Question
The velocity v ms−1 of a particle at time t seconds, is given by \(v = 2t + \cos 2t\) , for \(0 \le t \le 2\) .
Write down the velocity of the particle when \(t = 0\) .
When \(t = k\) , the acceleration is zero.
(i) Show that \(k = \frac{\pi }{4}\) .
(ii) Find the exact velocity when \(t = \frac{\pi }{4}\) .
When \(t < \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\) and when \(t > \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\) .
Sketch a graph of v against t .
Let d be the distance travelled by the particle for \(0 \le t \le 1\) .
(i) Write down an expression for d .
(ii) Represent d on your sketch.
Answer/Explanation
Markscheme
\(v = 1\) A1 N1
[1 mark]
(i) \(\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2\) A1
\(\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = – 2\sin 2t\) A1A1
Note: Award A1 for coefficient 2 and A1 for \( – \sin 2t\) .
evidence of considering acceleration = 0 (M1)
e.g. \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0\) , \(2 – 2\sin 2t = 0\)
correct manipulation A1
e.g. \(\sin 2k = 1\) , \(\sin 2t = 1\)
\(2k = \frac{\pi }{2}\) (accept \(2t = \frac{\pi }{2}\) ) A1
\(k = \frac{\pi }{4}\) AG N0
(ii) attempt to substitute \(t = \frac{\pi }{4}\) into v (M1)
e.g. \(2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)\)
\(v = \frac{\pi }{2}\) A1 N2
[8 marks]
A1A1A2 N4
Notes: Award A1 for y-intercept at \((0{\text{, }}1)\) , A1 for curve having zero gradient at \(t = \frac{\pi }{4}\) , A2 for shape that is concave down to the left of \(\frac{\pi }{4}\) and concave up to the right of \(\frac{\pi }{4}\) . If a correct curve is drawn without indicating \(t = \frac{\pi }{4}\) , do not award the second A1 for the zero gradient, but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essential features need to be clear.
[4 marks]
(i) correct expression A2
e.g. \(\int_0^1 {(2t + \cos 2t){\rm{d}}t} \) , \(\left[ {{t^2} + \frac{{\sin 2t}}{2}} \right]_0^1\) , \(1 + \frac{{\sin 2}}{2}\) , \(\int_0^1 {v{\rm{d}}t} \)
(ii)
A1
Note: The line at \(t = 1\) needs to be clearly after \(t = \frac{\pi }{4}\) .
[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 \(f(x) = 3\sin \left( {\frac{\pi }{2}x} \right)\), for \(0 \leqslant x \leqslant 4\).
(i) Write down the amplitude of \(f\).
(ii) Find the period of \(f\).
On the following grid sketch the graph of \(f\).
Answer/Explanation
Markscheme
(i) 3 A1 N1
(ii) valid attempt to find the period (M1)
eg\(\,\,\,\,\,\)\(\frac{{2\pi }}{b},{\text{ }}\frac{{2\pi }}{{\frac{\pi }{2}}}\)
period \( = 4\) A1 N2
[3 marks]
A1A1A1A1 N4
[4 marks]
Question
The following diagram shows the graph of a function \(f\), with domain \( – 2 \leqslant x \leqslant 4\).
The points \(( – 2,{\text{ }}0)\) and \((4,{\text{ }}7)\) lie on the graph of \(f\).
Write down the range of \(f\).
Write down \(f(2)\);
Write down \({f^{ – 1}}(2)\).
On the grid, sketch the graph of \({f^{ – 1}}\).
Answer/Explanation
Markscheme
correct range (do not accept \(0 \leqslant x \leqslant 7\)) A1 N1
eg\(\,\,\,\,\,\)\([0,{\text{ }}7],{\text{ }}0 \leqslant y \leqslant 7\)
[1 mark]
\(f(2) = 3\) A1 N1
[1 mark]
\({f^{ – 1}}(2) = 0\) A1 N1
[1 mark]
A1A1A1 N3
Notes: Award A1 for both end points within circles,
A1 for images of \((2,{\text{ }}3)\) and \((0,{\text{ }}2)\) within circles,
A1 for approximately correct reflection in \(y = x\), concave up then concave down shape (do not accept line segments).
