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 \(f(x) = \frac{1}{3}{x^3} + 2{x^2} – 5x\) . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.
a.Find \(f'(x)\) .[3]
b.Find the x-coordinate of M.[4]
c.Find the x-coordinate of N.[3]
d.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]
▶️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
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.
a.Find \(f'(x)\) .[3]
b.Find the x-coordinate of M.[4]
c.Find the x-coordinate of N.[3]
d.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]
▶️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
A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.
The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is \(\theta \) radians, where \(0 \le \theta \le \frac{\pi }{2}\) .
Write down an expression in terms of \(\theta \) for
a.(i) \(x\) ;
(ii) \(y\) .[2]
b.Let the area of the rectangle be A.
Show that \(A = 18\sin 2\theta \) .[3]
c.(i) Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) .
(ii) Hence, find the exact value of \(\theta \) which maximizes the area of the rectangle.
(iii) Use the second derivative to justify that this value of \(\theta \) does give a maximum.[8]
▶️Answer/Explanation
Markscheme
(i) \(x = 3\cos \theta \) A1 N1
(ii) \(y = 3\sin \theta \) A1 N1
[2 marks]
finding area (M1)
e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\)
substituting A1
e.g. \(A = 4 \times 3\sin \theta \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta \times 3\sin \theta \)
\(A = 18(2\sin \theta \cos \theta )\) A1
\(A = 18\sin 2\theta \) AG N0
[3 marks]
(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \) A2 N2
(ii) for setting derivative equal to 0 (M1)
e.g. \(36\cos 2\theta = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)
\(2\theta = \frac{\pi }{2}\) (A1)
\(\theta = \frac{\pi }{4}\) A1 N2
(iii) valid reason (seen anywhere) R1
e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f”(x) < 0\)
finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} = – 72\sin 2\theta \) A1
evidence of substituting \(\frac{\pi }{4}\) M1
e.g. \( – 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( – 72\sin \left( {\frac{\pi }{2}} \right)\) , \( – 72\)
\(\theta = \frac{\pi }{4}\) produces the maximum area AG N0
[8 marks]
Question
Let \(f(x) = {{\rm{e}}^{ – 3x}}\) and \(g(x) = \sin \left( {x – \frac{\pi }{3}} \right)\) .
Write down
a.(i) \(f'(x)\) ;
(ii) \(g'(x)\) .[2]
b.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]
▶️Answer/Explanation
Markscheme
(i) \( – 3{{\rm{e}}^{ – 3x}}\) A1 N1
(ii) \(\cos \left( {x – \frac{\pi }{3}} \right)\) A1 N1
[4 marks]
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]
Question
Consider \(f(x) = {x^2} + \frac{p}{x}\) , \(x \ne 0\) , where p is a constant.
a.Find \(f'(x)\) .[2]
b.There is a minimum value of \(f(x)\) when \(x = – 2\) . Find the value of \(p\) .[4]
▶️Answer/Explanation
Markscheme
\(f'(x) = 2x – \frac{p}{{{x^2}}}\) A1A1 N2
Note: Award A1 for \(2x\) , A1 for \( – \frac{p}{{{x^2}}}\) .
[2 marks]
evidence of equating derivative to 0 (seen anywhere) (M1)
evidence of finding \(f'( – 2)\) (seen anywhere) (M1)
correct equation A1
e.g. \( – 4 – \frac{p}{4} = 0\) , \( – 16 – p = 0\)
\(p = – 16\) A1 N3
[4 marks]
Question
Let \(f(x) = 3 + \frac{{20}}{{{x^2} – 4}}\) , for \(x \ne \pm 2\) . The graph of f is given below.
The y-intercept is at the point A.
a.(i) Find the coordinates of A.
(ii) Show that \(f'(x) = 0\) at A.[7]
b.The second derivative \(f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}\) . Use this to
(i) justify that the graph of f has a local maximum at A;
(ii) explain why the graph of f does not have a point of inflexion.[6]
c.Describe the behaviour of the graph of \(f\) for large \(|x|\) .[1]
d.Write down the range of \(f\) .[2]
▶️Answer/Explanation
Markscheme
(i) coordinates of A are \((0{\text{, }} – 2)\) A1A1 N2
(ii) derivative of \({x^2} – 4 = 2x\) (seen anywhere) (A1)
evidence of correct approach (M1)
e.g. quotient rule, chain rule
finding \(f'(x)\) A2
e.g. \(f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)\) , \(\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}\)
substituting \(x = 0\) into \(f'(x)\) (do not accept solving \(f'(x) = 0\) ) M1
at A \(f'(x) = 0\) AG N0
[7 marks]
(i) reference to \(f'(x) = 0\) (seen anywhere) (R1)
reference to \(f”(0)\) is negative (seen anywhere) R1
evidence of substituting \(x = 0\) into \(f”(x)\) M1
finding \(f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}\) \(\left( { = – \frac{5}{2}} \right)\) A1
then the graph must have a local maximum AG
(ii) reference to \(f”(x) = 0\) at point of inflexion (R1)
recognizing that the second derivative is never 0 A1 N2
e.g. \(40(3{x^2} + 4) \ne 0\) , \(3{x^2} + 4 \ne 0\) , \({x^2} \ne – \frac{4}{3}\) , the numerator is always positive
Note: Do not accept the use of the first derivative in part (b).
