Question
Let \(f(x) = 2{x^2} + 4x – 6\) .
Express \(f(x)\) in the form \(f(x) = 2{(x – h)^2} + k\) .[3]
Write down the equation of the axis of symmetry of the graph of f .[1]
Express \(f(x)\) in the form \(f(x) = 2(x – p)(x – q)\) .[2]
Answer/Explanation
Markscheme
evidence of obtaining the vertex (M1)
e.g. a graph, \(x = – \frac{b}{{2a}}\) , completing the square
\(f(x) = 2{(x + 1)^2} – 8\) A2 N3
[3 marks]
\(x = – 1\) (equation must be seen) A1 N1
[1 mark]
\(f(x) = 2(x – 1)(x + 3)\) A1A1 N2
[2 marks]
Question
Let \(f(x) = {x^3} – 4x + 1\) .
Expand \({(x + h)^3}\) .
Use the formula \(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) – f(x)}}{h}\) to show that the derivative of \(f(x)\) is \(3{x^2} – 4\) .
The tangent to the curve of f at the point \({\text{P}}(1{\text{, }} – 2)\) is parallel to the tangent at a point Q. Find the coordinates of Q.
The graph of f is decreasing for \(p < x < q\) . Find the value of p and of q.
Write down the range of values for the gradient of \(f\) .
Answer/Explanation
Markscheme
attempt to expand (M1)
\({(x + h)^3} = {x^3} + 3{x^2}h + 3x{h^2} + {h^3}\) A1 N2
[2 marks]
evidence of substituting \(x + h\) (M1)
correct substitution A1
e.g. \(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{{(x + h)}^3} – 4(x + h) + 1 – ({x^3} – 4x + 1)}}{h}\)
simplifying A1
e.g. \(\frac{{({x^3} + 3{x^2}h + 3x{h^2} + {h^3} – 4x – 4h + 1 – {x^3} + 4x – 1)}}{h}\)
factoring out h A1
e.g. \(\frac{{h(3{x^2} + 3xh + {h^2} – 4)}}{h}\)
\(f'(x) = 3{x^2} – 4\) AG N0
[4 marks]
\(f'(1) = – 1\) (A1)
setting up an appropriate equation M1
e.g. \(3{x^2} – 4 = – 1\)
at Q, \(x = – 1,y = 4\) (Q is \(( – 1{\text{, }}4)\)) A1 A1
[4 marks]
recognizing that f is decreasing when \(f'(x) < 0\) R1
correct values for p and q (but do not accept \(p = 1.15{\text{, }}q = – 1.15\) ) A1A1 N1N1
e.g. \(p = – 1.15{\text{, }}q = 1.15\) ; \( \pm \frac{2}{{\sqrt 3 }}\) ; an interval such as \( – 1.15 \le x \le 1.15\)
[3 marks]
\(f'(x) \ge – 4\) , \(y \ge – 4\) , \(\left[ { – 4,\infty } \right[\) A2 N2
[2 marks]
Question
Let \(f(x) = 5\cos \frac{\pi }{4}x\) and \(g(x) = – 0.5{x^2} + 5x – 8\) for \(0 \le x \le 9\) .
On the same diagram, sketch the graphs of f and g .
Consider the graph of \(f\) . Write down
(i) the x-intercept that lies between \(x = 0\) and \(x = 3\) ;
(ii) the period;
(iii) the amplitude.
Consider the graph of g . Write down
(i) the two x-intercepts;
(ii) the equation of the axis of symmetry.
Let R be the region enclosed by the graphs of f and g . Find the area of R.
Answer/Explanation
Markscheme
A1A1A1 N3
Note: Award A1 for f being of sinusoidal shape, with 2 maxima and one minimum, A1 for g being a parabola opening down, A1 for two intersection points in approximately correct position.
[3 marks]
(i) \((2{\text{, }}0)\) (accept \(x = 2\) ) A1 N1
(ii) \({\text{period}} = 8\) A2 N2
(iii) \({\text{amplitude}} = 5\) A1 N1
[4 marks]
(i) \((2{\text{, }}0)\) , \((8{\text{, }}0)\) (accept \(x = 2\) , \(x = 8\) ) A1A1 N1N1
(ii) \(x = 5\) (must be an equation) A1 N1
[3 marks]
METHOD 1
intersect when \(x = 2\) and \(x = 6.79\) (may be seen as limits of integration) A1A1
evidence of approach (M1)
e.g. \(\int {g – f} \) , \(\int {f(x){\rm{d}}x – \int {g(x){\rm{d}}x}}\) , \(\int_2^{6.79} {\left( {( – 0.5{x^2} + 5x – 8) – \left( {5\cos \frac{\pi }{4}x} \right)} \right)}\)
\({\text{area}} = 27.6\) A2 N3
METHOD 2
intersect when \(x = 2\) and \(x = 6.79\) (seen anywhere) A1A1
evidence of approach using a sketch of g and f , or \(g – f\) . (M1)
e.g. area = \(A + B – C\) , \(12.7324 + 16.0938 – 1.18129 \ldots \)
\({\text{area}} = 27.6\) A2 N3
[5 marks]
Question
Let \(f(x) = 2{x^2} – 8x – 9\) .
(i) Write down the coordinates of the vertex.
(ii) Hence or otherwise, express the function in the form \(f(x) = 2{(x – h)^2} + k\) .
Solve the equation \(f(x) = 0\) .
Answer/Explanation
Markscheme
(i) \((2{\text{, }} – 17)\) or \(x = 2\) , \(y = – 17\) A1A1 N2
(ii) evidence of valid approach (M1)
e.g. graph, completing the square, equating coefficients
\(f(x) = 2{(x – 2)^2} – 17\) A1 N2
[4 marks]
evidence of valid approach (M1)
e.g. graph, quadratic formula
\( – 0.9154759 \ldots \) , \(4.915475 \ldots \)
\(x = – 0.915\) , \(4.92\) A1A1 N3
[3 marks]
Question
[Maximum mark: 4] [without GDC]
The diagram represents the graph of the function \(f : x \mapsto (x – p)(x – q)\).
(a) Write down the values of \(p\) and \(q\).
(b) The function has a minimum value at the point C. Find the x-coordinate of C.
Answer/Explanation
Ans.
(a) \(p=-\frac{1}{2},q=2\) or vice versa
b) By symmetry C is midway between p, q ⇒ x-coordinate is \(\frac{-\frac{1}{2}+2}{2}=\frac{3}{4}\)