}{2}}} \times 4,{\text{ }}f'(x) = \frac{2}{{\sqrt {4x + 5} }}\)
\(f'(1) = \frac{2}{3}\) A1 N2
[4 marks]
recognize that \(g'(x)\) is the gradient of the tangent (M1)
eg\(\,\,\,\,\,\)\(g'(x) = m\)
\(g'(1) = 3\) A1 N2
[2 marks]
recognize that R is on the tangent (M1)
eg\(\,\,\,\,\,\)\(g(1) = 3 \times 1 + 6\), sketch
\(g(1) = 9\) A1 N2
[2 marks]
\(f(1) = \sqrt {4 + 5} {\text{ }}( = 3)\) (seen anywhere) A1
\(h(1) = 3 \times 9{\text{ }}( = 27)\) (seen anywhere) A1
choosing product rule to find \(h'(x)\) (M1)
eg\(\,\,\,\,\,\)\(uv’ + u’v\)
correct substitution to find \(h'(1)\) (A1)
eg\(\,\,\,\,\,\)\(f(1) \times g'(1) + f'(1) \times g(1)\)
\(h'(1) = 3 \times 3 + \frac{2}{3} \times 9{\text{ }}( = 15)\) A1
EITHER
attempt to substitute coordinates (in any order) into the equation of a straight line (M1)
eg\(\,\,\,\,\,\)\(y – 27 = h'(1)(x – 1),{\text{ }}y – 1 = 15(x – 27)\)
\(y – 27 = 15(x – 1)\) A1 N2
OR
attempt to substitute coordinates (in any order) to find the \(y\)-intercept (M1)
eg\(\,\,\,\,\,\)\(27 = 15 \times 1 + b,{\text{ }}1 = 15 \times 27 + b\)
\(y = 15x + 12\) A1 N2
[7 marks]
Question
The following diagram shows the graph of \(f(x) = 2x\sqrt {{a^2} – {x^2}} \), for \( – 1 \leqslant x \leqslant a\), where \(a > 1\).
The line \(L\) is the tangent to the graph of \(f\) at the origin, O. The point \({\text{P}}(a,{\text{ }}b)\) lies on \(L\).
The point \({\text{Q}}(a,{\text{ }}0)\) lies on the graph of \(f\). Let \(R\) be the region enclosed by the graph of \(f\) and the \(x\)-axis. This information is shown in the following diagram.
Let \({A_R}\) be the area of the region \(R\).
(i) Given that \(f'(x) = \frac{{2{a^2} – 4{x^2}}}{{\sqrt {{a^2} – {x^2}} }}\), for \( – 1 \leqslant x < a\), find the equation of \(L\).
(ii) Hence or otherwise, find an expression for \(b\) in terms of \(a\).
Show that \({A_R} = \frac{2}{3}{a^3}\).
Let \({A_T}\) be the area of the triangle OPQ. Given that \({A_T} = k{A_R}\), find the value of \(k\).
Answer/Explanation
Markscheme
(i) recognizing the need to find the gradient when \(x = 0\) (seen anywhere) R1
eg\(\,\,\,\,\,\)\(f'(0)\)
correct substitution (A1)
\(f'(0) = \frac{{2{a^2} – 4(0)}}{{\sqrt {{a^2} – 0} }}\)
\(f'(0) = 2a\) (A1)
correct equation with gradient 2\(a\) (do not accept equations of the form \(L = 2ax\)) A1 N3
eg\(\,\,\,\,\,\)\(y = 2ax,{\text{ }}y – b = 2a(x – a),{\text{ }}y = 2ax – 2{a^2} + b\)
(ii) METHOD 1
attempt to substitute \(x = a\) into their equation of \(L\) (M1)
eg\(\,\,\,\,\,\)\(y = 2a \times a\)
\(b = 2{a^2}\) A1 N2
METHOD 2
equating gradients (M1)
eg\(\,\,\,\,\,\)\(\frac{b}{a} = 2a\)
\(b = 2{a^2}\) A1 N2
[6 marks]
METHOD 1
recognizing that area \( = \int_0^a {f(x){\text{d}}x} \) (seen anywhere) R1
valid approach using substitution or inspection (M1)
