Question
Consider the graph of the function f (x) = x2 − \(\frac{k}{x}\) .
Write down f ′(x) . [3]
The equation of the tangent to the graph of y = f (x) at x = −2 is 2 y = 4 − 5x .
Write down the gradient of this tangent. [1]
Find the value of k . [2]
Answer/Explanation
Ans:
(a) \( 2x+ \frac{k}{x^{2}}\)
(b)-2.5 (\(\frac{-5}{2}\))
(c)
\(-2.5=2\times (-2)+\frac{k}{-2^{2}}\)
k=6
Question
Let \(f(x) = 1 + {{\text{e}}^{ – x}}\) and \(g(x) = 2x + b\), for \(x \in \mathbb{R}\), where \(b\) is a constant.
Find \((g \circ f)(x)\).
Given that \(\mathop {\lim }\limits_{x \to + \infty } (g \circ f)(x) = – 3\), find the value of \(b\).
Answer/Explanation
Markscheme
attempt to form composite (M1)
eg\(\,\,\,\,\,\)\(g(1 + {{\text{e}}^{ – x}})\)
correct function A1 N2
eg\(\,\,\,\,\,\)\((g \circ f)(x) = 2 + b + 2{{\text{e}}^{ – x}},{\text{ }}2(1 + {{\text{e}}^{ – x}}) + b\)
[2 marks]
evidence of \(\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ – x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ – x}})\) (M1)
eg\(\,\,\,\,\,\)\(2 + b + 2{{\text{e}}^{ – \infty }}\), graph with horizontal asymptote when \(x \to \infty \)
Note: Award M0 if candidate clearly has incorrect limit, such as \(x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}\).
evidence that \({{\text{e}}^{ – x}} \to 0\) (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\(\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ – x}}) = 0,{\text{ }}1 + {{\text{e}}^{ – x}} \to 1,{\text{ }}2(1) + b = – 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0\), graph of \(y = {{\text{e}}^{ – x}}\) or
\(y = 2{{\text{e}}^{ – x}}\) with asymptote \(y = 0\), graph of composite function with asymptote \(y = – 3\)
correct working (A1)
eg\(\,\,\,\,\,\)\(2 + b = – 3\)
\(b = – 5\) 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 \(g(x) = 2x\sin x\) .
Find \(g'(x)\) .
Find the gradient of the graph of g at \(x = \pi \) .
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]
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]
Question
Let \(f(x) = {x^3}\). The following diagram shows part of the graph of f .
The point \({\rm{P}}(a,f(a))\) , where \(a > 0\) , lies on the graph of f . The tangent at P crosses the x-axis at the point \({\rm{Q}}\left( {\frac{2}{3},0} \right)\) . This tangent intersects the graph of f at the point R(−2, −8) .
The equation of the tangent at P is \(y = 3x – 2\) . Let T be the region enclosed by the graph of f , the tangent [PR] and the line \(x = k\) , between \(x = – 2\) and \(x = k\) where \( – 2 < k < 1\) . This is shown in the diagram below.
(i) Show that the gradient of [PQ] is \(\frac{{{a^3}}}{{a – \frac{2}{3}}}\) .
(ii) Find \(f'(a)\) .
(iii) Hence show that \(a = 1\) .
Given that the area of T is \(2k + 4\) , show that k satisfies the equation \({k^4} – 6{k^2} + 8 = 0\) .
