Page 4 - 1202 Question Bank Mathematics Form 4

MUST

KNOW Common Mistakes

Number Bases Quadratic Expression
1
State the value of the underlined digit. Determine whether 7k + k is a quadratic expression in one
2
2
549
5 variable.
Correct Wrong Correct Wrong
= 1132 ©PAN ASIA PUBLICATIONS
Place 5 2 5 1 5 0 Place 5 3 5 2 5 1 Not a quadratic expression in A quadratic expression in
value value one variable because there is a one variable.
variable with a power which is
Number 4 Number 4 not a whole number.
4 × 5 = 100
2
Ensure the power of place Determine whether 2x + 5 is a quadratic expression in one
value is start with 0. variable.
4 × 5 = 20 Correct Wrong
1
Not a quadratic expression in A quadratic expression in
one variable because the highest one variable.
power of the variable is not two.

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Conversion of Number Bases Factorisation Method
2
Calculate the sum of value of digit 3 in base 10 for the number Solve the quadratic equation (3x – 2) = 9.
1302 and 2356 . Correct Wrong
6 9
2
Correct Wrong (3x – 2) = 9 3x – 2 = 3
2
For 1302 ˜ Digit value Sum of the value for digit 3 9x – 12x + 4 − 9 = 0 3x = 5
2
6 9x – 12x – 5 = 0 5
= 3 × 6 2 = 6 + 9 2 (3x – 5)(3x + 1) = 0 x = 3
2
= 108 10 = 117 10 5 1
For 2356 ˜ Digit value x = or x = –
3
3
9
= 3 × 9 2
2
= 243 10 Determine the roots of x + 7x + 6 = 14 by factorisation method.
Sum of the value for digit 3 Correct Wrong
= 108 + 243 10 x + 7x + 6 = 14 x + 7x + 6 = 14
2
2
10
= 351 10 x + 7x – 8 = 0 (x + 6)(x + 1) = 14
2
(x + 8)(x – 1) = 0 x + 6 = 14 or x + 1 = 14
x = –8 or x = 1 x = 8 x = 13
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Conversion of Number Bases Quadratic Functions
It is given 1m32 = 247 , find the place value for m. Diagram shows the graph of a y
8
5
Correct Wrong quadratic function. Determine (–2, 6)
the quadratic function in the
247 8 5 167 Remainder form of y = ax + bx + c.
2
= 2 × 8 + 4 × 8 × 7 × 8 0 5 33 – 2
2
1
= 167 10 5 6 – 3 –3 0 1 x
5 5 1 – 1 Correct Wrong
0 1 y = a(x + 3)(x – 1) y = (x + 3)(x – 1)
º 247 = 1132 5 y = a(x + 3x – x – 3) y = x + 2x – 3
2
2
8
2
Place value for m Place value for m = 1 y = a(x + 2x – 3)…a
= 5 2 Substitute (–2, 6) into a:
2
= 25 6 = a[(–2) + 2(–2) – 3]
a = –2
Thus, y = –2(x + 2x – 3)
2
y = –2x – 4x + 6
2
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Must Know 1202QB Maths Form 4.indd 2 21/02/2022 10:48 AM