eg\(\,\,\,\,\,\)\(\int {2x\sqrt u {\text{d}}x,{\text{ }}u = {a^2} – {x^2},{\text{ d}}u = – 2x{\text{d}}x,{\text{ }}\frac{2}{3}{{({a^2} – {x^2})}^{\frac{3}{2}}}} \)
correct working (A1)
eg\(\,\,\,\,\,\)\(\int {2x\sqrt {{a^2} – {x^2}} {\text{d}}x = \int { – \sqrt u {\text{d}}u} } \)
\(\int { – \sqrt u {\text{d}}u = – \frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \) (A1)
\(\int {f(x){\text{d}}x = – \frac{2}{3}{{({a^2} – {x^2})}^{\frac{3}{2}}} + c} \) (A1)
substituting limits and subtracting A1
eg\(\,\,\,\,\,\)\({A_R} = – \frac{2}{3}{({a^2} – {a^2})^{\frac{3}{2}}} + \frac{2}{3}{({a^2} – 0)^{\frac{3}{2}}},{\text{ }}\frac{2}{3}{({a^2})^{\frac{3}{2}}}\)
\({A_R} = \frac{2}{3}{a^3}\) AG N0
METHOD 2
recognizing that area \( = \int_0^a {f(x){\text{d}}x} \) (seen anywhere) R1
valid approach using substitution or inspection (M1)
eg\(\,\,\,\,\,\)\(\int {2x\sqrt u {\text{d}}x,{\text{ }}u = {a^2} – {x^2},{\text{ d}}u = – 2x{\text{d}}x,{\text{ }}\frac{2}{3}{{({a^2} – {x^2})}^{\frac{3}{2}}}} \)
correct working (A1)
eg\(\,\,\,\,\,\)\(\int {2x\sqrt {{a^2} – {x^2}} {\text{d}}x = \int { – \sqrt u {\text{d}}u} } \)
\(\int { – \sqrt u {\text{d}}u = – \frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \) (A1)
new limits for u (even if integration is incorrect) (A1)
eg\(\,\,\,\,\,\)\(u = 0{\text{ and }}u = {a^2},{\text{ }}\int_0^{{a^2}} {{u^{\frac{1}{2}}}{\text{d}}u,{\text{ }}\left[ { – \frac{2}{3}{u^{\frac{3}{2}}}} \right]} _{{a^2}}^0\)
substituting limits and subtracting A1
eg\(\,\,\,\,\,\)\({A_R} = – \left( {0 – \frac{2}{3}{a^3}} \right),{\text{ }}\frac{2}{3}{({a^2})^{\frac{3}{2}}}\)
\({A_R} = \frac{2}{3}{a^3}\) AG N0
[6 marks]
METHOD 1
valid approach to find area of triangle (M1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}({\text{OQ)(PQ), }}\frac{1}{2}ab\)
correct substitution into formula for \({A_T}\) (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\({A_T} = \frac{1}{2} \times a \times 2{a^2},{\text{ }}{a^3}\)
valid attempt to find \(k\) (must be in terms of \(a\)) (M1)
eg\(\,\,\,\,\,\)\({a^3} = k\frac{2}{3}{a^3},{\text{ }}k = \frac{{{a^3}}}{{\frac{2}{3}{a^3}}}\)
\(k = \frac{3}{2}\) A1 N2
METHOD 2
valid approach to find area of triangle (M1)
eg\(\,\,\,\,\,\)\(\int_0^a {(2ax){\text{d}}x} \)
correct working (A1)
eg\(\,\,\,\,\,\)\([a{x^2}]_0^a,{\text{ }}{a^3}\)
valid attempt to find \(k\) (must be in terms of \(a\)) (M1)
eg\(\,\,\,\,\,\)\({a^3} = k\frac{2}{3}{a^3},{\text{ }}k = \frac{{{a^3}}}{{\frac{2}{3}{a^3}}}\)
\(k = \frac{3}{2}\) A1 N2
[4 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 quadratic function \(f\) can be written in the form \(f(x) = a(x – p)(x – 3)\). The graph of \(f\) has axis of symmetry \(x = 2.5\) and \(y\)-intercept at \((0,{\text{ }} – 6)\)
Find the value of \(p\).
Find the value of \(a\).
The line \(y = kx – 5\) is a tangent to the curve of \(f\). Find the values of \(k\).
