Find the value of \(p\).

Find the value of \(a\).

The line \(y = kx - 5\) is a tangent to the curve of \(f\). Find the values of \(k\).

METHOD 1 (using x-intercept)

determining that 3 is an \(x\)-intercept     (M1)

eg\(\,\,\,\,\,\)\(x - 3 = 0\),

valid approach     (M1)

eg\(\,\,\,\,\,\)\(3 - 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5\)

\(p = 2\)     A1     N2

METHOD 2 (expanding f (x)) 

correct expansion (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a{x^2} - a(3 + p)x + 3ap,{\text{ }}{x^2} - (3 + p)x + 3p\)

valid approach involving equation of axis of symmetry     (M1)

eg\(\,\,\,\,\,\)\(\frac{{ - b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}\)

\(p = 2\)     A1     N2

METHOD 3 (using derivative)

correct derivative (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a(2x - 3 - p),{\text{ }}2x - 3 - p\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f’(2.5) = 0\)

\(p = 2\)     A1     N2

[3 marks]

attempt to substitute \((0,{\text{ }} - 6)\)     (M1)

eg\(\,\,\,\,\,\)\( - 6 = a(0 - 2)(0 - 3),{\text{ }}0 = a( - 8)( - 9),{\text{ }}a{(0)^2} - 5a(0) + 6a =  - 6\)

correct working     (A1)

eg\(\,\,\,\,\,\)\( - 6 = 6a\)

\(a =  - 1\)     A1     N2

[3 marks]

METHOD 1 (using discriminant)

recognizing tangent intersects curve once     (M1)

recognizing one solution when discriminant = 0     M1

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx - 5 =  - {x^2} + 5x - 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} - 5x + kx + 1 = 0\)

correct discriminant (if seen explicitly, not just in quadratic formula)     A1

eg\(\,\,\,\,\,\)\({(k - 5)^2} - 4,{\text{ }}25 - 10k + {k^2} - 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(k - 5 =  \pm 2,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

METHOD 2 (using derivatives)

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx - 5 =  - {x^2} + 5x - 6\)

recognizing derivative/slope are equal     (M1)

eg\(\,\,\,\,\,\)\(f’ = {m_T},{\text{ }}f' = k\)

correct derivative of \(f\)     (A1)

eg\(\,\,\,\,\,\)\( - 2x + 5\)

attempt to set up equation in terms of either \(x\) or \(k\)     M1

eg\(\,\,\,\,\,\)\(( - 2x + 5)x - 5 =  - {x^2} + 5x - 6,{\text{ }}k\left( {\frac{{5 - k}}{2}} \right) - 5 =  - {\left( {\frac{{5 - k}}{2}} \right)^2} + 5\left( {\frac{{5 - k}}{2}} \right) - 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} - 1 = 0,{\text{ }}{k^2} - 10k + 21 = 0\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(x =  \pm 1,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

[8 marks]