Algebra
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If for a non–zero x, 3x2 + 5x + 3 = 0, then the value of x3 + 1 is : x3
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3x2 + 5x + 3 = 0
⇒ 3x2 + 3 = –5x⇒ 3x2 + 3 = –5 x ⇒ 3x + 3 = –5 x ⇒ x + 1 = – 5 x 3
On cubing both sides,
x + 1 
3 = −5 3 = −125 x 3 27 ⇒ x3 + 1 + 3 x + 1 = −125 x3 x 27 ⇒ x3 + 1 + 3 × −5 = −125 x3 3 27 ⇒ x3 + 1 = −125 + 5 x3 27 = −125 + 135 = 10 27 27 Correct Option: A
3x2 + 5x + 3 = 0
⇒ 3x2 + 3 = –5x⇒ 3x2 + 3 = –5 x ⇒ 3x + 3 = –5 x ⇒ x + 1 = – 5 x 3
On cubing both sides,x + 1 3 = −5 3 = −125 x 3 27 ⇒ x3 + 1 + 3 x + 1 = −125 x3 x 27 ⇒ x3 + 1 + 3 × −5 = −125 x3 3 27 ⇒ x3 + 1 = −125 + 5 x3 27 = −125 + 135 = 10 27 27
- If a + b + c = 1, ab + bc + ca = – 1 and abc = –1, then the value of a3 + b3 + c3 is :
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a + b + c = 1
ab + bc + ca = –1
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ 1 = a2 + b2 + c2 + 2 (–1)
⇒ a2 + b2 + c2 = 3
∴ a3 + b3 + c3 = 3abc + 4
= –3 + 4 = 1Correct Option: A
a + b + c = 1
ab + bc + ca = –1
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ 1 = a2 + b2 + c2 + 2 (–1)
⇒ a2 + b2 + c2 = 3
∴ a3 + b3 + c3 = 3abc + 4
= –3 + 4 = 1
- If (2a – 3)2 + (3b + 4)2 + (6c + 1)2 = 0, then the value of
a3 + b3 + c3 − 3ac + 3 is : a2 + b2 + c2
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(2a – 3)2 + (3b + 4)2 + (6c + 1)2 = 0
∴ 2a – 3 = 0 ⇒ a = 3 , 2 3b + 4 = 0 ⇒ b = −4 3 6c + 1 = 0 ⇒ c = – 1 6 ∴ a + b + c = 3 – 4 – 1 2 3 6 = 9 − 8 − 1 = 0 6
∴ a3 + b3 + c3 − 3ac = 0∴ a3 + b3 + c3 − 3ac + 3 a2 + b2 + c2
= 0 + 3 = 3Correct Option: D
(2a – 3)2 + (3b + 4)2 + (6c + 1)2 = 0
∴ 2a – 3 = 0 ⇒ a = 3 , 2 3b + 4 = 0 ⇒ b = −4 3 6c + 1 = 0 ⇒ c = – 1 6 ∴ a + b + c = 3 – 4 – 1 2 3 6 = 9 − 8 − 1 = 0 6
∴ a3 + b3 + c3 − 3ac = 0∴ a3 + b3 + c3 − 3ac + 3 a2 + b2 + c2
= 0 + 3 = 3
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(where p is an even number and q is an odd number)If x + 1 = –2 then the vlaue of xp + xq is : x
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x + 1 = –2 x
⇒ x2 + 1 = –2x
⇒ x2 + 2x + 1 = 0
⇒ (x + 1)2 = 0
⇒ x = –1
∴ xp + xq
= (–1)p + (–1)q
= 1 – 1 = 0Correct Option: D
x + 1 = –2 x
⇒ x2 + 1 = –2x
⇒ x2 + 2x + 1 = 0
⇒ (x + 1)2 = 0
⇒ x = –1
∴ xp + xq
= (–1)p + (–1)q
= 1 – 1 = 0
- Equation of line √3x + y − 8 = 0 can be represented in normal form as
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Equation of line in normal form can be written as
ax + by + c = 0 √a² + b² √a² + b² √a² + b² ⇒ √3x + y − 8 = 0 √3² + 1² √√3² + 1¹ √√3² + 1¹ √3x + y − 8 = 0 √4 √4 √4 ⇒ √3x + y − 8 = 0 2 2 2 Correct Option: A
Equation of line in normal form can be written as
ax + by + c = 0 √a² + b² √a² + b² √a² + b² ⇒ √3x + y − 8 = 0 √3² + 1² √√3² + 1¹ √√3² + 1¹ √3x + y − 8 = 0 √4 √4 √4 ⇒ √3x + y − 8 = 0 2 2 2