Answer/Explanation
Markscheme
METHOD 1 (using x-intercept)
determining that 3 is an \(x\)-intercept (M1)
eg\(\,\,\,\,\,\)\(x – 3 = 0\), 
valid approach (M1)
eg\(\,\,\,\,\,\)\(3 – 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5\)
\(p = 2\) A1 N2
METHOD 2 (expanding f (x))
correct expansion (accept absence of \(a\)) (A1)
eg\(\,\,\,\,\,\)\(a{x^2} – a(3 + p)x + 3ap,{\text{ }}{x^2} – (3 + p)x + 3p\)
valid approach involving equation of axis of symmetry (M1)
eg\(\,\,\,\,\,\)\(\frac{{ – b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}\)
\(p = 2\) A1 N2
METHOD 3 (using derivative)
correct derivative (accept absence of \(a\)) (A1)
eg\(\,\,\,\,\,\)\(a(2x – 3 – p),{\text{ }}2x – 3 – p\)
valid approach (M1)
eg\(\,\,\,\,\,\)\(f’(2.5) = 0\)
\(p = 2\) A1 N2
[3 marks]
attempt to substitute \((0,{\text{ }} – 6)\) (M1)
eg\(\,\,\,\,\,\)\( – 6 = a(0 – 2)(0 – 3),{\text{ }}0 = a( – 8)( – 9),{\text{ }}a{(0)^2} – 5a(0) + 6a = – 6\)
correct working (A1)
eg\(\,\,\,\,\,\)\( – 6 = 6a\)
\(a = – 1\) A1 N2
[3 marks]
METHOD 1 (using discriminant)
recognizing tangent intersects curve once (M1)
recognizing one solution when discriminant = 0 M1
attempt to set up equation (M1)
eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6\)
rearranging their equation to equal zero (M1)
eg\(\,\,\,\,\,\)\({x^2} – 5x + kx + 1 = 0\)
correct discriminant (if seen explicitly, not just in quadratic formula) A1
eg\(\,\,\,\,\,\)\({(k – 5)^2} – 4,{\text{ }}25 – 10k + {k^2} – 4\)
correct working (A1)
eg\(\,\,\,\,\,\)\(k – 5 = \pm 2,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)
\(k = 3,{\text{ }}7\) A1A1 N0
METHOD 2 (using derivatives)
attempt to set up equation (M1)
eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6\)
recognizing derivative/slope are equal (M1)
eg\(\,\,\,\,\,\)\(f’ = {m_T},{\text{ }}f’ = k\)
correct derivative of \(f\) (A1)
eg\(\,\,\,\,\,\)\( – 2x + 5\)
attempt to set up equation in terms of either \(x\) or \(k\) M1
eg\(\,\,\,\,\,\)\(( – 2x + 5)x – 5 = – {x^2} + 5x – 6,{\text{ }}k\left( {\frac{{5 – k}}{2}} \right) – 5 = – {\left( {\frac{{5 – k}}{2}} \right)^2} + 5\left( {\frac{{5 – k}}{2}} \right) – 6\)
rearranging their equation to equal zero (M1)
eg\(\,\,\,\,\,\)\({x^2} – 1 = 0,{\text{ }}{k^2} – 10k + 21 = 0\)
correct working (A1)
eg\(\,\,\,\,\,\)\(x = \pm 1,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)
\(k = 3,{\text{ }}7\) A1A1 N0
[8 marks]
Question
Let \(f(x) = {x^2}\). The following diagram shows part of the graph of \(f\).
The line \(L\) is the tangent to the graph of \(f\) at the point \({\text{A}}( – k,{\text{ }}{k^2})\), and intersects the \(x\)-axis at point B. The point C is \(( – k,{\text{ }}0)\).
The region \(R\) is enclosed by \(L\), the graph of \(f\), and the \(x\)-axis. This is shown in the following diagram.
Write down \(f'(x)\).
Find the gradient of \(L\).
Show that the \(x\)-coordinate of B is \( – \frac{k}{2}\).
Find the area of triangle ABC, giving your answer in terms of \(k\).
Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).