[6 marks]
correct (informal) statement, including reference to approaching \(y = 3\) A1 N1
e.g. getting closer to the line \(y = 3\) , horizontal asymptote at \(y = 3\)
[1 mark]
correct inequalities, \(y \le – 2\) , \(y > 3\) , FT from (a)(i) and (c) A1A1 N2
[2 marks]
Question
Let \(f(x) = \frac{1}{2}{x^3} – {x^2} – 3x\) . Part of the graph of f is shown below.
There is a maximum point at A and a minimum point at B(3, − 9) .
a.Find the coordinates of A.[8]
b(i), (ii) and (iii).Write down the coordinates of
(i) the image of B after reflection in the y-axis;
(ii) the image of B after translation by the vector \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
5
\end{array}} \right)\) ;
(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor \(\frac{1}{2}\) .[6]
▶️Answer/Explanation
Markscheme
\(f(x) = {x^2} – 2x – 3\) A1A1A1
evidence of solving \(f'(x) = 0\) (M1)
e.g. \({x^2} – 2x – 3 = 0\)
evidence of correct working A1
e.g. \((x + 1)(x – 3)\) , \(\frac{{2 \pm \sqrt {16} }}{2}\)
\(x = – 1\) (ignore \(x = 3\) ) (A1)
evidence of substituting their negative x-value into \(f(x)\) (M1)
e.g. \(\frac{1}{3}{( – 1)^3} – {( – 1)^2} – 3( – 1)\) , \( – \frac{1}{3} – 1 + 3\)
\(y = \frac{5}{3}\) A1
coordinates are \(\left( { – 1,\frac{5}{3}} \right)\) N3
[8 marks]
(i) \(( – 3{\text{, }} – 9)\) A1 N1
(ii) \((1{\text{, }} – 4)\) A1A1 N2
(iii) reflection gives \((3{\text{, }}9)\) (A1)
stretch gives \(\left( {\frac{3}{2}{\text{, }}9} \right)\) A1A1 N3
[6 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
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.
a(i), (ii) and (iii).(i) Show that the gradient of [PQ] is \(\frac{{{a^3}}}{{a – \frac{2}{3}}}\) .
(ii) Find \(f'(a)\) .
(iii) Hence show that \(a = 1\) .[7]
b.Given that the area of T is \(2k + 4\) , show that k satisfies the equation \({k^4} – 6{k^2} + 8 = 0\) .[9]
▶️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 part of the graph of a quadratic function f .
The x-intercepts are at \(( – 4{\text{, }}0)\) and \((6{\text{, }}0)\) , and the y-intercept is at \((0{\text{, }}240)\) .
a.Write down \(f(x)\) in the form \(f(x) = – 10(x – p)(x – q)\) .[2]
b.Find another expression for \(f(x)\) in the form \(f(x) = – 10{(x – h)^2} + k\) .[4]
c.Show that \(f(x)\) can also be written in the form \(f(x) = 240 + 20x – 10{x^2}\) .[2]
d(i) and (ii).A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ – 1}}\) , at time t seconds is given by \(v = 240 + 20t – 10{t^2}\) , for \(0 \le t \le 6\) .
(i) Find the value of t when the speed of the particle is greatest.
(ii) Find the acceleration of the particle when its speed is zero.[7]
▶️Answer/Explanation
Markscheme
\(f(x) = – 10(x + 4)(x – 6)\) A1A1 N2
[2 marks]
METHOD 1
attempting to find the x-coordinate of maximum point (M1)
e.g. averaging the x-intercepts, sketch, \(y’ = 0\) , axis of symmetry
attempting to find the y-coordinate of maximum point (M1)
e.g. \(k = – 10(1 + 4)(1 – 6)\)
\(f(x) = – 10{(x – 1)^2} + 250\) A1A1 N4
METHOD 2
attempt to expand \(f(x)\) (M1)
e.g. \( – 10({x^2} – 2x – 24)\)
attempt to complete the square (M1)
e.g. \( – 10({(x – 1)^2} – 1 – 24)\)
\(f(x) = – 10{(x – 1)^2} + 250\) A1A1 N4
[4 marks]
attempt to simplify (M1)
e.g. distributive property, \( – 10(x – 1)(x – 1) + 250\)
correct simplification A1
e.g. \( – 10({x^2} – 6x + 4x – 24)\) , \( – 10({x^2} – 2x + 1) + 250\)
\(f(x) = 240 + 20x – 10{x^2}\) AG N0
[2 marks]
(i) valid approach (M1)
e.g. vertex of parabola, \(v'(t) = 0\)
\(t = 1\) A1 N2
(ii) recognizing \(a(t) = v'(t)\) (M1)
\(a(t) = 20 – 20t\) A1A1
speed is zero \( \Rightarrow t = 6\) (A1)
\(a(6) = – 100\) (\({\text{m}}{{\text{s}}^{ – 2}}\)) A1 N3
[7 marks]
Question
The following diagram shows part of the graph of a quadratic function f .
The x-intercepts are at \(( – 4{\text{, }}0)\) and \((6{\text{, }}0)\) , and the y-intercept is at \((0{\text{, }}240)\) .
a.Write down \(f(x)\) in the form \(f(x) = – 10(x – p)(x – q)\) .[2]
b.Find another expression for \(f(x)\) in the form \(f(x) = – 10{(x – h)^2} + k\) .[4]
c.Show that \(f(x)\) can also be written in the form \(f(x) = 240 + 20x – 10{x^2}\) .[2]
A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ – 1}}\) , at time t seconds is given by \(v = 240 + 20t – 10{t^2}\) , for \(0 \le t \le 6\) .
