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]

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