Question
Consider the function f with second derivative \(f”(x) = 3x – 1\) . The graph of f has a minimum point at A(2, 4) and a maximum point at \({\rm{B}}\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)\) .
Use the second derivative to justify that B is a maximum.
Given that \(f'(x) = \frac{3}{2}{x^2} – x + p\) , show that \(p = – 4\) .
Find \(f(x)\) .
Answer/Explanation
Markscheme
substituting into the second derivative M1
e.g. \(3 \times \left( { – \frac{4}{3}} \right) – 1\)
\(f”\left( { – \frac{4}{3}} \right) = – 5\) A1
since the second derivative is negative, B is a maximum R1 N0
[3 marks]
setting \(f'(x)\) equal to zero (M1)
evidence of substituting \(x = 2\) (or \(x = – \frac{4}{3}\) ) (M1)
e.g. \(f'(2)\)
correct substitution A1
e.g. \(\frac{3}{2}{(2)^2} – 2 + p\) , \(\frac{3}{2}{\left( { – \frac{4}{3}} \right)^2} – \left( { – \frac{4}{3}} \right) + p\)
correct simplification
e.g. \(6 – 2 + p = 0\) , \(\frac{8}{3} + \frac{4}{3} + p = 0\) , \(4 + p = 0\) A1
\(p = – 4\) AG N0
[4 marks]
evidence of integration (M1)
\(f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + c\) A1A1A1
substituting (2, 4) or \(\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)\) into their expression (M1)
correct equation A1
e.g. \(\frac{1}{2} \times {2^3} – \frac{1}{2} \times {2^2} – 4 \times 2 + c = 4\) , \(\frac{1}{2} \times 8 – \frac{1}{2} \times 4 – 4 \times 2 + c = 4\) , \(4 – 2 – 8 + c = 4\)
\(f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + 10\) A1 N4
[7 marks]
Question
The graph of the function \(y = f(x)\) passes through the point \(\left( {\frac{3}{2},4} \right)\) . The gradient function of f is given as \(f'(x) = \sin (2x – 3)\) . Find \(f(x)\) .
Answer/Explanation
Markscheme
evidence of integration
e.g. \(f(x) = \int {\sin (2x – 3){\rm{d}}x} \) (M1)
\( = – \frac{1}{2}\cos (2x – 3) + C\) A1A1
substituting initial condition into their expression (even if C is missing) M1
e.g. \(4 = – \frac{1}{2}\cos 0 + C\)
\(C = 4.5\) (A1)
\(f(x) = – \frac{1}{2}\cos (2x – 3) + 4.5\) A1 N5
[6 marks]
Question
Let \(f'(x) = 3{x^2} + 2\) . Given that \(f(2) = 5\) , find \(f(x)\) .
Answer/Explanation
Markscheme
evidence of anti-differentiation (M1)
e.g. \(\int {f'(x)} \) , \(\int {(3{x^2} + 2){\rm{d}}x} \)
\(f(x) = {x^3} + 2x + c\) (seen anywhere, including the answer) A1A1
attempt to substitute (2, 5) (M1)
e.g. \(f(2) = {(2)^3} + 2(2)\) , \(5 = 8 + 4 + c\)
finding the value of c (A1)
e.g. \(5 = 12 + c\) , \(c = – 7\)
\(f(x) = {x^3} + 2x – 7\) A1 N5
[6 marks]
Question
Let \(f(x) = \int {\frac{{12}}{{2x – 5}}} {\rm{d}}x\) , \(x > \frac{5}{2}\) . The graph of \(f\) passes through (\(4\), \(0\)) .
Find \(f(x)\) .
Answer/Explanation
Markscheme
attempt to integrate which involves \(\ln \) (M1)
eg \(\ln (2x – 5)\) , \(12\ln 2x – 5\) , \(\ln 2x\)
correct expression (accept absence of \(C\))
eg \(12\ln (2x – 5)\frac{1}{2} + C\) , \(6\ln (2x – 5)\) A2
attempt to substitute (4,0) into their integrated f (M1)
eg \(0 = 6\ln (2 \times 4 – 5)\) , \(0 = 6\ln (8 – 5) + C\)
\(C = – 6\ln 3\) (A1)
\(f(x) = 6\ln (2x – 5) – 6\ln 3\) \(\left( { = 6\ln \left( {\frac{{2x – 5}}{3}} \right)} \right)\) (accept \(6\ln (2x – 5) – \ln {3^6}\) ) A1 N5
Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve \(\ln\).
[6 marks]
Question
A rocket moving in a straight line has velocity \(v\) km s–1 and displacement \(s\) km at time \(t\) seconds. The velocity \(v\) is given by \(v(t) = 6{{\rm{e}}^{2t}} + t\) . When \(t = 0\) , \(s = 10\) .
Find an expression for the displacement of the rocket in terms of \(t\) .
Answer/Explanation
Markscheme
evidence of anti-differentiation (M1)
eg \(\int {(6{{\rm{e}}^{2t}} + t)} \)
\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + C\) A2A1
Note: Award A2 for \(3{{\rm{e}}^{2t}}\) , A1 for \(\frac{{{t^2}}}{2}\) .
attempt to substitute (\(0\), \(10\)) into their integrated expression (even if \(C\) is missing) (M1)
correct working (A1)
eg \(10 = 3 + C\) , \(C = 7\)
\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + 7\) A1 N6
Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration which must involve a power of \({\rm{e}}\).
[7 marks]