(i) Find the value of t when the speed of the particle is greatest.
(ii) Find the acceleration of the particle when its speed is zero.[7]
▶️Answer/Explanation
Markscheme
\(f(x) = – 10(x + 4)(x – 6)\) A1A1 N2
[2 marks]
METHOD 1
attempting to find the x-coordinate of maximum point (M1)
e.g. averaging the x-intercepts, sketch, \(y’ = 0\) , axis of symmetry
attempting to find the y-coordinate of maximum point (M1)
e.g. \(k = – 10(1 + 4)(1 – 6)\)
\(f(x) = – 10{(x – 1)^2} + 250\) A1A1 N4
METHOD 2
attempt to expand \(f(x)\) (M1)
e.g. \( – 10({x^2} – 2x – 24)\)
attempt to complete the square (M1)
e.g. \( – 10({(x – 1)^2} – 1 – 24)\)
\(f(x) = – 10{(x – 1)^2} + 250\) A1A1 N4
[4 marks]
attempt to simplify (M1)
e.g. distributive property, \( – 10(x – 1)(x – 1) + 250\)
correct simplification A1
e.g. \( – 10({x^2} – 6x + 4x – 24)\) , \( – 10({x^2} – 2x + 1) + 250\)
\(f(x) = 240 + 20x – 10{x^2}\) AG N0
[2 marks]
(i) valid approach (M1)
e.g. vertex of parabola, \(v'(t) = 0\)
\(t = 1\) A1 N2
(ii) recognizing \(a(t) = v'(t)\) (M1)
\(a(t) = 20 – 20t\) A1A1
speed is zero \( \Rightarrow t = 6\) (A1)
\(a(6) = – 100\) (\({\text{m}}{{\text{s}}^{ – 2}}\)) A1 N3
[7 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 .[3]
b.Show that \(b = \frac{\pi }{4}\) .[2]
c.Find \(f'(x)\) .[3]
d.At a point R, the gradient is \( – 2\pi \) . Find the x-coordinate of R.[6]
▶️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 \(f(x) = {{\rm{e}}^{6x}}\) .
a.Write down \(f'(x)\) .[1]
b(i) and (ii).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]
▶️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
Consider \(f(x) = {x^2}\sin x\) .
a.Find \(f'(x)\) .[4]
b.Find the gradient of the curve of \(f\) at \(x = \frac{\pi }{2}\) .[3]
▶️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
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 \(g(x) = a{(x – h)^2} + 3\) .
a.Find \(f'(x)\) .[3]
b.Find \(g(4)\) .[3]
c.(i) Write down the value of \(h\) .
(ii) Find the value of \(a\) .[4]
d.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]
▶️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
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 \(x = p\).
a.Show that \(f'(x) = \frac{{\ln x}}{x}\).[2]
b.There is a minimum on the graph of \(f\). Find the \(x\)-coordinate of this minimum.[3]
c.Write down the value of \(p\).[2]
d.The graph of \(g\) intersects the graph of \(f’\) when \(x = q\).
Find the value of \(q\).[3]
e.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]
▶️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]
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]
intercept when \(f'(x) = 0\) (M1)
\(p = 1\) A1 N2
[2 marks]
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]
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]
Question
Let \(f(x) = p{x^3} + p{x^2} + qx\).
a.Find \(f'(x)\).[2]
b.Given that \(f'(x) \geqslant 0\), show that \({p^2} \leqslant 3pq\).[5]
▶️Answer/Explanation
Markscheme
\(f'(x) = 3p{x^2} + 2px + q\) A2 N2
Note: Award A1 if only 1 error.
[2 marks]
evidence of discriminant (must be seen explicitly, not in quadratic formula) (M1)
eg \({b^2} – 4ac\)
correct substitution into discriminant (may be seen in inequality) A1
eg \({(2p)^2} – 4 \times 3p \times q,{\text{ }}4{p^2} – 12pq\)
\(f'(x) \geqslant 0\) then \(f’\) has two equal roots or no roots (R1)
recognizing discriminant less or equal than zero R1
eg \(\Delta \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0\)
correct working that clearly leads to the required answer A1
eg \({p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq\)
\({p^2} \leqslant 3pq\) AG N0
[5 marks]
Question
A function \(f\) has its derivative given by \(f'(x) = 3{x^2} – 2kx – 9\), where \(k\) is a constant.
a.Find \(f”(x)\).[2]
b.The graph of \(f\) has a point of inflexion when \(x = 1\).
Show that \(k = 3\).[3]
c.Find \(f'( – 2)\).[2]
d.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]
e.Given that \(f'( – 1) = 0\), explain why the graph of \(f\) has a local maximum when \(x = – 1\).[3]
▶️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 \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), for \(0 < x < 6\).
The graph of \(f\) has a maximum point at P.
The \(y\)-coordinate of P is \(\ln 27\).
a.Find the \(x\)-coordinate of P.[3]
b.Find \(f(x)\), expressing your answer as a single logarithm.[8]
c.The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).
Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).
