Q1. Use suitable identities to find the following products: (i) (-3x + 4)2 (ii) (p2 + 1/2)(p2 - 1/2) (v) (s - 2t)(s2 + 2st + 4t2) (vii) (-3m + 4k - l)2 (ix) (7/2 k - 2/3 m)3
Answer:
(i) (-3x + 4)2 = (4 - 3x)2 = 42 - 2(4)(3x) + (3x)2 = 16 - 24x + 9x2
(ii) (p2 + 1/2)(p2 - 1/2) = (p2)2 - (1/2)2 = p4 - 1/4
(v) (s - 2t)(s2 + 2st + 4t2) = (s - 2t)(s2 + s(2t) + (2t)2) = s3 - (2t)3 = s3 - 8t3
(vii) (-3m + 4k - l)2 = (-3m)2 + (4k)2 + (-l)2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m) = 9m2 + 16k2 + l2 - 24mk - 8kl + 6lm
(ix) (7/2 k - 2/3 m)3 = a3 - 3a2b + 3ab2 - b3 where a = 7/2 k and b = 2/3 m.
= (7/2 k)3 - 3(7/2 k)2(2/3 m) + 3(7/2 k)(2/3 m)2 - (2/3 m)3
= 343/8 k3 - 3(49/4 k2)(2/3 m) + 3(7/2 k)(4/9 m2) - 8/27 m3
= 343/8 k3 - (3 × 49 × 2)/(4 × 3) k2m + (3 × 7 × 4)/(2 × 9) km2 - 8/27 m3
= 343/8 k3 - 49/2 k2m + 14/3 km2 - 8/27 m3
Q2. Find the values using suitable identities: (i) 17 × 21 (ii) 104 × 96 (iii) 24 × 16 (iv) 1473 (v) 1993 (vi) 1273 (vii) (-107)3 (viii) (-299)3
Answer:
(i) 17 × 21 = (19 - 2)(19 + 2) = 192 - 22 = 361 - 4 = 357
(ii) 104 × 96 = (100 + 4)(100 - 4) = 1002 - 42 = 10000 - 16 = 9984
(iii) 24 × 16 = (20 + 4)(20 - 4) = 202 - 42 = 400 - 16 = 384
(iv) 1473 = (150 - 3)3 = 1503 - 3(150)2(3) + 3(150)(3)2 - 33 = 3375000 - 202500 + 4050 - 27 = 3176523
(v) 1993 = (200 - 1)3 = 2003 - 3(200)2(1) + 3(200)(1)2 - 13 = 8000000 - 120000 + 600 - 1 = 7880599
(vi) 1273 = (130 - 3)3 = 1303 - 3(130)2(3) + 3(130)(3)2 - 33 = 2197000 - 152100 + 3510 - 27 = 2048383
(vii) (-107)3 = -1073 = -(100 + 7)3 = -[1003 + 3(100)2(7) + 3(100)(7)2 + 73] = -[1000000 + 210000 + 14700 + 343] = -1225043
(viii) (-299)3 = -2993 = -(300 - 1)3 = -[3003 - 3(300)2(1) + 3(300)(1)2 - 13] = -[27000000 - 270000 + 300 - 1] = -26730299
Q3. Factor the following algebraic expressions: (i) 4y2 + 1 + 1/(16y2) (ii) 9/m2 - 25/n2 (iii) 27b3 - 1/64 - b3 (iv) x2 + 5x/6 + 1/6 (v) 27u3 - 1/125 - 27/5 u2 + 9/25 u (vi) 64y3 + 1/(125z3) (vii) p3 + q3 + 27r3 - 9pqr (viii) 9/m2 - 12/m + 4 (ix) 8x3 - y3 + 6xyz + 3z3 (x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy (xi) 27u3 - 1/216 - 9u2/2 + u/4
Answer:
(i) 4y2 + 1 + 1/(16y2) = (2y)2 + 12 + (1/(4y))2 + 2(2y)(1) + 2(1)(1/(4y)) + 2(1/(4y))(2y) = (2y + 1 + 1/(4y))2 - 2(2y)(1/(4y))? Wait, let's check. The given expression is 4y2 + 1 + 1/(16y2).
This is not a perfect square of three terms. It is: (2y)2 + 12 + (1/(4y))2 + 2(2y)(1) + 2(1)(1/(4y)) + 2(1/(4y))(2y) = (2y + 1 + 1/(4y))2 - 2(2y)(1/(4y)) = (2y + 1 + 1/(4y))2 - 1. That doesn't match. Actually, the given expression is not a perfect square. It might be: (2y - 1/(4y))2 = 4y2 - 1 + 1/(16y2). So, if the middle term was -1, it would be a perfect square. But the middle term is +1. So, it's not a perfect square. Let's treat it as a quadratic in y: 4y2 + 1 + 1/(16y2) = (4y4 + y2 + 1/16) / y2. Factor the numerator: 4y4 + y2 + 1/16 = (2y2 + 1/4)2? No, (2y2 + 1/4)2 = 4y4 + y2 + 1/16. Yes! So, the expression is (2y2 + 1/4)2 / y2 = (2y + 1/(2y))2? Wait, (2y + 1/(4y))2 = 4y2 + 1 + 1/(16y2). So, it's (2y + 1/(4y))2.
