IB Math Analysis & Approaches Question bank-Topic SL 5.4- Tangents and normals, and their equations SL Paper 1

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 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 \(f'(x)\) .

[3]
a.

Find the x-coordinate of M.

[4]
b.

Find the x-coordinate of N.

[3]
c.

The line L is the tangent to the curve of f at \((3{\text{, }}12)\). Find the equation of L in the form \(y = ax + b\) .

[4]
d.
Answer/Explanation

Markscheme

\(f'(x) = {x^2} + 4x – 5\)     A1A1A1     N3

[3 marks]

a.

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]

b.

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]

c.

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]

d.

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 \(y = – 4x + 18\) .

[4]
a.

Point A is the x-intercept of L . Find the x-coordinate of A.

[2]
b.

Find an expression for the area of R .

[3]
c.

The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .

[8]
d.
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]

a.

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]

b.

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]

c.

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]

d.

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 \(y = 4x – 2\) .

[5]
a.

Find the x-intercept of T .

[2]
b.

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 .

[9]
c(i) and (ii).
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]

a.

appropriate approach     (M1)

e.g. \(4x – 2 = 0\)

\(x = \frac{1}{2}\)     A1     N2

[2 marks]

b.

(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]

c(i) and (ii).

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 .

[4]
a.

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 \(a,b \in \mathbb{Z}\) .

[6]
b.

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 .

[3]
c.
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]

a.

\({\rm{area}} = \int_2^{12} {\frac{{90}}{{3x + 4}}} {\rm{d}}x\)

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]

b.

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]

c.

Question

Let \(f(x) = {{\rm{e}}^{6x}}\) .

Write down \(f'(x)\) .

[1]
a.

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 .

[4]
b(i) and (ii).

Hence, write down the equation of this tangent.

[1]
c.
Answer/Explanation

Markscheme

\(f'(x) = 6{{\rm{e}}^{6x}}\)     A1     N1

[1 mark]

a.

(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]

b(i) and (ii).

\(y = 6x + 1\)     A1     N1

[1 mark]

c.

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 \(f'(x)\) .

[3]
a.

Find \(g(4)\) .

[3]
b.

(i)     Write down the value of \(h\) .

(ii)     Find the value of \(a\) .

[4]
c.

Find the value of \(x\) for which the tangent to the graph of \(f\) is parallel to the tangent to the graph of \(g\) .

[6]
d.
Answer/Explanation

Markscheme

\(f'(x) = \cos x + x – 2\)     A1A1A1     N3

Note: Award A1 for each term.

[3 marks]

a.

recognizing \(g(0) = 5\) gives the point (\(0\), \(5\))     (R1)

recognize symmetry     (M1)

eg vertex, sketch

\(g(4) = 5\)     A1     N3

[3 marks]

b.

(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]

c.

\(g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5\)

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]

d.

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 \(h”(2)\) .

[3]
a.

Explain why P is a point of inflexion.

[2]
b.

find the \(y\)-coordinate of P.

[2]
c.

find the equation of the normal to the graph of \(h\) at P.

[7]
d.
Answer/Explanation

Markscheme

\(g(3) = – 18\) , \(f'(3) = 1\) , \(h”(2) = – 6\)     A1A1A1     N3

[3 marks]

a.

\(h”(3) = 0\)     (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]

b.

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]

c.

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]

d.

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 \(f”(x)\).

[2]
a.

The graph of \(f\) has a point of inflexion when \(x = 1\).

Show that \(k = 3\).

[3]
b.

Find \(f'( – 2)\).

[2]
c.

Find the equation of the tangent to the curve of \(f\) at \(( – 2,{\text{ }}1)\), giving your answer in the form \(y = ax + b\).

[4]
d.

Given that \(f'( – 1) = 0\), explain why the graph of \(f\) has a local maximum when \(x =  – 1\).

[3]
e.
Answer/Explanation

Markscheme

\(f”(x) = 6x – 2k\)     A1A1     N2

[2 marks]

a.

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]

b.

correct substitution into \(f'(x)\)     (A1)

eg\(\;\;\;3{( – 2)^2} – 6( – 2) – 9\)

\(f'( – 2) = 15\)     A1     N2

[2 marks]

c.

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]

d.

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]

e.

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 \(x = 5\).

[2]
a.

Find the set of values of \(x\) for which the graph of \(f\) is concave down.

[2]
b.

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 \(f(6)\).

[5]
c.

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 \(x = 6\).

