Algebra
-
If (2a – 1)2 + (4b – 3)2 + (4c + 5)2 = 0 ,then the value of a3 + b3 + c3 - 3abc = 0 a2 + b2 + c2
-
View Hint View Answer Discuss in Forum
(2a – 1)2 + (4b – 3)2 + (4c + 5)2 = 0
∴ 2a – 1 = 0 ⇒ 2a = 1 ⇒ a = 1 2 4b – 3 = 0 ⇒ 4b = 3 ⇒ b = 3 4 4c + 5 = 0 ⇒ 4c = –5 ⇒ c = -5 4
[ If x2 + y2 + z2 = 0, x = 0, y = 0, z = 0]∴ a + b + c = 1 + 3 - 5 2 4 4 a + b + c = 6 + 9 - 15 = 0 12
∴ a3 + b3 + c3 - 3abc = 0
∴ Required answer = 0Correct Option: D
(2a – 1)2 + (4b – 3)2 + (4c + 5)2 = 0
∴ 2a – 1 = 0 ⇒ 2a = 1 ⇒ a = 1 2 4b – 3 = 0 ⇒ 4b = 3 ⇒ b = 3 4 4c + 5 = 0 ⇒ 4c = –5 ⇒ c = -5 4
[ If x2 + y2 + z2 = 0, x = 0, y = 0, z = 0]∴ a + b + c = 1 + 3 - 5 2 4 4 a + b + c = 6 + 9 - 15 = 0 12
∴ a3 + b3 + c3 - 3abc = 0
∴ Required answer = 0
- If x4 + 2x3 + ax2 + bx + 9 is a perfect square, where a and b are positive real numbers, then the values of a and b are
-
View Hint View Answer Discuss in Forum
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (x2 + x + 3)2 = x4 + x2 + 9 + 2x3 + 6x + 6x2
= x4 + 2x3 + 7x2 + 6x + 9
On comparing with x4 + 2x3 + ax2 + bx + 9
a = 7, b = 6Correct Option: C
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
∴ (x2 + x + 3)2 = x4 + x2 + 9 + 2x3 + 6x + 6x2
= x4 + 2x3 + 7x2 + 6x + 9
On comparing with x4 + 2x3 + ax2 + bx + 9
a = 7, b = 6
-
If a2 + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax + by + cz = 20, then the value of a + b + c is x + y + z
-
View Hint View Answer Discuss in Forum
(ax + by + cz)2 = (a2 + b2 + c2)(x2 + y2 + z2) = 400
⇒ a2x2 + b2y2 + c2z2 + 2abxy + 2bcyz + 2acxz
= a2x2 + a2y2 + a2z2 + b2x2 + b2y2 + b2z2 +
c2x2 + c2y2 + c2z2
⇒ a2y2 + a2z2 + b2x2 + b2z2 +
c2x2 + c2y2 = 2abxy + 2bcyz + 2acxz
⇒ a2y2 - 2abxy + b2x2 + a2z2 + c2x2 - 2acxz + b2z2 + c2y2 - 2bcyz = 0
⇒ (ay – bx)2 + (az – cx)2 + (bz – cy)2 = 0⇒ ay – bx = 0 ⇒ ay = bx ⇒ a = x b y
az – cx = 0 ⇒ az = cx ⇒ a = x c z
∴ a = kx ; b = ky; c = kz
∴ a2 + b2 + c2 = 16
⇒ k2(x2 + y2 + z2) = 16
⇒ k2 × 25 = 16
⇒ k2 = 16 ⇒ k = 4 25 5 ∴ a + b + c = k = 4 x + y + z 5
Correct Option: C
(ax + by + cz)2 = (a2 + b2 + c2)(x2 + y2 + z2) = 400
⇒ a2x2 + b2y2 + c2z2 + 2abxy + 2bcyz + 2acxz
= a2x2 + a2y2 + a2z2 + b2x2 + b2y2 + b2z2 +
c2x2 + c2y2 + c2z2
⇒ a2y2 + a2z2 + b2x2 + b2z2 +
c2x2 + c2y2 = 2abxy + 2bcyz + 2acxz
⇒ a2y2 - 2abxy + b2x2 + a2z2 + c2x2 - 2acxz + b2z2 + c2y2 - 2bcyz = 0
⇒ (ay – bx)2 + (az – cx)2 + (bz – cy)2 = 0⇒ ay – bx = 0 ⇒ ay = bx ⇒ a = x b y
az – cx = 0 ⇒ az = cx ⇒ a = x c z
∴ a = kx ; b = ky; c = kz
∴ a2 + b2 + c2 = 16
⇒ k2(x2 + y2 + z2) = 16
⇒ k2 × 25 = 16
⇒ k2 = 16 ⇒ k = 4 25 5 ∴ a + b + c = k = 4 x + y + z 5
-
The value of x which satisfies the equation x + a2 + 2c2 + x + b2 + 2a2 + x + b2 + 2a2 = 0 is b + c c + a a + b
-
View Hint View Answer Discuss in Forum
Of the given options,
x = -(a2 + b2 + c2)∴ x + a2 + 2c2 = - a2 - b2 - c2 + a2 + 2c2 b + c b + c = c2 - b2 = c – b b + c x + b2 + 2a2 = - a2 - b2 - c2 + b2 + 2a2 c + a b + c = a2 - c2 = a – c c + a x + c2 + 2b2 = - a2 - b2 - c2 + c2 + 2b2 a + b b + c = b2 - a2 = b – a a + b
∴ c – b + a – c + b – a = 0
Correct Option: B
Of the given options,
x = -(a2 + b2 + c2)∴ x + a2 + 2c2 = - a2 - b2 - c2 + a2 + 2c2 b + c b + c = c2 - b2 = c – b b + c x + b2 + 2a2 = - a2 - b2 - c2 + b2 + 2a2 c + a b + c = a2 - c2 = a – c c + a x + c2 + 2b2 = - a2 - b2 - c2 + c2 + 2b2 a + b b + c = b2 - a2 = b – a a + b
∴ c – b + a – c + b – a = 0
- If a3 = 117 + b3 and a = 3 + b,then the value of (a + b) is :
-
View Hint View Answer Discuss in Forum
a3 – b3 = 117; a – b = 3
⇒ (a – b)(a2 + ab + b2) = 117
⇒ 3 × (a2 + ab + b2) = 117⇒ a2 + ab + b2 = 117 = 39 3
⇒ (a - b)2 + 3ab = 39
⇒ 32 + 3ab = 39
⇒ 3ab = 39 – 9 = 30⇒ ab = 30 = 10 3
⇒ (a + b)2 = (a - b)2 + 4ab
= 9 + 4 × 10 = 49
∴ a + b = √49 = ± 7
Correct Option: A
a3 – b3 = 117; a – b = 3
⇒ (a – b)(a2 + ab + b2) = 117
⇒ 3 × (a2 + ab + b2) = 117⇒ a2 + ab + b2 = 117 = 39 3
⇒ (a - b)2 + 3ab = 39
⇒ 32 + 3ab = 39
⇒ 3ab = 39 – 9 = 30⇒ ab = 30 = 10 3
⇒ (a + b)2 = (a - b)2 + 4ab
= 9 + 4 × 10 = 49
∴ a + b = √49 = ± 7
