Model Test – Solved ഉത്തരങ്ങൾ
Every question of the model paper, in the same order, with the full working, the marking scheme and a step-by-step animation for every longer question. Mark your own answers honestly.
The highest power of x that appears is x3, so the degree (കൃത്യങ്കം) is 3.
Marks: 1 mark.
Review: Page 8 · Degree
Put 2 in place of x: p(2) = 3 × 2 + 4 = 6 + 4 = 10.
Marks: 1 mark.
Review: Page 5 · The notation p(x)
In (c) x is only squared and multiplied by the fixed number √3. (a) and (d) take the reciprocal of x, (b) takes the square root of x.
Marks: 1 mark for (c).
Review: Page 8 · Polynomial or not?
The term with x2 is −7x2; its coefficient (ഗുണകം) is −7 – the sign belongs to the coefficient.
Marks: 1 mark (½ if the sign is missed).
Review: Page 8 · Coefficients
With x = 0 every term containing x becomes 0, and only the number without x is left. E.g. p(x) = 2x2 − 5x + 3 gives p(0) = 3.
Marks: 1 mark.
Review: Page 10 · p(0)
4 × (x + 1) = 4x + 4.
Marks: 1 mark.
Review: Page 5 · Perimeter as a rule
x × x + 3x + 2x + 2 × 3 = x2 + 5x + 6.
Marks: 1 mark.
Review: Page 3 · Multiplying brackets
a, b, c are fixed numbers (the coefficients) and a ≠ 0 (രണ്ടാംകൃതി ബഹുപദം).
Marks: 1 mark.
Review: Page 8 · General forms
p(0) = 0 − 0 + 1 = 1. p(1) = 2 − 3 + 1 = 0. p(−1) = 2 × 1 − 3 × (−1) + 1 = 2 + 3 + 1 = 6. p(2) = 2 × 4 − 6 + 1 = 3.
Marks: ½ mark for each value.
The old perimeter is 2 × (6 + 4) = 20 cm. Each of the four sides grows by x, so 4x is added: p(x) = 4x + 20. (Or 2((6 + x) + (4 + x)) = 4x + 20.) p(2½) = 4 × 2½ + 20 = 10 + 20 = 30 cm.
Marks: 1 mark for p(x), 1 mark for p(2½).
(i) By Pythagoras the diagonal is √(x2 + x2) = √(2x2) = √2 x. The only operation on x is multiplying by the fixed number √2, so it is a polynomial (degree 1). (ii) The area is (√2 x)2 = 2x2, a polynomial of degree 2.
Marks: 1 mark each (½ for the expression, ½ for the reason).
Review: Page 8 · ½x² + √2x is a polynomial · Page 4 · Pythagoras
(x + √x)(x − √x) = x2 − x√x + √x·x − (√x)2 = x2 − x. The square roots cancel, leaving only x2 and x: a polynomial of degree 2 (for x ≥ 0, where √x exists).
Marks: 1 mark for multiplying out, 1 mark for x2 − x and degree 2.
Review: Page 9 · Exercise 2, Q2(iii) · Page 3 · (a + b)(a − b)
p(1) = 1 + 2 + 5 + 3 = 11, the sum of the coefficients. p(10) = 1000 + 200 + 50 + 3 = 1253: the coefficients 1, 2, 5, 3 become the digits, because each power of 10 is a place value.
Marks: ½ mark for each value, 1 mark for the pattern.
(i) The new sides are x + 5 and x + 3: a(x) = (x + 5)(x + 3) = x2 + 3x + 5x + 15 = x2 + 8x + 15. (ii) a(1) = 1 + 8 + 15 = 24 (check 6 × 4); a(2) = 4 + 16 + 15 = 35 (check 7 × 5). (iii) The model has four parts: 15, 5x, 3x and x2.
Marks: 1 mark for a(x), ½ mark each for a(1), a(2), 1 mark for the diagram.
