IB Math Analysis & Approaches Question bank-Topic: SL 5.6 The chain rule for composite functions SL Paper 1

Question

Let  \(f:x \mapsto {\sin ^3}x\) .

(i) Write down the range of the function f .

(ii) Consider \(f(x) = 1\) , \(0 \le x \le 2\pi \) . Write down the number of solutions to this equation. Justify your answer.

[5]
a.

Find \(f'(x)\) , giving your answer in the form \(a{\sin ^p}x{\cos ^q}x\) where \(a{\text{, }}p{\text{, }}q \in \mathbb{Z}\) .

[2]
b.

Let \(g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}\) for \(0 \le x \le \frac{\pi }{2}\) . Find the volume generated when the curve of g is revolved through \(2\pi \) about the x-axis.

[7]
c.
Answer/Explanation

Markscheme

(i) range of f is \([ – 1{\text{, }}1]\) , \(( – 1 \le f(x) \le 1)\)     A2     N2

(ii) \({\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1\)     A1

justification for one solution on \([0{\text{, }}2\pi ]\)    R1

e.g. \(x = \frac{\pi }{2}\) , unit circle, sketch of \(\sin x\)

1 solution (seen anywhere)     A1     N1

[5 marks]

a.

\(f'(x) = 3{\sin ^2}x\cos x\)     A2     N2

[2 marks]

b.

using \(V = \int_a^b {\pi {y^2}{\rm{d}}x} \)     (M1)

\(V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x\)     (A1)

\( = \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x} \)     A1

\(V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}\) \(\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) – {{\sin }^3}0} \right)} \right)\)     A2

evidence of using \(\sin \frac{\pi }{2} = 1\) and \(\sin 0 = 0\)     (A1)

e.g. \(\pi \left( {1 – 0} \right)\)

\(V = \pi \)     A1     N1

[7 marks]

c.

Question

Let \(f(x) = {{\rm{e}}^{ – 3x}}\) and \(g(x) = \sin \left( {x – \frac{\pi }{3}} \right)\) .

Write down

(i)     \(f'(x)\) ;

(ii)    \(g'(x)\) .

[2]
a.

Let \(h(x) = {{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right)\) . Find the exact value of \(h’\left( {\frac{\pi }{3}} \right)\) .

[4]
b.
Answer/Explanation

Markscheme

(i) \( – 3{{\rm{e}}^{ – 3x}}\)     A1     N1

(ii) \(\cos \left( {x – \frac{\pi }{3}} \right)\)     A1     N1

[4 marks]

a.

evidence of choosing product rule     (M1)

e.g. \(uv’ + vu’\)

correct expression     A1

e.g. \( – 3{{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3x}}\cos \left( {x – \frac{\pi }{3}} \right)\)

complete correct substitution of \(x = \frac{\pi }{3}\)     (A1)

e.g. \( – 3{{\rm{e}}^{ – 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right)\)        

\(h’\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ – \pi }}\)     A1     N3

[4 marks]

b.

Question

Let \(g(x) = 2x\sin x\) .

Find \(g'(x)\) .

[4]
a.

Find the gradient of the graph of g at \(x = \pi \) .

[3]
b.
Answer/Explanation

Markscheme

evidence of choosing the product rule     (M1)

e.g. \(uv’ + vu’\)

correct derivatives \(\cos x\) , 2     (A1)(A1)

\(g'(x) = 2x\cos x + 2\sin x\)     A1     N4

[4 marks]

a.

attempt to substitute into gradient function     (M1)

e.g. \(g'(\pi )\)

correct substitution     (A1)

e.g. \(2\pi \cos \pi  + 2\sin \pi \)

\({\text{gradient}} = – 2\pi \)     A1     N2

[3 marks]

b.

Question

The following diagram shows the graph of \(f(x) = a\sin (b(x – c)) + d\) , for \(2 \le x \le 10\) .


There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of

(i)     a ;

(ii)    c ;

(iii)   d .

[3]
a(i), (ii) and (iii).

Show that \(b = \frac{\pi }{4}\) .

[2]
b.

Find \(f'(x)\) .

[3]
c.

At a point R, the gradient is \( – 2\pi \) . Find the x-coordinate of R.

[6]
d.
Answer/Explanation

Markscheme

(i) \(a = 8\)     A1     N1

(ii) \(c = 2\)     A1     N1

(iii) \(d = 4\)     A1     N1

[3 marks]

a(i), (ii) and (iii).

