Algebra
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If x is a rational number and (x + 1)3 - (x - 1)3 = 2 , then the sum of numerator and denominator of x is (x + 1)2 - (x - 1)2
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( x + 1 )3 - ( x - 1 )3 = 2 ( x + 1 )2 - ( x - 1 )2 ⇒ ( x3 + 3x2 + 3x + 1 ) - ( x3 - 3x2 + 3x - 1 ) = 2 ( x2 + 2x + 1 ) - ( x2 - 2x + 1 ) ⇒ x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 = 2 x2 + 2x + 1 - x2 + 2x - 1 ⇒ 6x2 + 2 = 2 4x ⇒ 3x2 + 1 = 1 ⇒ 3x2 + 1 = 4x 4x
⇒ 3x2 - 4x + 1 = 0
⇒ 3x2 - 3x - x + 1 = 0
⇒ 3x(x – 1) – 1(x – 1) = 0
⇒ (3x – 1)(x – 1) = 0
⇒ 3x – 1 = 0, or, x – 1 = 0⇒ x = 1 or 1 3
Hence, sum of the numerator and denominator = 1 + 3 = 4 or, 1 + 1 = 2
Correct Option: B
( x + 1 )3 - ( x - 1 )3 = 2 ( x + 1 )2 - ( x - 1 )2 ⇒ ( x3 + 3x2 + 3x + 1 ) - ( x3 - 3x2 + 3x - 1 ) = 2 ( x2 + 2x + 1 ) - ( x2 - 2x + 1 ) ⇒ x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 = 2 x2 + 2x + 1 - x2 + 2x - 1 ⇒ 6x2 + 2 = 2 4x ⇒ 3x2 + 1 = 1 ⇒ 3x2 + 1 = 4x 4x
⇒ 3x2 - 4x + 1 = 0
⇒ 3x2 - 3x - x + 1 = 0
⇒ 3x(x – 1) – 1(x – 1) = 0
⇒ (3x – 1)(x – 1) = 0
⇒ 3x – 1 = 0, or, x – 1 = 0⇒ x = 1 or 1 3
Hence, sum of the numerator and denominator = 1 + 3 = 4 or, 1 + 1 = 2
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If p + q + r = 1 and a + b + c = 0 a b c p q r where p, q, r and a, b, c are non-zero, then the value of p2 + q2 + r2 is a2 b2 c2
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Let p = x , q = y , r = z a b c
∴ x + y + z = 1and 1 + 1 + 1 = 0 x y z ⇒ yz + xz + xy = 0 xyz
⇒ yz + xz + xy = 0
∴ x + y + z = 1
On squaring both sides ,
x2 + y2 + z2 + 2xy + 2yz + 2zx = 1
⇒ x2 + y2 + z2 + 0 = 1
⇒ x2 + y2 + z2 = 1Correct Option: C
Let p = x , q = y , r = z a b c
∴ x + y + z = 1and 1 + 1 + 1 = 0 x y z ⇒ yz + xz + xy = 0 xyz
⇒ yz + xz + xy = 0
∴ x + y + z = 1
On squaring both sides ,
x2 + y2 + z2 + 2xy + 2yz + 2zx = 1
⇒ x2 + y2 + z2 + 0 = 1
⇒ x2 + y2 + z2 = 1
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If x4 + 1 = 119, and x > 1, then find the positive value of x3 - 1 . x4 x3
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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 ⇒ x - 1 2 + 2 = 11 x ⇒ x - 1 2 = 11 - 2 = 9 = 32 x ⇒ x - 1 = 3 x
On cubing both sides,⇒ x - 1 3 = 27 x ⇒ x3 - 1 - 3x. 1 x - 1 = 27 x3 x x ⇒ x3 - 1 - 3 × 3 = 27 x3 ⇒ x3 - 1 = 27 + 9 = 36 x3 Correct Option: C
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 ⇒ x - 1 2 + 2 = 11 x ⇒ x - 1 2 = 11 - 2 = 9 = 32 x ⇒ x - 1 = 3 x
On cubing both sides,⇒ x - 1 3 = 27 x ⇒ x3 - 1 - 3x. 1 x - 1 = 27 x3 x x ⇒ x3 - 1 - 3 × 3 = 27 x3 ⇒ x3 - 1 = 27 + 9 = 36 x3
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If x2 – 3x + 1 = 0, then the value of x6 + x4 + x2 + 1 will be x3
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x2 – 3 x + 1 = 0
⇒ x2 + 1 = 3 x⇒ x2 + 1 = 3 x ⇒ x + 1 = 3 ........... (i) x ∴ Expression = x6 + x4 + x2 + 1 x3 Expression = x6 + x4 + x2 + 1 x3 x3 x3 x3 Expression = x3 + x + 1 + 1 x x3 Expression = x3 + 1 + x + 1 = 2 × 2 = 4 x3 x Expression = x + 1 3 - 3 . x . 1 x + 1 + x + 1 x x x x
Expression = 33 – 3 × 3 + 3 = 27 – 9 + 3 = 21
Correct Option: C
x2 – 3 x + 1 = 0
⇒ x2 + 1 = 3 x⇒ x2 + 1 = 3 x ⇒ x + 1 = 3 ........... (i) x ∴ Expression = x6 + x4 + x2 + 1 x3 Expression = x6 + x4 + x2 + 1 x3 x3 x3 x3 Expression = x3 + x + 1 + 1 x x3 Expression = x3 + 1 + x + 1 = 2 × 2 = 4 x3 x Expression = x + 1 3 - 3 . x . 1 x + 1 + x + 1 x x x x
Expression = 33 – 3 × 3 + 3 = 27 – 9 + 3 = 21
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If x + 1 = 5, then x6 + 1 is x x6
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x + 1 = 5 x
On cubing both sides,⇒ x + 1 3 = (5)3 x ⇒ x3 + 1 + 3x. 1 x + 1 = 125 x3 x x ⇒ x3 + 1 + 3 × 5 = 125 x3 ⇒ x3 + 1 = 125 - 15 = 110 x3
On squaring both sides,x6 + 1 + 2.x3. 1 = 12100 x6 x3 ⇒ x6 + 1 = 12100 – 2 = 12098 x6 Correct Option: A
x + 1 = 5 x
On cubing both sides,⇒ x + 1 3 = (5)3 x ⇒ x3 + 1 + 3x. 1 x + 1 = 125 x3 x x ⇒ x3 + 1 + 3 × 5 = 125 x3 ⇒ x3 + 1 = 125 - 15 = 110 x3
On squaring both sides,x6 + 1 + 2.x3. 1 = 12100 x6 x3 ⇒ x6 + 1 = 12100 – 2 = 12098 x6