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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

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