SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 3 & 4)


Question 3:
(a) Find the value of p and q for the equation (√2 – 1)(p + q√2) = 5 – √8. [3 marks]

(b) Find the value of p which satisfies the following equation:
3 log p pq + 3 log q pq +2=5p [3 marks]


Solution:
(a)
( 2 1 )( p+q 2 )=5 8 p+q 2 = 5 8 2 1 p+q 2 = 5 8 2 1 × 2 +1 2 +1 p+q 2 = 5 2 +5 8×2 8 1 p+q 2 =5 2 +54 4×2 p+q 2 =5 2 +12 2 p+q 2 =1+3 2 p=1, q=3


(b)
3 log p pq + 3 log q pq +2=5p 3 log pq pq log pq p + 3 log pq pq log pq q =5p2 3 log pq p+3 log pq q log pq pq =5p2 3 log pq p+3 log pq q=5p2 log pq ( pq ) 3 =5p2 3 log pq ( pq )=5p2 3=5p2 5p=5 p=1


Question 4:
Determine whether the following equations are system of linear equations in three variables or not. Explain. [1 mark]
2x + y = z2
4x – 2y + 3z = 6
11y + 5xz = 3

Solve the following system of linear equations and explain the result of the findings. [4 marks]
y – 7z = -2
xy + 5z = 2
-2x + 2y – 10z = 6

Solution
:

(a) No, because the power for one of the equation is 2.

(b)
y – 7z = -2
y = -2 + 7z ………… (1)

xy + 5z = 2
2x – 2y + 10z = 4 ………… (2)

-2x + 2y – 10z = 6 ………… (3)

(2) + (3),
0 = 10

Since 0 ≠ 10, therefore there is no solution for the linear equations

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