▶️Answer/Explanation
Markscheme
recognizing \(f'(x) = 0\) (M1)
correct working (A1)
eg\(\,\,\,\,\,\)\(6 – 2x = 0\)
\(x = 3\) A1 N2
[3 marks]
evidence of integration (M1)
eg\(\,\,\,\,\,\)\(\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} } \)
using substitution (A1)
eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x – {x^2}\)
correct integral A1
eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x – {x^2})\)
substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\)) (M1)
eg\(\,\,\,\,\,\)\(\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27\)
correct working (A1)
eg\(\,\,\,\,\,\)\(c = \ln 27 – \ln 9\)
EITHER
\(c = \ln 3\) (A1)
attempt to substitute their value of \(c\) into \(f(x)\) (M1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 3\) A1 N4
OR
attempt to substitute their value of \(c\) into \(f(x)\) (M1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9\)
correct use of a log law (A1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9\)
\(f(x) = \ln \left( {3(6x – {x^2})} \right)\) A1 N4
[8 marks]
\(a = 3\) A1 N1
correct working A1
eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)
correct use of log law (A1)
eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)
\(b = 3\) A1 N2
[4 marks]
Question
Fred makes an open metal container in the shape of a cuboid, as shown in the following diagram.
The container has height \(x{\text{ m}}\), width \(x{\text{ m}}\) and length \(y{\text{ m}}\). The volume is \(36{\text{ }}{{\text{m}}^3}\).
Let \(A(x)\) be the outside surface area of the container.
a.Show that \(A(x) = \frac{{108}}{x} + 2{x^2}\).[4]
b.Find \(A'(x)\).[2]
c.Given that the outside surface area is a minimum, find the height of the container.[5]
d.Fred paints the outside of the container. A tin of paint covers a surface area of \({\text{10 }}{{\text{m}}^{\text{2}}}\) and costs $20. Find the total cost of the tins needed to paint the container.[5]
▶️Answer/Explanation
Markscheme
correct substitution into the formula for volume A1
eg\(\,\,\,\,\,\)\(36 = y \times x \times x\)
valid approach to eliminate \(y\) (may be seen in formula/substitution) M1
eg\(\,\,\,\,\,\)\(y = \frac{{36}}{{{x^2}}},{\text{ }}xy = \frac{{36}}{x}\)
correct expression for surface area A1
eg\(\,\,\,\,\,\)\(xy + xy + xy + {x^2} + {x^2},{\text{ area}} = 3xy + 2{x^2}\)
correct expression in terms of \(x\) only A1
eg\(\,\,\,\,\,\)\(3x\left( {\frac{{36}}{{{x^2}}}} \right) + 2{x^2},{\text{ }}{x^2} + {x^2} + \frac{{36}}{x} + \frac{{36}}{x} + \frac{{36}}{x},{\text{ }}2{x^2} + 3\left( {\frac{{36}}{x}} \right)\)
\(A(x) = \frac{{108}}{x} + 2{x^2}\) AG N0
[4 marks]
\(A'(x) = – \frac{{108}}{{{x^2}}} + 4x,{\text{ }}4x – 108{x^{ – 2}}\) A1A1 N2
Note: Award A1 for each term.
[2 marks]
recognizing that minimum is when \(A'(x) = 0\) (M1)
correct equation (A1)
eg\(\,\,\,\,\,\)\( – \frac{{108}}{{{x^2}}} + 4x = 0,{\text{ }}4x = \frac{{108}}{{{x^2}}}\)
correct simplification (A1)
eg\(\,\,\,\,\,\)\( – 108 + 4{x^3} = 0,{\text{ }}4{x^3} = 108\)
correct working (A1)
eg\(\,\,\,\,\,\)\({x^3} = 27\)
\({\text{height}} = 3{\text{ (m) }}({\text{accept }}x = 3)\) A1 N2
[5 marks]
attempt to find area using their height (M1)
eg\(\,\,\,\,\,\)\(\frac{{108}}{3} + 2{(3)^2},{\text{ }}9 + 9 + 12 + 12 + 12\)
minimum surface area \( = 54{\text{ }}{{\text{m}}^{\text{2}}}\) (may be seen in part (c)) A1
attempt to find the number of tins (M1)
eg\(\,\,\,\,\,\)\(\frac{{54}}{{10}},{\text{ }}5.4\)
6 (tins) (A1)
$120 A1 N3
[5 marks]
Question
Let \(f(x) = \cos x\).
Let \(g(x) = {x^k}\), where \(k \in {\mathbb{Z}^ + }\).
Let \(k = 21\) and \(h(x) = \left( {{f^{(19)}}(x) \times {g^{(19)}}(x)} \right)\).
a.(i) Find the first four derivatives of \(f(x)\).
(ii) Find \({f^{(19)}}(x)\).[4]
b.(i) Find the first three derivatives of \(g(x)\).
(ii) Given that \({g^{(19)}}(x) = \frac{{k!}}{{(k – p)!}}({x^{k – 19}})\), find \(p\).[5]
c.(i) Find \(h'(x)\).
