Algebra
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then the relation among x, y, z isIf x ≠ 0, y ≠ 0 and z ≠ 0 and 1 + 1 + 1 + 1 + 1 + 1 + 1 , x2 y2 z2 16 xy yz zx
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Check through options.
If x = y = z, then1 + 1 + 1 = 3 x2 y2 z2 x2
and1 + 1 + 1 xy yz zx = 1 + 1 + 1 = 3 x2 x2 x2 x2 Correct Option: D
Check through options.
If x = y = z, then1 + 1 + 1 = 3 x2 y2 z2 x2
and1 + 1 + 1 xy yz zx = 1 + 1 + 1 = 3 x2 x2 x2 x2
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If a = 4 and b = 15 , then 18c2 − 7a2 is equal to b 5 c 16 45c2 + 20a2
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a × b = 4 × 15 b c 5 16 a = 3 c 4 ⇒ a = 3 c 4
Put in the given equation,
Correct Option: D
a × b = 4 × 15 b c 5 16 a = 3 c 4 ⇒ a = 3 c 4
Put in the given equation,
- Number of solutions of the two equations 4x – y = 2 and 2y – 8x + 4 = 0 is
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4x – y = 2 ....(i)
2y – 8x + 4 = 0
⇒ 8x – 2y = 4 .....(ii)
For simultaneous linear equations
a1x + b1y = c1
a2x + b2y = c2 ifa1 + b1 + c1 , there are infinite solutions. a2 b2 c2 Correct Option: D
4x – y = 2 ....(i)
2y – 8x + 4 = 0
⇒ 8x – 2y = 4 .....(ii)
For simultaneous linear equations
a1x + b1y = c1
a2x + b2y = c2 ifa1 + b1 + c1 , there are infinite solutions. a2 b2 c2
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If x − a2 + x − b2 + x − c2 = 4(a + b + c), then x is equal to b + c c + a a + b
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Check through option
When x = (a + b + c)2,x − a2 + x − b2 + x − c2 b + c c + a a + b = (a + b + c)2 − a2 + (a + b + c)2 − b2 + (a + b + c)2 − c2 b + c c + a a + b = (2a + b + c)(b + c) + (a + 2b + c)(c + a) + (a + b + 2c)(a + b) b + c c + a a + b
= 2a + b + c + a + 2b + c + a + b + 2c
= 4a+4b + 4c = 4 (a + b + c) = RHS.Correct Option: A
Check through option
When x = (a + b + c)2,x − a2 + x − b2 + x − c2 b + c c + a a + b = (a + b + c)2 − a2 + (a + b + c)2 − b2 + (a + b + c)2 − c2 b + c c + a a + b = (2a + b + c)(b + c) + (a + 2b + c)(c + a) + (a + b + 2c)(a + b) b + c c + a a + b
= 2a + b + c + a + 2b + c + a + b + 2c
= 4a+4b + 4c = 4 (a + b + c) = RHS.
- If a2 + b2 + 4c2 = 2 (a + b – 2c) − 3 and a, b, c are real, then the value of (a2 + b2 + c2) is
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a2 + b2 + 4c2 = 2a + 2b – 4c − 3
⇒ a2 + b2 + 4c2 – 2a – 2b + 4c + 3 = 0
⇒ a2 – 2a + 1 + b2 – 2b + 1 + 4c2 + 4c + 1 = 0
⇒ (a – 1)2 +(b – 1)2 + (2c + 1)2 = 0
∴ a – 1 = 0 ⇒ a = 1;
b – 1 = 0 ⇒ b = 1;2c + 1 = 0 ⇒ c = − 1 2 ∴ a2 + b2 + c2 = 1 + 1 + 1 = 2 1 4 4 Correct Option: D
a2 + b2 + 4c2 = 2a + 2b – 4c − 3
⇒ a2 + b2 + 4c2 – 2a – 2b + 4c + 3 = 0
⇒ a2 – 2a + 1 + b2 – 2b + 1 + 4c2 + 4c + 1 = 0
⇒ (a – 1)2 +(b – 1)2 + (2c + 1)2 = 0
∴ a – 1 = 0 ⇒ a = 1;
b – 1 = 0 ⇒ b = 1;2c + 1 = 0 ⇒ c = − 1 2 ∴ a2 + b2 + c2 = 1 + 1 + 1 = 2 1 4 4
