Solve the inequality $f(x) > x$.

Find $\int {f(x){\text{d}}x}$.

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]