Algebra
-
If x4 + 1 = 119 , then the values of x3 + 1 are x4 x3
-
View Hint View Answer Discuss in Forum
x4 + 1 = 119 x4 ⇒ 
x2 + 1 
2 - 2 = 119 x2 ⇒ x2 + 1 2 = 119 + 2 = 121 x2 ⇒ x2 + 1 2 = (11)2 x2 ⇒ x2 + 1 = 11 x2 Again, x + 1 2 - 2 = 11 x ⇒ x + 1 2 = 11 + 2 = 13 x ⇒ x + 1 = ± √13 x
On cubing both sides,⇒ x + 1 3 = ( ± √13 )3 x ⇒ x3 + 1 + 3 . x . 1 x + 1 = ± √13 x3 x x ⇒ x3 + 1 + 3 × ( ± √13 ) = ± √13 x3 ⇒ x3 - 1 = ± (13√13 - 3√13) = ± 10√13 x3
Correct Option: A
x4 + 1 = 119 x4 ⇒ x2 + 1 2 - 2 = 119 x2 ⇒ x2 + 1 2 = 119 + 2 = 121 x2 ⇒ x2 + 1 2 = (11)2 x2 ⇒ x2 + 1 = 11 x2 Again, x + 1 2 - 2 = 11 x ⇒ x + 1 2 = 11 + 2 = 13 x ⇒ x + 1 = ± √13 x
On cubing both sides,⇒ x + 1 3 = ( ± √13 )3 x ⇒ x3 + 1 + 3 . x . 1 x + 1 = ± √13 x3 x x ⇒ x3 + 1 + 3 × ( ± √13 ) = ± √13 x3 ⇒ x3 - 1 = ± (13√13 - 3√13) = ± 10√13 x3
- If x = 2015, y = 2014 and z = 2013, then value of x2 + y2 + z2 – xy – yz – zx is
-
View Hint View Answer Discuss in Forum
x – y = 2015 – 2014 = 1
y – z = 2014 – 2013 = 1
z – x = 2013 – 2015 = –2∴ x2 + y2 + z2 – xy – yz – zx = 1 [ 2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ (x - y)2 + (y - z)2 + (z - x)2 ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ (-1)2 + (-1)2 + (2)2 ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ 1 + 1 + 4 ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 × 6 = 3 2 Correct Option: A
x – y = 2015 – 2014 = 1
y – z = 2014 – 2013 = 1
z – x = 2013 – 2015 = –2∴ x2 + y2 + z2 – xy – yz – zx = 1 [ 2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ (x - y)2 + (y - z)2 + (z - x)2 ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ (-1)2 + (-1)2 + (2)2 ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 [ 1 + 1 + 4 ] 2 ⇒ x2 + y2 + z2 – xy – yz – zx = 1 × 6 = 3 2
- If p3 + 3p2 + 3p = 7, then the value of p2 + 2p is
-
View Hint View Answer Discuss in Forum
Using Rule 8,
p3 + 3p2 + 3p = 7
⇒ p3 + 3p2 + 3p + 1 = 7 + 1 = 8
⇒ ( p + 1 )3 = (2)3
⇒ p + 1 = 2 ⇒ p = 2 – 1 = 1
∴ p2 + 2p = 1 + 2 × 1 = 3Correct Option: B
Using Rule 8,
p3 + 3p2 + 3p = 7
⇒ p3 + 3p2 + 3p + 1 = 7 + 1 = 8
⇒ ( p + 1 )3 = (2)3
⇒ p + 1 = 2 ⇒ p = 2 – 1 = 1
∴ p2 + 2p = 1 + 2 × 1 = 3
- If a + b + c = 0, then the value of (a + b – c)2 + (b + c – a)2 + (c + a – b)2 is
-
View Hint View Answer Discuss in Forum
a + b + c = 0 (Given)
∴ a + b = – c
b + c = – a
c + a = – b
∴ (a + b – c)2 + (b + c – a)2 + (c + a – b)2 = (-c – c)2 + (-a – a)2 + (-b – b)2
= (– 2c)2 + (- 2a)2 + (– 2b)2 = 4c2 + 4a2 + 4b2 = 4(c2 + a2 + b2)Correct Option: C
a + b + c = 0 (Given)
∴ a + b = – c
b + c = – a
c + a = – b
∴ (a + b – c)2 + (b + c – a)2 + (c + a – b)2 = (-c – c)2 + (-a – a)2 + (-b – b)2
= (– 2c)2 + (- 2a)2 + (– 2b)2 = 4c2 + 4a2 + 4b2 = 4(c2 + a2 + b2)
-
If x = p + 1 and y = p - 1 , then the value of x4 - 2x2y2 + y4 is : p p
-
View Hint View Answer Discuss in Forum
x = p + 1 p y = p - 1 p ∴ x + y = p + 1 + p - 1 = 2p p p x - y = p + 1 - p + 1 = 2 p p p
∴ x4 - 2x2y2 + y4 = ( x2 - y2 )2 = { (x - y)(x + y) }2x4 - 2x2y2 + y4 = 2p × 2 2 = 42 = 16 p Correct Option: C
x = p + 1 p y = p - 1 p ∴ x + y = p + 1 + p - 1 = 2p p p x - y = p + 1 - p + 1 = 2 p p p
∴ x4 - 2x2y2 + y4 = ( x2 - y2 )2 = { (x - y)(x + y) }2x4 - 2x2y2 + y4 = 2p × 2 2 = 42 = 16 p