(ii) 9/m2 - 25/n2 = (3/m)2 - (5/n)2 = (3/m - 5/n)(3/m + 5/n)
(iii) 27b3 - 1/64 - b3 = 26b3 - 1/64. This can be factored as (∛(26)b - 1/4)(...). Or if it's a typo and it's 27b3 - 1/64 - 9b2/4 + 3b/16, then it would be (3b - 1/4)3. As given, it is not a standard identity.
(iv) x2 + 5x/6 + 1/6 = (x + 1/2)(x + 1/3). Because 1/2 + 1/3 = 5/6 and (1/2)(1/3) = 1/6.
(v) 27u3 - 1/125 - 27/5 u2 + 9/25 u = (3u)3 - (1/5)3 - 3(3u)2(1/5) + 3(3u)(1/5)2 = (3u - 1/5)3
(vi) 64y3 + 1/(125z3) = (4y)3 + (1/(5z))3 = (4y + 1/(5z))(16y2 - 4y/(5z) + 1/(25z2))
(vii) p3 + q3 + 27r3 - 9pqr = p3 + q3 + (3r)3 - 3(p)(q)(3r) = (p + q + 3r)(p2 + q2 + 9r2 - pq - 3qr - 3rp)
(viii) 9/m2 - 12/m + 4 = (3/m)2 - 2(3/m)(2) + 22 = (3/m - 2)2
(ix) 8x3 - y3 + 6xyz + 3z3 = (2x)3 + (-y)3 + (∛3z)3 - 3(2x)(-y)(∛3z)? This seems incorrect. Let's use the identity a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca). We need to express 8x3 - y3 + 3z3 + 6xyz as a3 + b3 + c3 - 3abc. Let a = 2x, b = -y, c = z. Then a3 + b3 + c3 - 3abc = 8x3 - y3 + z3 - 3(2x)(-y)(z) = 8x3 - y3 + z3 + 6xyz. This matches the given expression. So, the factor is (2x - y + z)(4x2 + y2 + z2 + 2xy + yz - 2zx).
(x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy = (2x)2 + (3y)2 + (6z)2 + 2(2x)(3y) + 2(3y)(6z) + 2(6z)(2x) = (2x + 3y + 6z)2
(xi) 27u3 - 1/216 - 9u2/2 + u/4 = (3u)3 - (1/6)3 - 3(3u)2(1/6) + 3(3u)(1/6)2 = (3u - 1/6)3
Q4. Simplify the following: (i) (4x2 + 4x + 1) / (4x2 - 1) (ii) (9a3 - 3b) / (24a2 - 36) (iii) (s3 + 125t3) / (s2 - 2st - 35t2) Assume that the denominators are not equal to 0.
Answer:
(i) Numerator: 4x2 + 4x + 1 = (2x + 1)2
Denominator: 4x2 - 1 = (2x - 1)(2x + 1)
So, the expression simplifies to: (2x + 1)2 / ((2x - 1)(2x + 1)) = (2x + 1) / (2x - 1).
(ii) Numerator: 9a3 - 3b = 3(3a3 - b)
Denominator: 24a2 - 36 = 12(2a2 - 3)
So, the expression simplifies to: 3(3a3 - b) / [12(2a2 - 3)] = (3a3 - b) / [4(2a2 - 3)].
(iii) Numerator: s3 + 125t3 = s3 + (5t)3 = (s + 5t)(s2 - 5st + 25t2)
Denominator: s2 - 2st - 35t2 = s2 - 7st + 5st - 35t2 = s(s - 7t) + 5t(s - 7t) = (s - 7t)(s + 5t)
So, the expression simplifies to: (s + 5t)(s2 - 5st + 25t2) / [(s - 7t)(s + 5t)] = (s2 - 5st + 25t2) / (s - 7t).
Q5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units. (i) 25a2 – 30ab + 9b2 (ii) 36s2 – 49t2
Answer:
(i) 25a2 – 30ab + 9b2 = (5a)2 - 2(5a)(3b) + (3b)2 = (5a - 3b)2.
So, length = breadth = (5a - 3b) units. Or length = (5a - 3b) and breadth = (5a - 3b).
(ii) 36s2 – 49t2 = (6s)2 - (7t)2 = (6s - 7t)(6s + 7t).