METHOD 1

recognizing that period \( = 8\)     (A1)

correct working     A1

e.g. \(8 = \frac{{2\pi }}{b}\) , \(b = \frac{{2\pi }}{8}\)

\(b = \frac{\pi }{4}\)     AG     N0

METHOD 2

attempt to substitute     M1

e.g. \(12 = 8\sin (b(4 – 2)) + 4\)

correct working     A1

e.g. \(\sin 2b = 1\)

\(b = \frac{\pi }{4}\)     AG     N0

[2 marks]

b.

evidence of attempt to differentiate or choosing chain rule     (M1)

e.g. \(\cos \frac{\pi }{4}(x – 2)\) , \(\frac{\pi }{4} \times 8\)

\(f'(x) = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)\) (accept \(2\pi \cos \frac{\pi }{4}(x – 2)\) )     A2     N3

[3 marks]

c.

recognizing that gradient is \(f'(x)\)     (M1)

e.g. \(f'(x) = m\)

correct equation     A1

e.g. \( – 2\pi  = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)\) , \( – 1 = \cos \left( {\frac{\pi }{4}(x – 2)} \right)\)

correct working     (A1)

e.g. \({\cos ^{ – 1}}( – 1) = \frac{\pi }{4}(x – 2)\)

using \({\cos ^{ – 1}}( – 1) = \pi \) (seen anywhere)     (A1)

e.g. \(\pi  = \frac{\pi }{4}(x – 2)\)

simplifying     (A1)

e.g. \(4 = (x – 2)\)

\(x = 6\)     A1     N4

[6 marks]

d.

Question

Let \(f(x) = \frac{{6x}}{{x + 1}}\) , for \(x > 0\) .

Find \(f'(x)\) .

[5]
a.

Let \(g(x) = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\) , for \(x > 0\) .

Show that \(g'(x) = \frac{1}{{x(x + 1)}}\) .

[4]
b.

Let \(h(x) = \frac{1}{{x(x + 1)}}\) . The area enclosed by the graph of h , the x-axis and the lines \(x = \frac{1}{5}\)  and \(x = k\) is \(\ln 4\) . Given that \(k > \frac{1}{5}\) , find the value of k .

[7]
c.
Answer/Explanation

Markscheme

METHOD 1

evidence of choosing quotient rule     (M1)

e.g. \(\frac{{u’v – uv’}}{{{v^2}}}\)

evidence of correct differentiation (must be seen in quotient rule)     (A1)(A1)

e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}(x + 1) = 1\)

correct substitution into quotient rule     A1

e.g. \(\frac{{(x + 1)6 – 6x}}{{{{(x + 1)}^2}}}\) , \(\frac{{6x + 6 – 6x}}{{{{(x + 1)}^2}}}\)

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)    A1     N4

[5 marks]

METHOD 2

evidence of choosing product rule     (M1)

e.g. \(6x{(x + 1)^{ – 1}}\) , \(uv’ + vu’\)

evidence of correct differentiation (must be seen in product rule)     (A1)(A1)

e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}{(x + 1)^{ – 1}} = – 1{(x + 1)^{ – 2}} \times 1\)

correct working     A1

e.g. \(6x \times – {(x + 1)^{ – 2}} + {(x + 1)^{ – 1}} \times 6\) , \(\frac{{ – 6x + 6(x + 1)}}{{{{(x + 1)}^2}}}\)

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)   A1     N4

[5 marks]

a.

METHOD 1

evidence of choosing chain rule     (M1)

e.g. formula, \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \left( {\frac{{6x}}{{x + 1}}} \right)\)

correct reciprocal of \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}}\) is \(\frac{{x + 1}}{{6x}}\) (seen anywhere)     A1

correct substitution into chain rule     A1

e.g. \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \frac{6}{{{{(x + 1)}^2}}}\) , \(\left( {\frac{6}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{{6x}}} \right)\)

working that clearly leads to the answer     A1

e.g. \(\left( {\frac{6}{{(x + 1)}}} \right)\left( {\frac{1}{{6x}}} \right)\) , \(\left( {\frac{1}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{x}} \right)\) , \(\frac{{6(x + 1)}}{{6x{{(x + 1)}^2}}}\)

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

METHOD 2

attempt to subtract logs     (M1)

e.g. \(\ln a – \ln b\) , \(\ln 6x – \ln (x + 1)\)

correct derivatives (must be seen in correct expression)     A1A1

e.g. \(\frac{6}{{6x}} – \frac{1}{{x + 1}}\) , \(\frac{1}{x} – \frac{1}{{x + 1}}\)

working that clearly leads to the answer     A1

e.g. \(\frac{{x + 1 – x}}{{x(x + 1)}}\) , \(\frac{{6x + 6 – 6x}}{{6x(x + 1)}}\) , \(\frac{{6(x + 1 – x)}}{{6x(x + 1)}}\)

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

b.

valid method using integral of  h(x) (accept missing/incorrect limits or missing \({\text{d}}x\) )     (M1)

e.g. \({\rm{area}} = \int_{\frac{1}{5}}^k {h(x){\rm{d}}x} \) , \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} \) 

recognizing that integral of derivative will give original function     (R1)

e.g. \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} {\rm{d}}x = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\)

correct substitution and subtraction     A1

e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln \left( {\frac{{6 \times \frac{1}{5}}}{{\frac{1}{5} + 1}}} \right)\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1)\)

setting their expression equal to \(\ln 4\)     (M1) 

e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1) = \ln 4\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) = \ln 4\) , \(\int_{\frac{1}{5}}^k {h(x){\rm{d}}x = \ln 4} \)

correct equation without logs     A1

e.g.\(\frac{{6k}}{{k + 1}} = 4\) , \(6k = 4(k + 1)\) 

correct working     (A1)

e.g. \(6k = 4k + 4\) , \(2k = 4\)

\(k = 2\)    A1     N4

[7 marks]

c.

