SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 1 & 2)


Question 1:
Given g( x )= 2x5 3xp , xq

(a)(i)
Express p in terms of q and find g-1 (x) in term of p.
(a)(ii)   Given  g n1 ( x )= px5 3x2 , xk ,
determine the value of n by comparing the g-1 (x) above with the given gn-1 (x).
Next, find the value of k. [5 marks]

(b)
What is the condition on p so that g = g-1 ? [1 mark]


Solution:
(a)(i)
3xp0 3xp x p 3 as xq q= p 3 p=3q g(x)= 2x5 3xp Let  g 1 (x)=y g(y)=x 2y5 3yp =x 2y5=3xypx y( 23x )=px+5 y( 3x2 )=px5 y= px5 3x2 g 1 (x)= px5 3x2


(a)(ii)
g n1 (x)= g 1 (x)= px5 3x2 n1=1 n=0 3x20 x 2 3 k= 2 3

(b)
g(x)= 2x5 3xp g 1 (x)= px5 3x2 g(x)= g 1 (x) if p=2

Question 2:
Find the range of values of x for 5 < 2x2 + x + 4 and 2x2 + x + 4 ≤ 10.
Hence, solve the inequality 5 < 2x2+ x + 4 ≤ 10.
[4 marks]

Solution:
5 < 2x2 + x + 4
0 < 2x2 + x – 1
2x2 + x – 1 > 0
(2x – 1)(x + 1) > 0
x < -1 or x > ½

2x2 + x + 4 ≤ 10
2x2 + x – 6 ≤ 0
(2x – 3)(x + 2) ≤ 0
-2 ≤ x ≤ 3/2

x < -1, x > ½ or -2 ≤ x ≤ 3/2
-2 ≤ x < -1   or  ½ < x ≤ 3/2

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