Algebra
- Given (a – b) = 2, (a3 – b3) = 26 then (a + b)2 is
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(a – b)3 = 23
⇒ a3 – b3 – 3ab (a – b) = 8
⇒ 26 – 3ab × 2 = 8
⇒ 6ab = 26 – 8 = 18⇒ 18 = 3 6
⇒ (a + b)2 = (a – b)2 + 4ab
= (2)2 + 4 × 3 = 4 + 12 = 16Correct Option: C
(a – b)3 = 23
⇒ a3 – b3 – 3ab (a – b) = 8
⇒ 26 – 3ab × 2 = 8
⇒ 6ab = 26 – 8 = 18⇒ 18 = 3 6
⇒ (a + b)2 = (a – b)2 + 4ab
= (2)2 + 4 × 3 = 4 + 12 = 16
- If p = 2 – a, then a3 + 6 ap + p3 – 8 = ?
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p = 2 – a ⇒ a + p – 2 = 0
∴ a3 + 6ap + p3 – 8
= a3 + p3 + (–2)3 – 3ap (–2)
= (a + p – 2) {a2 + p2 + (–2)2 – ap – p (–2) – a (–2)}
= (a + p – 2) (a2 + p2 + 4 – ap + 2p + 2a) = 0Correct Option: A
p = 2 – a ⇒ a + p – 2 = 0
∴ a3 + 6ap + p3 – 8
= a3 + p3 + (–2)3 – 3ap (–2)
= (a + p – 2) {a2 + p2 + (–2)2 – ap – p (–2) – a (–2)}
= (a + p – 2) (a2 + p2 + 4 – ap + 2p + 2a) = 0
- If 4r = h +√r2 + h2 then r : h is ? (r ≠ 0)
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4r = h + √r2 + h2
⇒ 4r − h = √r2 + h2
On squaring both sides,
(4r – h)2 = (√r2 + h2)2
⇒ 16r2 + h2 – 8rh = r2 + h2
⇒ 16r2 – r2 = 8 rh ⇒ 15r2 = 8rh⇒ 15r = 8h ⇒ r = 8 h 15 Correct Option: C
4r = h + √r2 + h2
⇒ 4r − h = √r2 + h2
On squaring both sides,
(4r – h)2 = (√r2 + h2)2
⇒ 16r2 + h2 – 8rh = r2 + h2
⇒ 16r2 – r2 = 8 rh ⇒ 15r2 = 8rh⇒ 15r = 8h ⇒ r = 8 h 15
- If x² – xy + y² = 2 and x4 + x²y² + y4 = 6, then the value of (x² + xy + y²) is :
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x4 + x²y² + y4 = 6
⇒ (x² – xy + y²) (x² + xy + y²) = 6
⇒ 2 × (x² + xy + y²) = 6⇒ x² + xy + y² = 6 = 3 2 Correct Option: C
x4 + x²y² + y4 = 6
⇒ (x² – xy + y²) (x² + xy + y²) = 6
⇒ 2 × (x² + xy + y²) = 6⇒ x² + xy + y² = 6 = 3 2
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If x 
3 − 2 
= 3 , then the value of x2 + 1 will be x x x2
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x 3 − 2 = 3 x x ⇒ 3x − 2 = 3 x ⇒ 3x − 3 = 2 x
On dividing by 3,⇒ x − 1 = 2 x 3
On squaring both sides,⇒ x2 + 1 − 2 = 4 x2 9 ⇒ x2 + 1 = 2 + 4 x2 9 = 2 4 9 Correct Option: D
x 3 − 2 = 3 x x ⇒ 3x − 2 = 3 x ⇒ 3x − 3 = 2 x
On dividing by 3,⇒ x − 1 = 2 x 3
On squaring both sides,⇒ x2 + 1 − 2 = 4 x2 9 ⇒ x2 + 1 = 2 + 4 x2 9 = 2 4 9
