Page 16 - TOP ONE ADDMATHS TG4

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2
(a) Given f(x) = 5x – 9x – 2, find the range of values (b) (i) Find the range of values of x for
2
of x such that f(x) is always (x + 2) , 11x + 12.
2
Diberi bahawa f(x) = 5x – 9x – 2, cari julat nilai x dengan Cari julat nilai x bagi (x + 2) , 11x + 12.
2
keadaan f(x) sentiasa (ii) Given the quadratic equation x(3k – x) = 4
(i) positive / positif, does not have any real roots, find the range
(ii) negative / negatif. of values of p.
Diberi persamaan kuadratik x(3k – x) = 4 tidak
2
(i) 5x – 9x – 2 . 0 mempunyai punca nyata, cari julat nilai p.
(x – 2)(5x + 1) . 0
2
When (x – 2)(5x + 1) = 0, x = 2 or x = – 1 (i) x + 4x + 4 , 11x + 12
5 x – 7x – 8 , 0
2
(x – 8)(x + 1) , 0
x When (x – 8)(x + 1) = 0, x = 8 or x = –1
–1 1
4
1
∴ For 5x – 9x – 2 . 0, x , – or x . 2. x
2
5 –1 8
2
(ii) 5x – 9x – 2 , 0 2
(x – 2)(5x + 1) , 0 ∴ For x – 7x – 8 , 0, –1 , x , 8.
When (x – 2)(5x + 1) = 0, x = 2 or x = – 1 (ii) 3kx – x = 4
2
5
x – 3kx + 4 = 0
2
No real roots, b – 4ac , 0
2
x (–3k) – 4(1)(4) , 0
2
1 2
– 9k – 16 , 0
2
5
(3k – 4)(3k + 4) , 0
1
2
∴ For 5x – 9x – 2 , 0, – , x , 2.
5 4 4
When (3k – 4)(3k + 4) = 0, k = or k = –
3 3
x
4 4
–
3 3
4
4
∴ For 9k – 16 , 0, – , k , .
2
3 3
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(c) Hashim threw a ball. The height of the ball is given by the function h(x) = –x + 6x + 1, where h is the
height of the ball from the ground, in m, and x is the horizontal distance of the ball from Hashim’s
position, in m. Find Daily Application
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Hashim membaling sebiji bola. Tinggi bola itu diberi oleh fungsi h(x) = –x + 6x + 1, dengan keadaan h ialah tinggi bola
dari permukaan tanah, dalam m, dan x ialah jarak mengufuk bola dari kedudukan Hashim, dalam m. Cari
(i) the maximum height, in m, of the ball thrown by Hashim,
tinggi maksimum, dalam m, bola yang dibaling oleh Hashim,
(ii) the horizontal distance between Hashim and the ball, in m, when the ball touches the ground.
jarak mengufuk di antara Hashim dan bola, dalam m, apabila bola itu menyentuh permukaan tanah.
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(i) h(x) = –x + 6x + 1 (ii) h(x) = 0
2
= –(x – 6x – 1) –x + 6x + 1 = 0 2
2
2 

–6
–6
= – x – 6x +   2 – 1 –   2 x = –b ± √b – 4ac
2
2a
2

2
= –[x – 6x + (–3) – 1 – (–3) ] = –6 ± √6 – 4(–1)(1)
2
2
2
= –[(x – 3) – 10] 2(–1)
2
2
= –(x – 3) + 10 = –6 ± √40
–2
–6 + √40 –6 – √40
a = –1 , 0, so the graph has a maximum point. x = or x =
–2 –2
The maximum value is 10 when x = 3. = –0.16 (ignore) = 6.16
Hence, the maximum height is 10 m.
Hence, the horizontal distance is 6.16 m.
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