[3 marks]
Question
Consider a function f (x) , for −2 ≤ x ≤ 2 . The following diagram shows the graph of f.
Write down the value of f (0).
Write down the value of f −1 (1).
Write down the range of f −1.
On the grid above, sketch the graph of f −1.
Answer/Explanation
Markscheme
\(f\left( 0 \right) = – \frac{1}{2}\) A1 N1
[1 mark]
f −1 (1) = 2 A1 N1
[1 mark]
−2 ≤ y ≤ 2, y∈ [−2, 2] (accept −2 ≤ x ≤ 2) A1 N1
[1 mark]
A1A1A1A1 N4
Note: Award A1 for evidence of approximately correct reflection in y = x with correct curvature.
(y = x does not need to be explicitly seen)
Only if this mark is awarded, award marks as follows:
A1 for both correct invariant points in circles,
A1 for the three other points in circles,
A1 for correct domain.
[4 marks]
Question
The following diagram shows the graph of a function \(f\), for −4 ≤ x ≤ 2.
On the same axes, sketch the graph of \(f\left( { – x} \right)\).
Another function, \(g\), can be written in the form \(g\left( x \right) = a \times f\left( {x + b} \right)\). The following diagram shows the graph of \(g\).
Write down the value of a and of b.
Answer/Explanation
Markscheme
A2 N2
[2 marks]
recognizing horizontal shift/translation of 1 unit (M1)
eg b = 1, moved 1 right
recognizing vertical stretch/dilation with scale factor 2 (M1)
eg a = 2, y ×(−2)
a = −2, b = −1 A1A1 N2N2
[4 marks]
Question
Consider a function \(f\). The line L1 with equation \(y = 3x + 1\) is a tangent to the graph of \(f\) when \(x = 2\)
Let \(g\left( x \right) = f\left( {{x^2} + 1} \right)\) and P be the point on the graph of \(g\) where \(x = 1\).
Write down \(f’\left( 2 \right)\).
Find \(f\left( 2 \right)\).
Show that the graph of g has a gradient of 6 at P.
Let L2 be the tangent to the graph of g at P. L1 intersects L2 at the point Q.
Find the y-coordinate of Q.
Answer/Explanation
Markscheme
recognize that \(f’\left( x \right)\) is the gradient of the tangent at \(x\) (M1)
eg \(f’\left( x \right) = m\)
\(f’\left( 2 \right) = 3\) (accept m = 3) A1 N2
[2 marks]
recognize that \(f\left( 2 \right) = y\left( 2 \right)\) (M1)
eg \(f\left( 2 \right) = 3 \times 2 + 1\)
\(f\left( 2 \right) = 7\) A1 N2
[2 marks]
recognize that the gradient of the graph of g is \(g’\left( x \right)\) (M1)
choosing chain rule to find \(g’\left( x \right)\) (M1)
eg \(\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u’ = 2x\)
\(g’\left( x \right) = f’\left( {{x^2} + 1} \right) \times 2x\) A2
\(g’\left( 1 \right) = 3 \times 2\) A1
\(g’\left( 1 \right) = 6\) AG N0
[5 marks]
at Q, L1 = L2 (seen anywhere) (M1)
recognize that the gradient of L2 is g’(1) (seen anywhere) (M1)
eg m = 6
finding g (1) (seen anywhere) (A1)
eg \(g\left( 1 \right) = f\left( 2 \right),\,\,g\left( 1 \right) = 7\)
attempt to substitute gradient and/or coordinates into equation of a straight line M1
eg \(y – g\left( 1 \right) = 6\left( {x – 1} \right),\,\,y – 1 = g’\left( 1 \right)\left( {x – 7} \right),\,\,7 = 6\left( 1 \right) + {\text{b}}\)
correct equation for L2
eg \(y – 7 = 6\left( {x – 1} \right),\,\,y = 6x + 1\) A1
correct working to find Q (A1)
eg same y-intercept, \(3x = 0\)
\(y = 1\) A1 N2
[7 marks]