Algebra
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If x = 1 , then the value of 
x + 1 
is : 2x2 + 5x + 2 6 x
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x = 1 2x2 + 5x + 2 6
⇒ 2x2 + 5x + 2 = 6x
⇒ 2x2 + 2 = 6x – 5x = x⇒ x2 + 1 = x 2
On dividing by x,⇒ x + 1 = 1 x 2 Correct Option: B
x = 1 2x2 + 5x + 2 6
⇒ 2x2 + 5x + 2 = 6x
⇒ 2x2 + 2 = 6x – 5x = x⇒ x2 + 1 = x 2
On dividing by x,⇒ x + 1 = 1 x 2
- If (3a + 1)2 + (b – 1)2 + (2c – 3)2 = 0, then the value of (3a + b + 2c) is equal to :
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(3a + 1)2 + (b – 1)2 + (2c–3)2 = 0
⇒ 3a + 1 = 0
⇒ 3a = –1
b – 1 = 0
⇒ b = 1
2c – 3 = 0
⇒ 2c = 3
∴ 3a + b + 2c = –1 + 1 + 3 = 3Correct Option: A
(3a + 1)2 + (b – 1)2 + (2c–3)2 = 0
⇒ 3a + 1 = 0
⇒ 3a = –1
b – 1 = 0
⇒ b = 1
2c – 3 = 0
⇒ 2c = 3
∴ 3a + b + 2c = –1 + 1 + 3 = 3
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The value of the expression (a − b)2 + (b − c)2 + (c − a)2 is : (b − c)(c − a) (a − b)(c − a) (a − b)(b − c)
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(a − b)2 + (b − c)2 + (c − a)2 (b − c)(c − a) (a − b)(c − a) (a − b)(b − c) = (a − b)3 + (b − c)3 + (c − a)3 (a − b)(b − c)(c − a) (a − b)(b − c)(c − a) (a − b)(b − c)(c − a) = (a − b)3 +(b − c)3 + (c − a)3 (a − b)(b − c)(c − a)
[∵ [a – b + b – c + c – a = 0]]= 3(a − b)(b − c) (c − a) = 3 (a − b)(b − c)(c − a)
[a + b + x = 0,
∵ a3 + b3 + c3 = 3abc]Correct Option: B
(a − b)2 + (b − c)2 + (c − a)2 (b − c)(c − a) (a − b)(c − a) (a − b)(b − c) = (a − b)3 + (b − c)3 + (c − a)3 (a − b)(b − c)(c − a) (a − b)(b − c)(c − a) (a − b)(b − c)(c − a) = (a − b)3 +(b − c)3 + (c − a)3 (a − b)(b − c)(c − a)
[∵ [a – b + b – c + c – a = 0]]= 3(a − b)(b − c) (c − a) = 3 (a − b)(b − c)(c − a)
[a + b + x = 0,
∵ a3 + b3 + c3 = 3abc]
- Out of the given responses, one of the factors of (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 is
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a2 - b2 + b2 - c2 + c2 - a2 = 0
[ If x + y + z = 0 , x3 + y3 + z3 = 3xyz ]
∴ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2)(b2 - c2)(c2 - a2)
⇒ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3 (a + b) (a – b) (b + c) (b – c) (c + a) (c – a)Correct Option: A
a2 - b2 + b2 - c2 + c2 - a2 = 0
[ If x + y + z = 0 , x3 + y3 + z3 = 3xyz ]
∴ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2)(b2 - c2)(c2 - a2)
⇒ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3 (a + b) (a – b) (b + c) (b – c) (c + a) (c – a)
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If the sum of a and its reciprocal is 1 and a ≠ 0, b ≠ 0, then the value of a3 + b3 is b
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According to question,
a + b = 1 a a
⇒ a2 + b2 = ab
⇒ a2 – ab + b2 = 0
∴ a3 + b3 = (a + b)( a2 – ab + b2 ) = 0
Correct Option: C
According to question,
a + b = 1 a a
⇒ a2 + b2 = ab
⇒ a2 – ab + b2 = 0
∴ a3 + b3 = (a + b)( a2 – ab + b2 ) = 0