Answer/Explanation
Markscheme
\(f'(x) = 2x\) A1 N1
[1 mark]
attempt to substitute \(x = – k\) into their derivative (M1)
gradient of \(L\) is \( – 2k\) A1 N2
[2 marks]
METHOD 1
attempt to substitute coordinates of A and their gradient into equation of a line (M1)
eg\(\,\,\,\,\,\)\({k^2} = – 2k( – k) + b\)
correct equation of \(L\) in any form (A1)
eg\(\,\,\,\,\,\)\(y – {k^2} = – 2k(x + k),{\text{ }}y = – 2kx – {k^2}\)
valid approach (M1)
eg\(\,\,\,\,\,\)\(y = 0\)
correct substitution into \(L\) equation A1
eg\(\,\,\,\,\,\)\( – {k^2} = – 2kx – 2{k^2},{\text{ }}0 = – 2kx – {k^2}\)
correct working A1
eg\(\,\,\,\,\,\)\(2kx = – {k^2}\)
\(x = – \frac{k}{2}\) AG N0
METHOD 2
valid approach (M1)
eg\(\,\,\,\,\,\)\({\text{gradient}} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}},{\text{ }} – 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}\)
recognizing \(y = 0\) at B (A1)
attempt to substitute coordinates of A and B into slope formula (M1)
eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}},{\text{ }}\frac{{ – {k^2}}}{{x + k}}\)
correct equation A1
eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}} = – 2k,{\text{ }}\frac{{ – {k^2}}}{{x + k}} = – 2k,{\text{ }} – {k^2} = – 2k(x + k)\)
correct working A1
eg\(\,\,\,\,\,\)\(2kx = – {k^2}\)
\(x = – \frac{k}{2}\) AG N0
[5 marks]
valid approach to find area of triangle (M1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)\)
area of \({\text{ABC}} = \frac{{{k^3}}}{4}\) A1 N2
[2 marks]
METHOD 1 (\(\int {f – {\text{triangle}}} \))
valid approach to find area from \( – k\) to 0 (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ – k} f } \)
correct integration (seen anywhere, even if M0 awarded) A1
eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ – k}^0\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(0 – \frac{{{{( – k)}^3}}}{3}\), area from \( – k\) to 0 is \(\frac{{{k^3}}}{3}\)
Note: Award M0 for substituting into original or differentiated function.
attempt to find area of \(R\) (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {f(x){\text{d}}x – {\text{ triangle}}} \)
correct working for \(R\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}\)
correct substitution into \({\text{triangle}} = pR\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)
\(p = 3\) A1 N2
METHOD 2 (\(\int {(f – L)} \))
valid approach to find area of \(R\) (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^{ – \frac{k}{2}} {{x^2} – ( – 2kx – {k^2}){\text{d}}x + \int_{ – \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ – k}^{ – \frac{k}{2}} {(f – L) + \int_{ – \frac{k}{2}}^0 f } } } \)
correct integration (seen anywhere, even if M0 awarded) A2
eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ – k}^{ – \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ – \frac{k}{2}}^0\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(\left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { – \frac{k}{2}} \right)}^2} + {k^2}\left( { – \frac{k}{2}} \right)} \right) – \left( {\frac{{{{( – k)}^3}}}{3} + k{{( – k)}^2} + {k^2}( – k)} \right) + (0) – \left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3}} \right)\)
Note: Award M0 for substituting into original or differentiated function.
correct working for \(R\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} – \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} – \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} – {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}\)
correct substitution into \({\text{triangle}} = pR\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)
\(p = 3\) A1 N2
[7 marks]
Question
Let \(f(x) = {x^2} – x\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\).
The graph of \(f\) crosses the \(x\)-axis at the origin and at the point \({\text{P}}(1,{\text{ }}0)\).
The line L is the normal to the graph of f at P.
The line \(L\) intersects the graph of \(f\) at another point Q, as shown in the following diagram.
Show that \(f’(1) = 1\).
Find the equation of \(L\) in the form \(y = ax + b\).
Find the \(x\)-coordinate of Q.
Find the area of the region enclosed by the graph of \(f\) and the line \(L\).