So, length = (6s + 7t) and breadth = (6s - 7t) units, or vice versa.
Q6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units. (i) 6a2 – 24b2 (ii) 3ps2 – 15ps + 12p
Answer:
(i) 6a2 – 24b2 = 6(a2 - 4b2) = 6(a - 2b)(a + 2b).
So, possible dimensions: length = 6, breadth = (a - 2b), height = (a + 2b). Or any combination of these factors.
(ii) 3ps2 – 15ps + 12p = 3p(s2 - 5s + 4) = 3p(s - 1)(s - 4).
So, possible dimensions: length = 3p, breadth = (s - 1), height = (s - 4). Or any combination.
Q7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Answer:
The playground is a square of side 40 m. The path of width s m is around it. The outer square (including the path) has side (40 + 2s) m.
Area of the path = Area of outer square - Area of playground = (40 + 2s)2 - 402 = (1600 + 160s + 4s2) - 1600 = 160s + 4s2 = 4s(40 + s) square metres.
Q8. If a number plus its reciprocal equals 10/3, find the number.
Answer:
Let the number be x. Then, x + 1/x = 10/3.
Multiplying both sides by 3x: 3x2 + 3 = 10x ⇒ 3x2 - 10x + 3 = 0.
Factorising: 3x2 - 9x - x + 3 = 3x(x - 3) - 1(x - 3) = (3x - 1)(x - 3) = 0.
So, x = 3 or x = 1/3.
Thus, the number is 3 or 1/3.
Q9. A rectangular pool has area 2x2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Answer:
Area = Length × Width
Given area = 2x2 + 7x + 3 and width = 2x + 1.
So, Length = Area / Width = (2x2 + 7x + 3) / (2x + 1).
Factorising the numerator: 2x2 + 7x + 3 = 2x2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
So, Length = [(2x + 1)(x + 3)] / (2x + 1) = x + 3 hastas.
Q10. If both x – 2 and x – 1/2 are factors of px2 + 5x + r, show that p = r.
Answer:
Let f(x) = px2 + 5x + r.
If (x - 2) is a factor, then f(2) = 0.
So, p(2)2 + 5(2) + r = 0 ⇒ 4p + 10 + r = 0 ⇒ 4p + r = -10 ... (1)
If (x - 1/2) is a factor, then f(1/2) = 0.
So, p(1/2)2 + 5(1/2) + r = 0 ⇒ p/4 + 5/2 + r = 0 ⇒ p/4 + r = -5/2.
Multiplying by 4: p + 4r = -10 ... (2)
From (1): 4p + r = -10
From (2): p + 4r = -10
Subtracting (2) from (1): 3p - 3r = 0 ⇒ 3(p - r) = 0 ⇒ p = r.
Hence proved.
Q11. If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = – 25.
Answer:
We know the identity: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca).
We are given a + b + c = 5 and ab + bc + ca = 10.
We need to find a2 + b2 + c2.
Using (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca):
52 = a2 + b2 + c2 + 2(10)
25 = a2 + b2 + c2 + 20
a2 + b2 + c2 = 5
Now, a2 + b2 + c2 – ab – bc – ca = 5 – 10 = –5.
So, a3 + b3 + c3 – 3abc = (5)(–5) = –25.
Hence proved.
Q12. By factoring the expression, check that n3 – n is always divisible by 6 for all natural numbers n. Give reasons.
Answer:
n3 – n = n(n2 – 1) = n(n – 1)(n + 1).
This is the product of three consecutive integers: (n – 1), n, (n + 1).
Among any three consecutive integers, there is always at least one multiple of 2 and one multiple of 3.
So, the product is divisible by 2 and by 3.
Since 2 and 3 are coprime, the product is divisible by 6.
Thus, n3 – n is always divisible by 6 for all natural numbers n.
Q13. Find the value of x3 + y3 – 12xy + 64, when x + y = –4 x3 – 8y3 – 36xy – 216, when x = 2y + 6
Answer:
Part 1: Given x + y = –4, so x + y + 4 = 0.
We know the identity: if a + b + c = 0, then a3 + b3 + c3 = 3abc.
Here, take a = x, b = y, c = 4. Then x + y + 4 = 0.
So, x3 + y3 + 43 = 3(x)(y)(4) = 12xy.
Thus, x3 + y3 – 12xy + 64 = 0. (Because x3 + y3 + 64 = 12xy).
So, the value is 0.
Part 2: Given x = 2y + 6 ⇒ x – 2y – 6 = 0.
Take a = x, b = –2y, c = –6. Then a + b + c = x – 2y – 6 = 0.
So, x3 + (–2y)3 + (–6)3 = 3(x)(–2y)(–6) = 36xy.
x3 – 8y3 – 216 = 36xy.
Thus, x3 – 8y3 – 36xy – 216 = 0.
So, the value is 0.