Date | May 2014 | Marks available | 5 | Reference code | 14M.2.hl.TZ1.6 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Let \(f(x) = x{(x + 2)^6}\).
Solve the inequality \(f(x) > x\).
Find \(\int {f(x){\text{d}}x} \).
Markscheme
METHOD 1
sketch showing where the lines cross or zeros of \(y = x{(x + 2)^6} - x\) (M1)
\(x = 0\) (A1)
\(x = - 1\) and \(x = - 3\) (A1)
the solution is \( - 3 < x < - 1\) or \(x > 0\) A1A1
Note: Do not award either final A1 mark if strict inequalities are not given.
METHOD 2
separating into two cases \(x > 0\) and \(x < 0\) (M1)
if \(x > 0\) then \({(x + 2)^6} > 1 \Rightarrow \) always true (M1)
if \(x < 0\) then \({(x + 2)^6} < 1 \Rightarrow - 3 < x < - 1\) (M1)
so the solution is \( - 3 < x < - 1\) or \(x > 0\) A1A1
Note: Do not award either final A1 mark if strict inequalities are not given.
METHOD 3
\(f(x) = {x^7} + 12{x^6} + 60{x^5} + 160{x^4} + 240{x^3} + 192{x^2} + 64x\) (A1)
solutions to \({x^7} + 12{x^6} + 60{x^5} + 160{x^4} + 240{x^3} + 192{x^2} + 63x = 0\) are (M1)
\(x = 0,{\text{ }}x = - 1\) and \(x = - 3\) (A1)
so the solution is \( - 3 < x < - 1\) or \(x > 0\) A1A1
Note: Do not award either final A1 mark if strict inequalities are not given.
METHOD 4
\(f(x) = x\) when \(x{(x + 2)^6} = x\)
either \(x = 0\) or \({(x + 2)^6} = 1\) (A1)
if \({(x + 2)^6} = 1\) then \(x + 2 = \pm 1\) so \(x = - 1\) or \(x = - 3\) (M1)(A1)
the solution is \( - 3 < x < - 1\) or \(x > 0\) A1A1
Note: Do not award either final A1 mark if strict inequalities are not given.
[5 marks]
METHOD 1 (by substitution)
substituting \(u = x + 2\) (M1)
\({\text{d}}u = {\text{d}}x\)
\(\int {(u - 2){u^6}{\text{d}}u} \) M1A1
\( = \frac{1}{8}{u^8} - \frac{2}{7}{u^7}( + c)\) (A1)
\( = \frac{1}{8}{(x + 2)^8} - \frac{2}{7}{(x + 2)^7}( + c)\) A1
METHOD 2 (by parts)
\(u = x \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = 1,{\text{ }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = {(x + 2)^6} \Rightarrow v = \frac{1}{7}{(x + 2)^7}\) (M1)(A1)
\(\int {x{{(x + 2)}^6}{\text{d}}x = \frac{1}{7}x{{(x + 2)}^7} - \frac{1}{7}\int {{{(x + 2)}^7}{\text{d}}x} } \) M1
\( = \frac{1}{7}x{(x + 2)^7} - \frac{1}{{56}}{(x + 2)^8}( + c)\) A1A1
METHOD 3 (by expansion)
\(\int {f(x){\text{d}}x = \int {\left( {{x^7} + 12{x^6} + 60{x^5} + 160{x^4} + 240{x^3} + 192{x^2} + 64x} \right){\text{d}}x} } \) M1A1
\( = \frac{1}{8}{x^8} + \frac{{12}}{7}{x^7} + 10{x^6} + 32{x^5} + 60{x^4} + 64{x^3} + 32{x^2}( + c)\) M1A2
Note: Award M1A1 if at least four terms are correct.
[5 marks]