Answer/Explanation
Markscheme
\(f’(x) = 2x – 1\) A1A1
correct substitution A1
eg\(\,\,\,\,\,\)\(2(1) – 1,{\text{ }}2 – 1\)
\(f’(1) = 1\) AG N0
[3 marks]
correct approach to find the gradient of the normal (A1)
eg\(\,\,\,\,\,\)\(\frac{{ – 1}}{{f'(1)}},{\text{ }}{m_1}{m_2} = – 1,{\text{ slope}} = – 1\)
attempt to substitute correct normal gradient and coordinates into equation of a line (M1)
eg\(\,\,\,\,\,\)\(y – 0 = – 1(x – 1),{\text{ }}0 = – 1 + b,{\text{ }}b = 1,{\text{ }}L = – x + 1\)
\(y = – x + 1\) A1 N2
[3 marks]
equating expressions (M1)
eg\(\,\,\,\,\,\)\(f(x) = L,{\text{ }} – x + 1 = {x^2} – x\)
correct working (must involve combining terms) (A1)
eg\(\,\,\,\,\,\)\({x^2} – 1 = 0,{\text{ }}{x^2} = 1,{\text{ }}x = 1\)
\(x = – 1\,\,\,\,\,\left( {{\text{accept }}Q( – 1,{\text{ }}2)} \right)\) A2 N3
[4 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(\int {L – f,{\text{ }}\int_{ – 1}^1 {(1 – {x^2}){\text{d}}x} } \), splitting area into triangles and integrals
correct integration (A1)(A1)
eg\(\,\,\,\,\,\)\(\left[ {x – \frac{{{x^3}}}{3}} \right]_{ – 1}^1,{\text{ }} – \frac{{{x^3}}}{3} – \frac{{{x^2}}}{2} + \frac{{{x^2}}}{2} + x\)
substituting their limits into their integrated function and subtracting (in any order) (M1)
eg\(\,\,\,\,\,\)\(1 – \frac{1}{3} – \left( { – 1 – \frac{{ – 1}}{3}} \right)\)
Note: Award M0 for substituting into original or differentiated function.
area \( = \frac{4}{3}\) A2 N3
[6 marks]
Question
Consider a function \(f\). The line L1 with equation \(y = 3x + 1\) is a tangent to the graph of \(f\) when \(x = 2\)
Let \(g\left( x \right) = f\left( {{x^2} + 1} \right)\) and P be the point on the graph of \(g\) where \(x = 1\).
Write down \(f’\left( 2 \right)\).
Find \(f\left( 2 \right)\).
Show that the graph of g has a gradient of 6 at P.
Let L2 be the tangent to the graph of g at P. L1 intersects L2 at the point Q.
Find the y-coordinate of Q.
Answer/Explanation
Markscheme
recognize that \(f’\left( x \right)\) is the gradient of the tangent at \(x\) (M1)
eg \(f’\left( x \right) = m\)
\(f’\left( 2 \right) = 3\) (accept m = 3) A1 N2
[2 marks]
recognize that \(f\left( 2 \right) = y\left( 2 \right)\) (M1)
eg \(f\left( 2 \right) = 3 \times 2 + 1\)
\(f\left( 2 \right) = 7\) A1 N2
[2 marks]
recognize that the gradient of the graph of g is \(g’\left( x \right)\) (M1)
choosing chain rule to find \(g’\left( x \right)\) (M1)
eg \(\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u’ = 2x\)
\(g’\left( x \right) = f’\left( {{x^2} + 1} \right) \times 2x\) A2
\(g’\left( 1 \right) = 3 \times 2\) A1
\(g’\left( 1 \right) = 6\) AG N0
[5 marks]
at Q, L1 = L2 (seen anywhere) (M1)
recognize that the gradient of L2 is g’(1) (seen anywhere) (M1)
eg m = 6
finding g (1) (seen anywhere) (A1)
eg \(g\left( 1 \right) = f\left( 2 \right),\,\,g\left( 1 \right) = 7\)
attempt to substitute gradient and/or coordinates into equation of a straight line M1
eg \(y – g\left( 1 \right) = 6\left( {x – 1} \right),\,\,y – 1 = g’\left( 1 \right)\left( {x – 7} \right),\,\,7 = 6\left( 1 \right) + {\text{b}}\)
correct equation for L2
eg \(y – 7 = 6\left( {x – 1} \right),\,\,y = 6x + 1\) A1
correct working to find Q (A1)
eg same y-intercept, \(3x = 0\)
\(y = 1\) A1 N2
[7 marks]