(ii) Hence, show that \(h'(\pi ) = \frac{{ – 21!}}{2}{\pi ^2}\).[7]
▶️Answer/Explanation
Markscheme
(i) \(f'(x) = – \sin x,{\text{ }}f”(x) = – \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x\) A2 N2
(ii) valid approach (M1)
eg\(\,\,\,\,\,\)recognizing that 19 is one less than a multiple of 4, \({f^{(19)}}(x) = {f^{(3)}}(x)\)
\({f^{(19)}}(x) = \sin x\) A1 N2
[4 marks]
(i) \(g'(x) = k{x^{k – 1}}\)
\(g”(x) = k(k – 1){x^{k – 2}},{\text{ }}{g^{(3)}}(x) = k(k – 1)(k – 2){x^{k – 3}}\) A1A1 N2
(ii) METHOD 1
correct working that leads to the correct answer, involving the correct expression for the 19th derivative A2
eg\(\,\,\,\,\,\)\(k(k – 1)(k – 2) \ldots (k – 18) \times \frac{{(k – 19)!}}{{(k – 19)!}},{{\text{ }}_k}{P_{19}}\)
\(p = 19\) (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\)) A1 N1
METHOD 2
correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient A2
eg\(\,\,\,\,\,\)\(g” = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k – 1)(k – 2) = \frac{{k!}}{{(k – 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k – 3}})\)
\({g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k – 19)! \times 19!}},{{\text{ }}_k}{P_{19}}\)
\(p = 19\) (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\)) A1 N1
[5 marks]
(i) valid approach using product rule (M1)
eg\(\,\,\,\,\,\)\(uv’ + vu’,{\text{ }}{f^{(19)}}{g^{(20)}} + {f^{(20)}}{g^{(19)}}\)
correct 20th derivatives (must be seen in product rule) (A1)(A1)
eg\(\,\,\,\,\,\)\({g^{(20)}}(x) = \frac{{21!}}{{(21 – 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x\)
\(h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)\) A1 N3
(ii) substituting \(x = \pi \) (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\({f^{(19)}}(\pi ){g^{(20)}}(\pi ) + {f^{(20)}}(\pi ){g^{(19)}}(\pi ),{\text{ }}\sin \pi \frac{{21!}}{{1!}}\pi + \cos \pi \frac{{21!}}{{2!}}{\pi ^2}\)
evidence of one correct value for \(\sin \pi \) or \(\cos \pi \) (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\(\sin \pi = 0,{\text{ }}\cos \pi = – 1\)
evidence of correct values substituted into \(h'(\pi )\) A1
eg\(\,\,\,\,\,\)\(21!(\pi )\left( {0 – \frac{\pi }{{2!}}} \right),{\text{ }}21!(\pi )\left( { – \frac{\pi }{2}} \right),{\text{ }}0 + ( – 1)\frac{{21!}}{2}{\pi ^2}\)
Note: If candidates write only the first line followed by the answer, award A1A0A0.
\(\frac{{ – 21!}}{2}{\pi ^2}\) AG N0
[7 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.
a.i.Write down \(f'(x)\).[1]
a.ii.Find the gradient of \(L\).[2]
b.Show that the \(x\)-coordinate of B is \( – \frac{k}{2}\).[5]
c.Find the area of triangle ABC, giving your answer in terms of \(k\).[2]
d.Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).[7]
▶️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.
a.Show that \(f’(1) = 1\).[3]
b.Find the equation of \(L\) in the form \(y = ax + b\).[3]
c.Find the \(x\)-coordinate of Q.[4]
d.Find the area of the region enclosed by the graph of \(f\) and the line \(L\).[6]
▶️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
A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20\(\pi \) cm3.
The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.
a.Express h in terms of r.[2]
b.Show that \(C = 20\pi {r^2} + \frac{{320\pi }}{r}\).[4]
c.Given that there is a minimum value for C, find this minimum value in terms of \(\pi \).[9]
▶️Answer/Explanation
Markscheme
correct equation for volume (A1)
eg \(\pi {r^2}h = 20\pi \)
\(h = \frac{{20}}{{{r^2}}}\) A1 N2
[2 marks]
attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side
correct expression for cost of two circles in terms of r (seen anywhere) A1
eg \(2\pi {r^2} \times 10\)
correct expression for cost of curved side (seen anywhere) (A1)
eg \(2\pi r \times h \times 8\)
correct expression for cost of curved side in terms of r A1
eg \(8 \times 2\pi r \times \frac{{20}}{{{r^2}}},\,\,\frac{{320\pi }}{{{r^2}}}\)
\(C = 20\pi {r^2} + \frac{{320\pi }}{r}\) AG N0
[4 marks]
recognize \(C’ = 0\) at minimum (R1)
eg \(C’ = 0,\,\,\frac{{{\text{d}}C}}{{{\text{d}}r}} = 0\)
correct differentiation (may be seen in equation)
\(C’ = 40\pi r – \frac{{320\pi }}{{{r^2}}}\) A1A1
correct equation A1
eg \(40\pi r – \frac{{320\pi }}{{{r^2}}} = 0,\,\,40\pi r\frac{{320\pi }}{{{r^2}}}\)
correct working (A1)
eg \(40{r^3} = 320,\,\,{r^3} = 8\)
r = 2 (m) A1
attempt to substitute their value of r into C
eg \(20\pi \times 4 + 320 \times \frac{\pi }{2}\) (M1)
correct working
eg \(80\pi + 160\pi \) (A1)
\(240\pi \) (cents) A1 N3
Note: Do not accept 753.6, 753.98 or 754, even if 240\(\pi \) is seen.
[9 marks]
Question
Given that \(f(x) = \frac{1}{x}\) , answer the following.
Find the first four derivatives of \(f(x)\) .
Write an expression for \({f^{(n)}}(x)\) in terms of x and n .
