Algebra
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If x2 + 1 = 2 , then the value of x − 1 is x2 x
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⇒ x2 + 1 = 2 x2 ⇒ 
x − 1 
2 + 2x × 1 = 2 x x ⇒ x − 1 2 = 2 – 2 = 0 x ⇒ x − 1 = 0 x Correct Option: B
⇒ x2 + 1 = 2 x2 ⇒ x − 1 2 + 2x × 1 = 2 x x ⇒ x − 1 2 = 2 – 2 = 0 x ⇒ x − 1 = 0 x
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If x + 1 = 4, then the value 9x2 + 1 is 9x 9x2
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x + 1 = 4 9x
On multiplying by 3,3x + 1 = 12 3x
On squaring both sides,3x + 1 2 = (12)2 3x ⇒ 9x2 + 1 + 2 × 3x × 1 9x2 3x
= 144⇒ 9x2 + 1 = 144 – 2 = 142 9x2 Correct Option: B
x + 1 = 4 9x
On multiplying by 3,3x + 1 = 12 3x
On squaring both sides,3x + 1 2 = (12)2 3x ⇒ 9x2 + 1 + 2 × 3x × 1 9x2 3x
= 144⇒ 9x2 + 1 = 144 – 2 = 142 9x2
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If x = 6pq , then the value of x + 3p + x + 3q is p + q x − 3p x − 3q
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x = 6pq = 3p × 2q p + q p + q ⇒ x = 2q 3p p + q ⇒ x + 3p = 2q + p + q x − 3p 2q − p − q
(By componendo and dividendo)⇒ x + 3p = 3q + p .....(i) x − 3p q − p Again, x = 6pq = 2p × 3q p + q p + q ⇒ x = 2p 3q p + q ⇒ x + 3q = 2p + p + q x − 3q 2p − p − q
(By componendo and dividendo)⇒ x + 3q = 3p + q .....(ii) x − 3q p − q ∴ x + 3p + x + 3q = 3q + p + 3p + q x − 3p x − 3q q − p p − q = 3q + p − 3p + q q − p q − p = 3q + p − 3p − q = 2q − 2p q − p q − p = 2(q − p) = 2 q − p Correct Option: C
x = 6pq = 3p × 2q p + q p + q ⇒ x = 2q 3p p + q ⇒ x + 3p = 2q + p + q x − 3p 2q − p − q
(By componendo and dividendo)⇒ x + 3p = 3q + p .....(i) x − 3p q − p Again, x = 6pq = 2p × 3q p + q p + q ⇒ x = 2p 3q p + q ⇒ x + 3q = 2p + p + q x − 3q 2p − p − q
(By componendo and dividendo)⇒ x + 3q = 3p + q .....(ii) x − 3q p − q ∴ x + 3p + x + 3q = 3q + p + 3p + q x − 3p x − 3q q − p p − q = 3q + p − 3p + q q − p q − p = 3q + p − 3p − q = 2q − 2p q − p q − p = 2(q − p) = 2 q − p
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If a + 1 = 4, then the value
of (a − 2)2 +1 2 is : a − 2 a − 2
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a + 1 = 4 a − 2 ⇒ (a − 2) + 1 = 4 − 2 = 2 (a − 2)
On squaring both sides,
(a − 2) + 1 
2 = 4 (a − 2) ⇒ (a − 2)2 + 1 + 2 × (a − 2) × 1 = 4 (a − 2)2 (a − 2) ⇒ (a − 2)2 + 1 = 4 − 2 = 2 (a − 2)2 Correct Option: B
a + 1 = 4 a − 2 ⇒ (a − 2) + 1 = 4 − 2 = 2 (a − 2)
On squaring both sides,(a − 2) + 1 2 = 4 (a − 2) ⇒ (a − 2)2 + 1 + 2 × (a − 2) × 1 = 4 (a − 2)2 (a − 2) ⇒ (a − 2)2 + 1 = 4 − 2 = 2 (a − 2)2
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If ab = 21 and (a + b)2 = 25 then the value of a2 + b2 + 3ab is (a − b)2 4
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(a + b)2 = 25 (a − b)2 4
By componendo and dividendo,(a + b)2 + (a − b)2 = 25 + 4 (a + b)2 − (a − b)2 25 − 4 ⇒ 2(a2 + b2) = 29 4ab 21 ⇒ a2 + b2 = 29 2 × 21 21
⇒ a2 + b2 = 2 × 29 = 58
∴ a2 + b2 + 3ab = 58 + 3 × 21
= 58 + 63 = 121Correct Option: B
(a + b)2 = 25 (a − b)2 4
By componendo and dividendo,(a + b)2 + (a − b)2 = 25 + 4 (a + b)2 − (a − b)2 25 − 4 ⇒ 2(a2 + b2) = 29 4ab 21 ⇒ a2 + b2 = 29 2 × 21 21
⇒ a2 + b2 = 2 × 29 = 58
∴ a2 + b2 + 3ab = 58 + 3 × 21
= 58 + 63 = 121