Answer/Explanation
Markscheme
(i) substitute into gradient \( = \frac{{{y_1} – {y_2}}}{{{x_1} – {x_2}}}\) (M1)
e.g. \(\frac{{f(a) – 0}}{{a – \frac{2}{3}}}\)
substituting \(f(a) = {a^3}\)
e.g. \(\frac{{{a^3} – 0}}{{a – \frac{2}{3}}}\) A1
gradient \(\frac{{{a^3}}}{{a – \frac{2}{3}}}\) AG N0
(ii) correct answer A1 N1
e.g. \(3{a^2}\) , \(f'(a) = 3\) , \(f'(a) = \frac{{{a^3}}}{{a – \frac{2}{3}}}\)
(iii) METHOD 1
evidence of approach (M1)
e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{{a^3}}}{{a – \frac{2}{3}}}\)
simplify A1
e.g. \(3{a^2}\left( {a – \frac{2}{3}} \right) = {a^3}\)
rearrange A1
e.g. \(3{a^3} – 2{a^2} = {a^3}\)
evidence of solving A1
e.g. \(2{a^3} – 2{a^2} = 2{a^2}(a – 1) = 0\)
\(a = 1\) AG N0
METHOD 2
gradient RQ \( = \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}\) A1
simplify A1
e.g. \(\frac{{ – 8}}{{ – \frac{8}{3}}},3\)
evidence of approach (M1)
e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}\) , \(\frac{{{a^3}}}{{a – \frac{2}{3}}} = 3\)
simplify A1
e.g. \(3{a^2} = 3\) , \({a^2} = 1\)
\(a = 1\) AG N0
[7 marks]
approach to find area of T involving subtraction and integrals (M1)
e.g. \(\int {f – (3x – 2){\rm{d}}x} \) , \(\int_{ – 2}^k {(3x – 2) – \int_{ – 2}^k {{x^3}} } \) , \(\int {({x^3} – 3x + 2)} \)
correct integration with correct signs A1A1A1
e.g. \(\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x\) , \(\frac{3}{2}{x^2} – 2x – \frac{1}{4}{x^4}\)
correct limits \( – 2\) and k (seen anywhere) A1
e.g. \(\int_{ – 2}^k {({x^3} – 3x + 2){\rm{d}}x} \) , \(\left[ {\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x} \right]_{ – 2}^k\)
attempt to substitute k and \( – 2\) (M1)
correct substitution into their integral if 2 or more terms A1
e.g. \(\left( {\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2k} \right) – (4 – 6 – 4)\)
setting their integral expression equal to \(2k + 4\) (seen anywhere) (M1)
simplifying A1
e.g. \(\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2 = 0\)
\({k^4} – 6{k^2} + 8 = 0\) AG N0
[9 marks]
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 .
Show that \(b = \frac{\pi }{4}\) .
Find \(f'(x)\) .
At a point R, the gradient is \( – 2\pi \) . Find the x-coordinate of R.
Answer/Explanation
Markscheme
(i) \(a = 8\) A1 N1
(ii) \(c = 2\) A1 N1
(iii) \(d = 4\) A1 N1
[3 marks]
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]
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]
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]
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
Part of the graph of \(f(x) = a{x^3} – 6{x^2}\) is shown below.
The point P lies on the graph of \(f\) . At P, x = 1.
Find \(f'(x)\) .
The graph of \(f\) has a gradient of \(3\) at the point P. Find the value of \(a\) .
Answer/Explanation
Markscheme
\(f'(x) = 3a{x^2} – 12x\) A1A1 N2
Note: Award A1 for each correct term.
[2 marks]
setting their derivative equal to 3 (seen anywhere) A1
e.g. \(f'(x) = 3\)
attempt to substitute \(x = 1\) into \(f'(x)\) (M1)
e.g. \(3a{(1)^2} – 12(1)\)
correct substitution into \(f'(x)\) (A1)
e.g. \(3a – 12\) , \(3a = 15\)
\(a = 5\) A1 N2
[4 marks]
Question
Consider \(f(x) = {x^2}\sin x\) .
Find \(f'(x)\) .
Find the gradient of the curve of \(f\) at \(x = \frac{\pi }{2}\) .
Answer/Explanation
Markscheme
evidence of choosing product rule (M1)
eg \(uv’ + vu’\)
correct derivatives (must be seen in the product rule) \(\cos x\) , \(2x\) (A1)(A1)
\(f'(x) = {x^2}\cos x + 2x\sin x\) A1 N4
[4 marks]
substituting \(\frac{\pi }{2}\) into their \(f'(x)\) (M1)
eg \(f’\left( {\frac{\pi }{2}} \right)\) , \({\left( {\frac{\pi }{2}} \right)^2}\cos \left( {\frac{\pi }{2}} \right) + 2\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)\)
correct values for both \(\sin \frac{\pi }{2}\) and \(\cos \frac{\pi }{2}\) seen in \(f'(x)\) (A1)
eg \(0 + 2\left( {\frac{\pi }{2}} \right) \times 1\)
\(f’\left( {\frac{\pi }{2}} \right) = \pi \) A1 N2
[3 marks]
Question
Consider \(f(x) = \ln ({x^4} + 1)\) .