▶️Answer/Explanation
Markscheme
\(f'(x) = – {x^{ – 2}}\) (or \( – \frac{1}{{{x^2}}}\) ) A1 N1
\(f”(x) = 2{x^{ – 3}}\) (or \(\frac{2}{{{x^3}}}\) ) A1 N1
\(f”'(x) = – 6{x^{ – 4}}\) (or \( – \frac{6}{{{x^4}}}\) ) A1 N1
\({f^{(4)}}(x) = 24{x^{ – 5}}\) (or \(\frac{{24}}{{{x^5}}}\) ) A1 N1
[4 marks]
\({f^{(n)}}(x) = \frac{{{{( – 1)}^n}n!}}{{{x^{n + 1}}}}\) or \({( – 1)^n}n!({x^{ – (n + 1)}})\) A1A1A1 N3
[3 marks]
Question
Let \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . The following diagram shows the graph of \(f\) .
The \(y\)-intercept is at (\(0\), \(1\)) , there is a minimum point at A (\(p\), \(q\)) and a maximum point at B.
a.Find \(f'(x)\) .[2]
Hence
b(i) and (ii).(i) show that \(q = – 2\) ;
(ii) verify that A is a minimum point.[10]
c.Find the maximum value of \(f(x)\) .[3]
d.The function \(f(x)\) can be written in the form \(r\cos (x – a)\) .
Write down the value of r and of a .[2]
▶️Answer/Explanation
Markscheme
\(f'(x) = – \sin x + \sqrt 3 \cos x\) A1A1 N2
[2 marks]
(i) at A, \(f'(x) = 0\) R1
correct working A1
e.g. \(\sin x = \sqrt 3 \cos x\)
\(\tan x = \sqrt 3 \) A1
\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\) A1
attempt to substitute their x into \(f(x)\) M1
e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)
correct substitution A1
e.g. \( – \frac{1}{2} + \sqrt 3 \left( { – \frac{{\sqrt 3 }}{2}} \right)\)
correct working that clearly leads to \( – 2\) A1
e.g. \( – \frac{1}{2} – \frac{3}{2}\)
\(q = – 2\) AG N0
(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\) A1A1
e.g. \(f'(\pi ) = 0 – \sqrt 3 \) , \(f'(2\pi ) = 0 + \sqrt 3 \)
\(f'(x)\) changes sign from negative to positive R1
so A is a minimum AG N0
[10 marks]
max when \(x = \frac{\pi }{3}\) R1
correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\) A1
e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)
max value is 2 A1 N1
[3 marks]
\(r = 2\) , \(a = \frac{\pi }{3}\) A1A1 N2
[2 marks]
Question
Find the derivatives of the following functions
▶️Answer/Explanation
Ans
Question
Find the derivatives of the following functions
▶️Answer/Explanation
Ans
Question
The following table shows the values of two functions \(f\) and \(g\) and their derivatives when \(x = 1\) and \(x = 0\).
Find the derivatives of the following functions when \(x = 1\).
▶️Answer/Explanation
Ans
For the last 2:
\(f'(-g(x))\times (-g'(x))\). Hence, at \(x=1\), \(f'(-g(1))\times (-g'(1))=f'(1)\times (-2)=-6\)
\(g'(f(x)+2)\times (f'(x))\). Hence,at \(x=1\), \(g'(f(1)+2)\times (f'(1))=g'(0)\times 3=15\)
Question
If \(y=\ln (2x-1)\), find \(\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}\).
▶️Answer/Explanation
Ans
\(y=\ln (2x-1)\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{2x-1}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=2(2x-1)^{-1}\)
\(\Rightarrow \frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=-2(2x-1)^{-2}(2)\)
\(\Rightarrow \frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=\frac{-4}{(2x-1)^{2}}\)
Question
If \(f(x)=\ln (2x-1),x>\frac{1}{2}\), find \(f’\) and the value of \(x\) where the gradient of \(f (x)\) is equal to \(x\).
▶️Answer/Explanation
Ans
If \(f(x)=\ln (2x-1), Then f'(x)=\frac{2}{2x-1}\)
Put \(\frac{2}{2x-1}=x\Rightarrow x=1.28\) (using a GDC or the quadratic formula)
Question
Let \(y=\sin (kx)-kx\cos (kx)\), where \(k\) is a constant. Show that \(\frac{\mathrm{d} y}{\mathrm{d} x}=k^{2}x\sin (kx)\)
▶️Answer/Explanation
Ans
\(y=\sin (kx)-kx\cos (kx)\)
\(\frac{\mathrm{d} y}{\mathrm{d} x}=k\cos(kx)-k\left \{\cos (kx)+x[-k\ sin (kx)] \right \}\)
\(=k \cos (kx)-k \cos(kx)+k^{2}x\ sin (kx)=k^{2}x\ sin (kx)\)
Question
Consider the function \(f(x)=\ln \sqrt{x^{2}+4}\). Find (a) \(f'(x)\) (b) \({f}”(x)\)
▶️Answer/Explanation
Ans
(a) METHOD 1
\(f(x)=\frac{1}{2}\ln (x^{2}+4)\)
\(f'(x)=\frac{1}{2}\left ( \frac{2x}{x^{2}+4} \right )\)
\(f'(x)=\frac{x}{x^{2}+4}\)
METHOD 2
\(f'(x)=\frac{1}{\sqrt{x^{2}+4}}\times \frac{1}{2}\left ( x^{2}+4 \right )^{-\frac{1}{2}}(2x)\)
\(=\frac{x}{x^{2}+4}\)
(b) \({f}”(x)=\frac{x^{2}+4-x(2x)}{(x^{2}+4)^{2}}\)
\(=\frac{4-x^{2}}{(x^{2}+4)^{2}}\)
Question
Let \(f(x)=\frac{x^{2}+5x+5}{x+2},x\neq -2\).