[6]
d.
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]

a.

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]

b.

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]

c.

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]

d.

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 \(f'(1)\).

[4]
a.

Write down \(g'(1)\).

[2]
b.

Find \(g(1)\).

[2]
c.

Let \(h(x) = f(x) \times g(x)\). Find the equation of the tangent to the graph of \(h\) at the point where \(x = 1\).

[7]
d.
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]

a.

recognize that \(g'(x)\) is the gradient of the tangent     (M1)

eg\(\,\,\,\,\,\)\(g'(x) = m\)

\(g'(1) = 3\)    A1     N2

[2 marks]

b.

recognize that R is on the tangent     (M1)

eg\(\,\,\,\,\,\)\(g(1) = 3 \times 1 + 6\), sketch

\(g(1) = 9\)    A1     N2

[2 marks]

c.

\(f(1) = \sqrt {4 + 5} {\text{ }}( = 3)\) (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]

d.

Question

The following diagram shows the graph of \(f(x) = 2x\sqrt {{a^2} – {x^2}} \), for \( – 1 \leqslant x \leqslant a\), where \(a > 1\).

M16/5/MATME/SP1/ENG/TZ2/10

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.

M16/5/MATME/SP1/ENG/TZ2/10.b+c

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 \(a\).

[6]
a.

Show that \({A_R} = \frac{2}{3}{a^3}\).

[6]
b.

Let \({A_T}\) be the area of the triangle OPQ. Given that \({A_T} = k{A_R}\), find the value of \(k\).

[4]
c.
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]

a.

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]

b.

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]

c.

Question

The following diagram shows the graph of \(f’\), the derivative of \(f\).

M17/5/MATME/SP1/ENG/TZ1/06

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.

[1]
a.i.

Find the equation of the normal to the curve of \(f\) at P.

[3]
a.ii.

Determine the concavity of the graph of \(f\) when \(4 < x < 5\) and justify your answer.

[2]
b.
Answer/Explanation

Markscheme

\( – 2\)     A1     N1

[1 mark]

a.i.

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]

a.ii.

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)

M17/5/MATME/SP1/ENG/TZ1/06.b/M

Note:     The reason given must refer to a specific function/graph. Referring to “the graph” or “it” is not sufficient.

[2 marks]

b.

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 \(p\).

[3]
a.

Find the value of \(a\).

[3]
b.

The line \(y = kx – 5\) is a tangent to the curve of \(f\). Find the values of \(k\).

[8]
c.
Answer/Explanation

Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an \(x\)-intercept     (M1)

eg\(\,\,\,\,\,\)\(x – 3 = 0\), M17/5/MATME/SP1/ENG/TZ1/09.a/M

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]

a.

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]

b.

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]

c.

Question

Let \(f(x) = {x^2}\). The following diagram shows part of the graph of \(f\).

M17/5/MATME/SP1/ENG/TZ2/10

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.

M17/5/MATME/SP1/ENG/TZ2/10.d

Write down \(f'(x)\).

[1]
a.i.

Find the gradient of \(L\).

[2]
a.ii.

Show that the \(x\)-coordinate of B is \( – \frac{k}{2}\).

[5]
b.

Find the area of triangle ABC, giving your answer in terms of \(k\).

[2]
c.

Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).

[7]
d.
Answer/Explanation

Markscheme

\(f'(x) = 2x\)     A1     N1

[1 mark]

a.i.

attempt to substitute \(x =  – k\) into their derivative     (M1)

gradient of \(L\) is \( – 2k\)     A1     N2

[2 marks]

a.ii.

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]

b.

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]

c.

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]

d.

Question

Let \(f(x) = {x^2} – x\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\).

N17/5/MATME/SP1/ENG/TZ0/08

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.

N17/5/MATME/SP1/ENG/TZ0/08.c.d

Show that \(f’(1) = 1\).

[3]
a.

Find the equation of \(L\) in the form \(y = ax + b\).

[3]
b.

Find the \(x\)-coordinate of Q.

[4]
c.

Find the area of the region enclosed by the graph of \(f\) and the line \(L\).

[6]
d.
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]

a.

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]

b.

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]

c.

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]

d.

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 \(f’\left( 2 \right)\).

[2]
a.i.

Find \(f\left( 2 \right)\).

[2]
a.ii.

Show that the graph of g has a gradient of 6 at P.

[5]
b.

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.

[7]
c.
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]

a.i.

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]

a.ii.

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]

b.

 at Q, L1L2 (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]

c.

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