Algebra
-
If 4a - 4 + 3 = 0 then the value of : a3 - 1 + 3 = ? a a3
-
View Hint View Answer Discuss in Forum
4a - 4 = -3 a
On dividing by 4,⇒ a - 1 = -3 a 4 ∴ a3 - 1 = 
a - 1 
3 + 3 . a . 1 a - 1 a3 a a a ⇒ a3 - 1 = -3 3 + 3 × -3 a3 4 4 ⇒ a3 - 1 = - 27 - 9 a3 64 4 ⇒ a3 - 1 = -27 - 144 = -171 a3 64 64 ∴ a3 - 1 + 3 = -171 + 3 = -171 + 192 a3 64 64 a3 - 1 + 3 = 21 a3 64
Correct Option: C
4a - 4 = -3 a
On dividing by 4,⇒ a - 1 = -3 a 4 ∴ a3 - 1 = a - 1 3 + 3 . a . 1 a - 1 a3 a a a ⇒ a3 - 1 = -3 3 + 3 × -3 a3 4 4 ⇒ a3 - 1 = - 27 - 9 a3 64 4 ⇒ a3 - 1 = -27 - 144 = -171 a3 64 64 ∴ a3 - 1 + 3 = -171 + 3 = -171 + 192 a3 64 64 a3 - 1 + 3 = 21 a3 64
-
If x2 + x = 5 then the value of (x + 3)3 + 1 is (x + 3)3
-
View Hint View Answer Discuss in Forum
x2 + x = 5 (Given)
Let, x + 3 = a
∴ 1 = 1 x + 3 a Now , a + 1 = (x + 3) + 1 a (x + 3) a + 1 = (x + 3)2 + 1 a (x + 3) a + 1 = x2 + 6x + 9 + 1 a x + 3 a + 1 = x2 + x + 5x + 10 a x + 3 a + 1 = 5 + 5x + 10 a x + 3 a + 1 = 5x + 15 = 5(x + 3) = 5 a x + 3 x + 3 ∴ a3 + 1 = a + 1 3 - 3 . a . 1 a + 1 a3 a a a
Required answer = (5)3 – 3 × 5 = 125 – 15 = 110
Correct Option: B
x2 + x = 5 (Given)
Let, x + 3 = a
∴ 1 = 1 x + 3 a Now , a + 1 = (x + 3) + 1 a (x + 3) a + 1 = (x + 3)2 + 1 a (x + 3) a + 1 = x2 + 6x + 9 + 1 a x + 3 a + 1 = x2 + x + 5x + 10 a x + 3 a + 1 = 5 + 5x + 10 a x + 3 a + 1 = 5x + 15 = 5(x + 3) = 5 a x + 3 x + 3 ∴ a3 + 1 = a + 1 3 - 3 . a . 1 a + 1 a3 a a a
Required answer = (5)3 – 3 × 5 = 125 – 15 = 110
- If x2 + y2 + z2 = 2 (x + z – 1), then the value of :
x3 + y3 + z3 = ?
-
View Hint View Answer Discuss in Forum
x2 + y2 + z2 = 2(x + z – 1)
⇒ x2 + y2 + z2 = 2x + 2z – 2
⇒ x2 - 2x + y2 + z2 - 2z + 2 = 0
⇒ x2 - 2x + 1 + y2 + z2 - 2z + 1 = 0
⇒ (x - 1)2 + y2 + (z - 1)2 = 0
∴ a2 + b2 + c2 = 0 ⇒ a = 0, b = 0, c = 0]
∴ x – 1 = 0 ⇒ x = 1
y = 0
z – 1 = 0 ⇒ z = 1
∴ x3 + y3 + z3 = 1 + 0 + 1 = 2Correct Option: A
x2 + y2 + z2 = 2(x + z – 1)
⇒ x2 + y2 + z2 = 2x + 2z – 2
⇒ x2 - 2x + y2 + z2 - 2z + 2 = 0
⇒ x2 - 2x + 1 + y2 + z2 - 2z + 1 = 0
⇒ (x - 1)2 + y2 + (z - 1)2 = 0
∴ a2 + b2 + c2 = 0 ⇒ a = 0, b = 0, c = 0]
∴ x – 1 = 0 ⇒ x = 1
y = 0
z – 1 = 0 ⇒ z = 1
∴ x3 + y3 + z3 = 1 + 0 + 1 = 2
- The HCF of x8 – 1 and x4 + 2x3 – 2x – 1 is :
-
View Hint View Answer Discuss in Forum
x8 – 1 = (x4)2 - 12
= (x4 + 1)(x4 - 1)
= (x4 + 1)(x2 - 1) (x2 + 1)
= (x4 + 1)(x2 + 1)(x + 1)(x - 1)
[ ∴ a2 - b2 = (a + b)(a - b) ]
x4 + 2x3 – 2x – 1 = (x4 - 1) + 2x3 – 2x
= (x2 - 1) (x2 + 1) + 2x(x2 - 1)
= (x2 + 1 + 2x)(x2 - 1)
= (x + 1)2(x + 1)(x - 1)
∴ H.C.F = (x + 1) (x – 1) = (x2 - 1)Correct Option: B
x8 – 1 = (x4)2 - 12
= (x4 + 1)(x4 - 1)
= (x4 + 1)(x2 - 1) (x2 + 1)
= (x4 + 1)(x2 + 1)(x + 1)(x - 1)
[ ∴ a2 - b2 = (a + b)(a - b) ]
x4 + 2x3 – 2x – 1 = (x4 - 1) + 2x3 – 2x
= (x2 - 1) (x2 + 1) + 2x(x2 - 1)
= (x2 + 1 + 2x)(x2 - 1)
= (x + 1)2(x + 1)(x - 1)
∴ H.C.F = (x + 1) (x – 1) = (x2 - 1)
-
If x24 + 1 = 7 , then the value of x72 + 1 is x12 x36
-
View Hint View Answer Discuss in Forum
x24 + 1 = 7 x12 ⇒ x24 + 1 x12 x12 ⇒ x12 + 1 = 7 x12 ∴ x72 + 1 = x72 + 1 x36 x36 x36 = x36 + 1 x36 = x12 + 1 3 - 3 . x12 . 1 x12 + 1 x12 x12 x12
[ ∴ a3 + b3 = (a + b)3 - 3ab(a + b)]⇒ x36 + 1 = 73 - 3 × 7 = 343 – 21 = 322 x36
Correct Option: B
x24 + 1 = 7 x12 ⇒ x24 + 1 x12 x12 ⇒ x12 + 1 = 7 x12 ∴ x72 + 1 = x72 + 1 x36 x36 x36 = x36 + 1 x36 = x12 + 1 3 - 3 . x12 . 1 x12 + 1 x12 x12 x12
[ ∴ a3 + b3 = (a + b)3 - 3ab(a + b)]⇒ x36 + 1 = 73 - 3 × 7 = 343 – 21 = 322 x36