The second derivative is given by \(f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}\) .
The equation \(f”(x) = 0\) has only three solutions, when \(x = 0\) , \( \pm \sqrt[4]{3}\) \(( \pm 1.316 \ldots )\) .
Find the value of \(f(0)\) .
Find the set of values of \(x\) for which \(f\) is increasing.
(i) Find \(f”(1)\) .
(ii) Hence, show that there is no point of inflexion on the graph of \(f\) at \(x = 0\) .
There is a point of inflexion on the graph of \(f\) at \(x = \sqrt[4]{3}\) \((x = 1.316 \ldots )\) .
Sketch the graph of \(f\) , for \(x \ge 0\) .
Answer/Explanation
Markscheme
substitute \(0\) into \(f\) (M1)
eg \(\ln (0 + 1)\) , \(\ln 1\)
\(f(0) = 0\) A1 N2
[2 marks]
\(f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}\) (seen anywhere) A1A1
Note: Award A1 for \(\frac{1}{{{x^4} + 1}}\) and A1 for \(4{x^3}\) .
recognizing \(f\) increasing where \(f'(x) > 0\) (seen anywhere) R1
eg \(f'(x) > 0\) , diagram of signs
attempt to solve \(f'(x) > 0\) (M1)
eg \(4{x^3} = 0\) , \({x^3} > 0\)
\(f\) increasing for \(x > 0\) (accept \(x \ge 0\) ) A1 N1
[5 marks]
(i) substituting \(x = 1\) into \(f”\) (A1)
eg \(\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}\) , \(\frac{{4 \times 2}}{4}\)
\(f”(1) = 2\) A1 N2
(ii) valid interpretation of point of inflexion (seen anywhere) R1
eg no change of sign in \(f”(x)\) , no change in concavity,
\(f’\) increasing both sides of zero
attempt to find \(f”(x)\) for \(x < 0\) (M1)
eg \(f”( – 1)\) , \(\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}\) , diagram of signs
correct working leading to positive value A1
eg \(f”( – 1) = 2\) , discussing signs of numerator and denominator
there is no point of inflexion at \(x = 0\) AG N0
[5 marks]
A1A1A1 N3
Notes: Award A1 for shape concave up left of POI and concave down right of POI.
Only if this A1 is awarded, then award the following:
A1 for curve through (\(0\), \(0\)) , A1 for increasing throughout.
Sketch need not be drawn to scale. Only essential features need to be clear.
[3 marks]
Question
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 \(f(x) = 1 + {{\text{e}}^{ – x}}\) and \(g(x) = 2x + b\), for \(x \in \mathbb{R}\), where \(b\) is a constant.
Find \((g \circ f)(x)\).
Given that \(\mathop {\lim }\limits_{x \to + \infty } (g \circ f)(x) = – 3\), find the value of \(b\).
Answer/Explanation
Markscheme
attempt to form composite (M1)
eg\(\,\,\,\,\,\)\(g(1 + {{\text{e}}^{ – x}})\)
correct function A1 N2
eg\(\,\,\,\,\,\)\((g \circ f)(x) = 2 + b + 2{{\text{e}}^{ – x}},{\text{ }}2(1 + {{\text{e}}^{ – x}}) + b\)
[2 marks]
evidence of \(\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ – x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ – x}})\) (M1)
eg\(\,\,\,\,\,\)\(2 + b + 2{{\text{e}}^{ – \infty }}\), graph with horizontal asymptote when \(x \to \infty \)
Note: Award M0 if candidate clearly has incorrect limit, such as \(x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}\).
evidence that \({{\text{e}}^{ – x}} \to 0\) (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\(\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ – x}}) = 0,{\text{ }}1 + {{\text{e}}^{ – x}} \to 1,{\text{ }}2(1) + b = – 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0\), graph of \(y = {{\text{e}}^{ – x}}\) or
\(y = 2{{\text{e}}^{ – x}}\) with asymptote \(y = 0\), graph of composite function with asymptote \(y = – 3\)
correct working (A1)
eg\(\,\,\,\,\,\)\(2 + b = – 3\)
\(b = – 5\) A1 N2
[4 marks]