Review: Page 6 · a(x) = x² + 5x + 6
(i) Two sides together make half the perimeter, 30 cm, so the other side is 30 − x: a(x) = x(30 − x) = 30x − x2. (ii) a(10) = 10 × 20 = 200 and a(20) = 20 × 10 = 200: a side of 10 makes the other side 20 and vice versa – it is the same rectangle turned around. (iii) Values x and y give the same area when x + y = 30.
Marks: 1 mark for a(x), 1 mark for the values with the reason, 1 mark for x + y = 30.
Review: Page 7 · Exercise 1, Q3
p(0) = d; p(1) = a + b + c + d; p(−1) = a(−1) + b(1) + c(−1) + d = −a + b − c + d. Here a = 2, b = −1, c = 4, d = −5: p(0) = −5; p(1) = 2 − 1 + 4 − 5 = 0; p(−1) = −2 − 1 − 4 − 5 = −12.
Marks: 1½ marks for the general values, 1½ marks for the three numbers.
Review: Page 10 · p(0), p(1), p(−1) · Page 11 · Exercise 2, Q4(vi)
(i) Yes, degree 2: √5 and ½ are fixed numbers. (ii) No: 2/x divides by the changing number (reciprocal). (iii) No: the square root of an expression containing x. (iv) No: (√x + 1)2 = x + 2√x + 1 and the √x does not go away.
Marks: ¾ mark each (decision + reason).
Review: Page 8 · Polynomial or not? · Page 9 · Exercise 2, Q1–Q2
(i) Each end loses x twice: length 12 − 2x, width 8 − 2x, height x. (ii) (12 − 2x)(8 − 2x) = 96 − 24x − 16x + 4x2 = 4x2 − 40x + 96, so v(x) = x(4x2 − 40x + 96) = 4x3 − 40x2 + 96x. (iii) Degree 3. (iv) v(1) = 4 − 40 + 96 = 60 (check 10 × 6 × 1); v(2) = 32 − 160 + 192 = 64 (check 8 × 4 × 2). (v) The width 8 − 2x must be positive, so 0 < x < 4.
Marks: (i) 1, (ii) 2, (iii) ½, (iv) 1, (v) ½ mark.
Review: Page 7 · Exercise 1, Q2 (the box) · Page 12 · Box from a card
(i) (x + 2)(x + 3) = x2 + 5x + 6; (x + 1)(x2 + 5x + 6) = x3 + 5x2 + 6x + x2 + 5x + 6 = x3 + 6x2 + 11x + 6. (ii) Edges 2, 3, 4: 2 × 3 × 4 = 24; and 1 + 6 + 11 + 6 = 24. (iii) s(x) = 2[(x + 1)(x + 2) + (x + 2)(x + 3) + (x + 1)(x + 3)] = 2[(x2 + 3x + 2) + (x2 + 5x + 6) + (x2 + 4x + 3)] = 2(3x2 + 12x + 11) = 6x2 + 24x + 22. (iv) s(1) = 6 + 24 + 22 = 52 (check 2(2·3 + 3·4 + 2·4) = 2 × 26 = 52).
Marks: (i) 2, (ii) 1, (iii) 1½, (iv) ½ mark.
Review: Page 6 · v(x) = x³ + 6x² + 11x + 6 · Page 3 · Multiplying brackets
Marks summary
| Section | Questions | Marks each | Total |
|---|---|---|---|
| A | 1–8 | 1 | 8 |
| B | 9–13 | 2 | 10 |
| C | 14–17 | 3 | 12 |
| D | 18–19 | 5 | 10 |
| Total | 40 | ||
- Section A – about 10 minutes. One-line answers: the degree, a coefficient, one substitution.
- Section B – about 20 minutes. Write each substitution in full, with brackets around negative numbers: 3 × (−1), not 3 × −1.
- Section C – about 25 minutes. Show how you multiply brackets, and give a reason for every “polynomial or not” decision.
- Section D – about 25 minutes. Draw the card or block, label every length with x, and check each value by multiplying the lengths directly.
- Checking – keep 10 minutes: put a small number like x = 1 into your polynomial and into the original lengths; both must agree.
Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 5, 6, 8 and 10 with the animations. Then return to the chapter home and tick off what you have learned.