(a) Find \(f'(x)\).
(b) Solve \(f'(x) > 2\).
▶️Answer/Explanation
Ans
METHOD 1
(a) \(f'(x)=\frac{(x+2)(2x+5)-(x^{2}+5x+5)}{(x+2)^{2}}=\frac{x^{2}+4x+5}{(x+2)^{2}}\)
(b) \(\frac{x^{2}+4x+5}{(x+2)^{2}}>2\)
\(\Rightarrow x^{2}+4x+5>2x^{2}+8x+8\)
\(\Rightarrow x^{2}+4x+3<0\Rightarrow -3<x<-1\) (and \(x\neq 2)\)
METHOD 2
(a) \(f(x)=x+3-\frac{1}{x+2} f'(x)=1+\frac{1}{(x+2)^{2}}\)
(b) \(1+\frac{1}{(x+2)^{2}}>2 \Rightarrow {(x+2)^{2}}<1\)
\(-1<x+2<1\)
\(-3<x<-1\) (and \(x\neq 2\))
Question
For the function \(f:x\mapsto \frac{1}{2}\sin 2x+\cos x\), find the possible values of \(\sin x\) for which \(f'(x)=0\).
▶️Answer/Explanation
Ans
\(f(x)=\frac{1}{2}\sin 2x+\cos x\)
\(f'(x)=\cos 2x-\sin x\)
\(=1-2\sin^{2}x-\sin x=(1+\sin x)(1-2\sin x)\)
\(=0\) when \(\sin x=-1\) or \(\frac{1}{2}\)
Extra question : Solve the equation \(f'(x)=0\) for \(-\pi \leq x\leq 2\pi \)
Ans
\(x=\frac{-\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}\)
Question
Consider the function \(y=\tan x-8\sin x\).
(a) Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\). (b) Find the value of \(\cos x\) for which \(\frac{\mathrm{d} y}{\mathrm{d} x}=0\).
▶️Answer/Explanation
Ans
(a) \(\frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x-8\cos x\)
(b) \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-8\cos ^{3}x}{\cos ^{2}x}\)
\(\frac{\mathrm{d} y}{\mathrm{d} x}=0 \Rightarrow \cos x=\frac{1}{2}\)
Extra question : Solve the equation \(\frac{\mathrm{d} y}{\mathrm{d} x}=0\). for \(-\pi \leq x\leq 2\pi \)
Ans
\(x=\frac{-\pi }{3},\frac{\pi }{3},\frac{5\pi }{3}\)
Question
Consider the function \(f(t)=3\sec ^{2}t+5t\). Find (a) \(f'(t)\). (b) (ⅰ) \(f(\pi )\); (ⅱ) \(f'(\pi )\);
▶️Answer/Explanation
Ans
(a) METHOD 1
\(f(t)=3\sec ^{2}t+5t=3(\cos t)^{-2}+5t\)
\(f(t)=-6(\cos t)^{-3}(-\sin t)+5=\frac{6\sin t}{cos^{3}t}+5\)
METHOD 2
\(f'(t)=3\times 2\sec t(\sec t\tan t)+5\)
\(=6\sec ^{2}t \tan t+5(=6\tan ^{2}t+6 \tan t+5)\)
(b) \(f(\pi )=\frac{3}{(\cos \pi )^{2}}+5\pi =3+5\pi\)
\(f'(\pi )=\frac{6\sin \pi }{(\cos \pi )^{3}}+5=5\)
Question
The function \(f\) is defined by \(f:x\mapsto 3^{x}\). Find the solution of the equation \({f}”(x)=2\).
▶️Answer/Explanation
Ans
\(f(x)=3^{x}\Rightarrow f(x)=3^{x}\ln 3\)
\(\Rightarrow f'(x)=3^{x}(\ln 3)^{2}\)
\(3^{x}(\ln 3)^{2}=2\)
\(3^{x}=\frac{2}{(\ln 3)^{2}}\)
\(x\ln 3=\ln \left ( \frac{2}{(\ln 3)^{2}} \right )\)
\(x=\frac{\ln \left ( \frac{2}{(\ln 3)^{2}} \right )}{\ln 3}\)
\(=0.460\)
Question
Let \(y=e^{3x}\sin (\pi x)\). Find (a) \(\frac{\mathrm{d} y}{\mathrm{d} x}\). (b) the smallest positive value of \(x\) for which \(\frac{\mathrm{d} y}{\mathrm{d} x}=0\)
▶️Answer/Explanation
Ans
\(y=e^{3x}\sin (\pi x)\)
(a) \(\frac{\mathrm{d} y}{\mathrm{d} x}=3e^{3x}\sin (\pi x)+\pi e^{3x}\cos (\pi x)\)
(b) \(0=e^{3x}(3\sin(\pi x)+\pi \cos (\pi x))\)
\(tan (\pi x)=-\frac{\pi }{3}\)
\(\pi x=-0.80845+\pi\)
\(x=0.7426… (0.743 to 3s.f.)\)
Question
Let \(f(x)=\cos ^{3}(4x+1),0\leq x\leq 1\). Find
(a) \(f'(x)\) (b) the exact values of the three roots of \(f'(x)=0\) .
