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2.12.6 Quadratic Functions, SPM Practice (Paper 1 Question 26 – 30)


Question 26 (SPM 2018 – 3 marks):
Faizal has a rectangular plywood with a dimension 3x metre in length and 2x metre in width. He cuts part of the plywood into a square shape with sides of x metre to make a table surface.
Find the range of values of x if the remaining area of the plywood is at least (x2 + 4) metre2.

Solution:



Area of plywood – area of square ≥ (x2 + 4)
3x(2x) – x2x2 + 4
6x2x2x2 ≥ 4
4x2 ≥ 4
x2 – 1 ≥ 0
(x + 1)(x – 1) ≥ 0
x ≤ –1 or x ≥ 1
Thus, x ≥ 1 (length is > 0)





Question 27 (SPM 2018 – 3 marks):
It is given that the curve y = (p – 2)x2x + 7, where p is a constant, intersects with the straight line y = 3x + 5 at two points.
Find the range of values of p.

Solution:
y=( p2 ) x 2 x+7 ……… ( 1 ) y=3x+5 ……………………… ( 2 ) Substitute ( 1 ) into ( 2 ): ( p2 ) x 2 x+7=3x+5 ( p2 ) x 2 4x+2=0 a=( p2 ), b=4, c=2 b 2 4ac>0 ( 4 ) 2 4( p2 )( 2 )>0 168p+16>0 8p>32 8p<32 p<4


Question 28 (SPM 2018 – 3 marks):
It is given that the quadratic equation hx2 – 3x + k = 0, where h and k are constants has roots β and 2β.
Express h in terms of k.

Solution:
h x 2 3x+k=0 a=h, b=3, c=k SOR= b a = ( 3 ) h = 3 h POR= c a = k h Given roots=β and 2β. SOR=β+2β=3β;  POR=β( 2β )=2 β 2 3 h =3β β= 1 h  ………….. ( 1 ) k h =2 β 2  ……….. ( 2 ) Substitute ( 1 ) into ( 2 ): k h =2 ( 1 h ) 2 k h = 2 h 2 h 2 h = 2 k h= 2 k


Question 29  (SPM 2015 – 3 marks):
The graph of a quadratic function f(x) =px2 – 8x + q, where p and q are constants, has a maximum point.
(a) Given p is an integer such that -2 < p < 2, state the value of p.
(b) Using the answer from 3(a), find the value of q when the graph touches the x-axis at one point.
[3 marks]


Answer:
(a) p = -1

(b)
$$ \begin{aligned} & f(x)=-x^2-8 x+q \\ & a=-1, b=-8, c=q \\ & b^2-4 a c=0 \\ & (-8)^2-4(-1) q=0 \\ & 64+4 q=0 \\ & 4 q=-64 \\ & q=-16 \end{aligned} $$


Question 30 (SPM 2015 – 2 marks):
It is given -7 is one of the roots of the quadratic equation (x + k)2 = 16, where k is a constant.
Find the values of k.
[2 marks]

Answer:
$$ \begin{aligned} (x+k)^2= & 16 \\ \sqrt{(-7+k)^2} & = \pm \sqrt{16} \\ -7+k & = \pm 4 \\ -7+k & =4 \\ k & =4+7=11 \\ -7+k & =-4 \\ k & =-4+7=3 \end{aligned} $$

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