Question

Let \(f(x) = \frac{{{{(\ln x)}^2}}}{2}\), for \(x > 0\).

Let \(g(x) = \frac{1}{x}\). The following diagram shows parts of the graphs of \(f’\) and g.

The graph of \(f’\) has an x-intercept at \(x = p\).

Show that \(f'(x) = \frac{{\ln x}}{x}\).

[2]
a.

There is a minimum on the graph of \(f\). Find the \(x\)-coordinate of this minimum.

[3]
b.

Write down the value of \(p\).

[2]
c.

The graph of \(g\) intersects the graph of \(f’\) when \(x = q\).

Find the value of \(q\).

[3]
d.

The graph of \(g\) intersects the graph of \(f’\) when \(x = q\).

Let \(R\) be the region enclosed by the graph of \(f’\), the graph of \(g\) and the line \(x = p\).

Show that the area of \(R\) is \(\frac{1}{2}\).

[5]
e.
Answer/Explanation

Markscheme

METHOD 1

correct use of chain rule     A1A1

eg     \(\frac{{2\ln x}}{2} \times \frac{1}{x},{\text{ }}\frac{{2\ln x}}{{2x}}\)

Note: Award A1 for \(\frac{{2\ln x}}{{2x}}\), A1 for \( \times \frac{1}{x}\).

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

METHOD 2

correct substitution into quotient rule, with derivatives seen     A1

eg     \(\frac{{2 \times 2\ln x \times \frac{1}{x} – 0 \times {{(\ln x)}^2}}}{4}\)

correct working     A1

eg     \(\frac{{4\ln x \times \frac{1}{x}}}{4}\)

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

a.

setting derivative \( = 0\)     (M1)

eg     \(f'(x) = 0,{\text{ }}\frac{{\ln x}}{x} = 0\)

correct working     (A1)

eg     \(\ln x = 0,{\text{ }}x = {{\text{e}}^0}\)

\(x = 1\)     A1     N2

[3 marks] 

b.

intercept when \(f'(x) = 0\)     (M1)

\(p = 1\)     A1     N2

[2 marks]

c.

equating functions     (M1)

eg     \(f’ = g,{\text{ }}\frac{{\ln x}}{x} = \frac{1}{x}\)

correct working     (A1)

eg     \(\ln x = 1\)

\(q = {\text{e   (accept }}x = {\text{e)}}\)     A1     N2

[3 marks]

d.

evidence of integrating and subtracting functions (in any order, seen anywhere)     (M1)

eg     \(\int_q^e {\left( {\frac{1}{x} – \frac{{\ln x}}{x}} \right){\text{d}}x{\text{, }}\int {f’ – g} } \)

correct integration \(\ln x – \frac{{{{(\ln x)}^2}}}{2}\)     A2

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     \((\ln {\text{e}} – \ln 1) – \left( {\frac{{{{(\ln {\text{e}})}^2}}}{2} – \frac{{{{(\ln 1)}^2}}}{2}} \right)\)

Note: Do not award M1 if the integrated function has only one term.

correct working     A1

eg     \((1 – 0) – \left( {\frac{1}{2} – 0} \right),{\text{ }}1 – \frac{1}{2}\)

\({\text{area}} = \frac{1}{2}\)     AG     N0

Notes: Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the markscheme.

[5 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

Consider f(x), g(x) and h(x), for x∈\(\mathbb{R}\) where h(x) = \(\left( {f \circ g} \right)\)(x).

Given that g(3) = 7 , g′ (3) = 4 and f ′ (7) = −5 , find the gradient of the normal to the curve of h at x = 3.

Answer/Explanation

Markscheme

recognizing the need to find h′      (M1)

recognizing the need to find h′ (3) (seen anywhere)      (M1)

evidence of choosing chain rule        (M1)

eg   \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,f’\left( {g\left( 3 \right)} \right) \times g’\left( 3 \right),\,\,f’\left( g \right) \times g’\)

correct working       (A1)

eg  \(f’\left( 7 \right) \times 4,\,\, – 5 \times 4\)

\(h’\left( 3 \right) =  – 20\)      (A1)

evidence of taking their negative reciprocal for normal       (M1)

eg  \( – \frac{1}{{h’\left( 3 \right)}},\,\,{m_1}{m_2} =  – 1\)

gradient of normal is \(\frac{1}{{20}}\)      A1 N4

[7 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 \(f’\left( 2 \right)\).

[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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