▶️Answer/Explanation
Ans
(a) \(f'(x)=3\cos ^{2}(4x+1)\times (-\sin (4x+1))\times 4\)
\(f'(x)=-12\cos ^{2}(4x+1)(\sin (4x+1))\)
(b) \(f'(x)=0\Rightarrow cos^{2}(4x+1)=0 or \sin (4x+1)=0\)
\(\Rightarrow x=\frac{\pi }{8}-\frac{1}{4},x=\frac{3\pi }{8}-\frac{1}{4} or x=\frac{\pi-1 }{4}\)
Question
An experiment is carried out in which the number \(n\) of bacteria in a liquid, is given by the formula \(n=650\ e^{kt}\) , where \(t\) is the time in minutes after the beginning of the
experiment and \(k\) is a constant. The number of bacteria doubles every \(20\) minutes. Find
(a) the exact value of \(k\);
(b) the rate at which the number of bacteria is increasing when \(t = 90\).
▶️Answer/Explanation
Ans
(a) \(1300=650e^{20k}\)
\(k=\frac{\ln 2}{20}\)
(b) \(\frac{\mathrm{d} n}{\mathrm{d} t}=650ke^{kt}\)
when \(t=90, \frac{\mathrm{d} n}{\mathrm{d} t}=509.734=510\) to \(3\) s.f.
Question
Differentiate \(y=arccos(1-2x^{2})\) with respect to \(x\), and simplify your answer.
▶️Answer/Explanation
Ans
Given \(y=arccos(1-2x^{2})\)
then \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-1}{(1-(1-2x^{2})^{2})^{1/2}} \times -4x\)
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{4x}{(1-(1-4x^{2}+4x^{4}))^{1/2}}\)
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{4x}{(4x^{2}-4x^{4})^{1/2}}\)
OR
\(\cos y=1-2x^{2}\)
\(-\sin y\frac{\mathrm{d} y}{\mathrm{d} x}=-4x\)
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-4x}{-\sin y}=\frac{4x}{\sqrt{1-(1-2x^{2})^{2}}}\)
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{4x}{\sqrt{4x^{2}-{4x^{4}}}}\)
Question
Let \(y=x\ arcsin x,x\epsilon\ ]-1,1[\). Show that \(\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=\frac{2-x^{2}}{(1-x^{2})\tfrac{3}{2}}\).
▶️Answer/Explanation
Ans
\(y=x\ arcsin\ x\)
\(\frac{\mathrm{d} y}{\mathrm{d} x}=arcsin\ x+\frac{x}{\sqrt{1-x^{2}}}\)
\(\frac{\mathrm{d}}{\mathrm{d} x}\left ( \frac{x}{\sqrt{1-x^{2}}} \right )=\frac{1(\sqrt{1-x^{2}}+x^{2}(1-x^{2})^{-\frac{1}{2}})}{1-x^{2}}\)
\(\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1(\sqrt{1-x^{2}}+x^{2}(1-x^{2})^{-\frac{1}{2}})}{1-x^{2}}\)
\(=\frac{1}{(1-x^{2})^{\frac{1}{2}}}+\frac{1}{(1-x^{2})^{\frac{1}{2}}}+\frac{x^{2}}{(1-x^{2})^{\frac{3}{2}}}\)
\(=\frac{2}{(1-x^{2})^{\frac{1}{2}}}+\frac{x^{2}}{(1-x^{2})^{\frac{3}{2}}}\)
\(=\frac{2(1-x^{2})+x^{2}}{(1-x^{2})^{\frac{3}{2}}}\)
\(=\frac{2-x^{2}}{(1-x^{2})^{\frac{3}{2}}}\)
Question
Let \(f\) be a cubic polynomial function. Given that \(f(0)=2, f'(0)=-3, f(1)=f'(1)\) and \({f}”(-1)=6\), find \(f(x)\).
▶️Answer/Explanation
Ans
\(f(x)=ax^{3}+bx^{2}+cx+d\)
\(f'(x)=3ax^{2}+2bx+c\)
\({f}”(x)=6ax+2b\)
\(f(0)=2=d\)
\(f'(1)=f(1)\rightarrow a+b+c+2=3a+2b+c\)
\(2=2a+b\)
\(f'(0)=-3=c\)
\({f}”(-1)=6=-6a+2b\)
\(b=\frac{12}{5}, a=-\frac{1}{5}\)
\(f(x)=-\frac{1}{5}x^{3}+\frac{12}{5}x^{2}-3x+2\) (Accept \(a=-\frac{1}{5}, b=\frac{12}{5}, c=-3, d=2\))
Question
The following table shows the values of two functions \(f\) and \(g\) and their derivatives when \(x = 1\) and \(x = 0\).
(a) Find the derivative of \(\frac{3f(x)}{g(x)-1}\) when \(x=0\).
(b) Find the derivative of \(f(g(x)+2x)\) when \(x=1\).
▶️Answer/Explanation
Ans
(a) derivative= \(\frac{3f'(x)[g(x)-1]-3f(x)g'(x)}{[g(x)-1]^{2}}\)
when \(x=0\Rightarrow derivative = \frac{3(1)(-4-1)-3(4)(5)}{(-4-1)^{2}}\)
=\(-3\)
(b) derivative = \(f'(g(x)+2x)(g'(x)+2)\)
when \(x=1\) derivative =\(f'(-1+2)(2+2)\)
=\((3)(4)\)
=\(12\)