Algebra
- If x = b + c – 2a, y = c + a – 2b, z = a + b – 2c, then the value of x2 + y2 – z2 + 2xy is
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x2 + y2 – z2 + 2xy
= x2 + y2 + 2xy – z2
= (x +y)2 – z2 = (x + y + z) (x + y – z)
= (b + c– 2a + c + a – 2b + a + b – 2c) (x + y– z) = 0Correct Option: A
x2 + y2 – z2 + 2xy
= x2 + y2 + 2xy – z2
= (x +y)2 – z2 = (x + y + z) (x + y – z)
= (b + c– 2a + c + a – 2b + a + b – 2c) (x + y– z) = 0
- If x2 + y2 – 4x – 4y + 8 = 0, then the value of x – y is
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x2 + y2 – 4x – 4y + 8 = 0
⇒ x2 – 4x + 4 + y2 – 4y + 4 = 0
⇒ (x – 2)2 + (y – 2)2 = 0
⇒ x = 2 and y = 2
∴ x – y = 2 – 2 = 0Correct Option: C
x2 + y2 – 4x – 4y + 8 = 0
⇒ x2 – 4x + 4 + y2 – 4y + 4 = 0
⇒ (x – 2)2 + (y – 2)2 = 0
⇒ x = 2 and y = 2
∴ x – y = 2 – 2 = 0
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is a perfect square ?For what valsue (s) of k the expression p + 1 √p + k2 4
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p + 1 √p + k2 4 = ( √p)2 + 2.√p. 1 + 
1 
2 − 1 2 + k2 8 8 8 ⇒ k2 = 1 2 ⇒ k = ± 1 8 8 Correct Option: C
p + 1 √p + k2 4 = ( √p)2 + 2.√p. 1 + 1 2 − 1 2 + k2 8 8 8 ⇒ k2 = 1 2 ⇒ k = ± 1 8 8
- If a = 23 and b = –29 then the value of 25a2 + 40ab + 16b2 is :
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25a2 + 40ab + 16b2
= (5a + 4b)2
= (5 × 23 – 29 × 4)2
= (115 – 116)2 = 1Correct Option: A
25a2 + 40ab + 16b2
= (5a + 4b)2
= (5 × 23 – 29 × 4)2
= (115 – 116)2 = 1
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If a = 0.25, b = – 0.05, c = 0.5, then the value of a2 − b2 − c2 − 2bc is a2 + b2 − 2ab − c2
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a2 − b2 − c2 − 2bc a2 + b2 − 2ab − c2 = a2 − (b2 + c2 + 2bc) (a2 + b2 − 2ab) − c2 = a2 − (b + c)2 (a − b)2 − c2 = (a + b + c)(a − b − c) (a − b + c)(a − b − c) = a + b + c = 0.25 − 0.05 + 0.5 a − b + c 0.25 + 0.05 + 0.5 = 0.7 = 7 0.8 8 Correct Option: A
a2 − b2 − c2 − 2bc a2 + b2 − 2ab − c2 = a2 − (b2 + c2 + 2bc) (a2 + b2 − 2ab) − c2 = a2 − (b + c)2 (a − b)2 − c2 = (a + b + c)(a − b − c) (a − b + c)(a − b − c) = a + b + c = 0.25 − 0.05 + 0.5 a − b + c 0.25 + 0.05 + 0.5 = 0.7 = 7 0.8 8
