Algebra
- If a2 + 1 = a, then the value of a3 is
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a2 + 1 = a
⇒ a2 – a + 1 = 0
⇒ (a + 1) (a2 – a + 1) = 0
⇒ a3 + 1 = 0
⇒ a3 = – 1Correct Option: C
a2 + 1 = a
⇒ a2 – a + 1 = 0
⇒ (a + 1) (a2 – a + 1) = 0
⇒ a3 + 1 = 0
⇒ a3 = – 1
- If x + y = √3 and x – y = √2 , then the value of 8xy (x2 + y2) is :
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x + y = √3
x – y = √2
∴ (x + y)2 + (x – y)2 = 3 + 2
⇒ 2 (x2 + y2) = 5 ...(i)
Again,
(x + y)2 – (x – y)2 = 3 – 2
⇒ 4xy = 1 ...(ii)
∴ (x2 + y2) = 5 × 1 = 5Correct Option: C
x + y = √3
x – y = √2
∴ (x + y)2 + (x – y)2 = 3 + 2
⇒ 2 (x2 + y2) = 5 ...(i)
Again,
(x + y)2 – (x – y)2 = 3 – 2
⇒ 4xy = 1 ...(ii)
∴ (x2 + y2) = 5 × 1 = 5
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If a = 1 (a > 0), then the value of a + 1 is a − 5 a
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a = 1 a – 5
⇒ a2 – 5a = 1
⇒ a2 – 5a – 1 = 0∴ a = 5 ± √(−5)² − 4 × 1 × (−1) 2 
If ax² + bx + c = 0, then x = −b ± √(b)² − 4ac 
2a = 5 ± √25 + 4 2 = 5 ± √29 2 If a = 5 ± √29 , then 2 1 = 2 a 5 + √29 2 × 5 − √29 5 + √29 5 − √29 = 2(5 − √29) = 5 − √29 25 − 29 −2 ∴ a + 1 = 5 + √29 − 5 − √29 a 2 2 = 5 + √29 − 5 + √29 = √29 2 Correct Option: A
a = 1 a – 5
⇒ a2 – 5a = 1
⇒ a2 – 5a – 1 = 0∴ a = 5 ± √(−5)² − 4 × 1 × (−1) 2 If ax² + bx + c = 0, then x = −b ± √(b)² − 4ac 2a = 5 ± √25 + 4 2 = 5 ± √29 2 If a = 5 ± √29 , then 2 1 = 2 a 5 + √29 2 × 5 − √29 5 + √29 5 − √29 = 2(5 − √29) = 5 − √29 25 − 29 −2 ∴ a + 1 = 5 + √29 − 5 − √29 a 2 2 = 5 + √29 − 5 + √29 = √29 2
- If a + b + c = 5, ab + bc + ca = 7 and abc = 3, find the value of
a + b + b + c + c + a b a c b a c
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a + b + c = 5;
ab + bc + ca = 7
abc = 3
a² + b² + c² = (a + b + c)² – 2 (ab + bc + ca)
= 25 – 2 × 7 = 11.
Clearly,
a = b = 1, c = 3
a = c = 1, b = 3
b = c = 1, a = 3∴ a + b + b + c + c + a b a c b a c = (1 + 1) + 1 + 3 + 3 + 1 3 3 = 8 + 1 + 1 = 8 2 3 3 3 Correct Option: A
a + b + c = 5;
ab + bc + ca = 7
abc = 3
a² + b² + c² = (a + b + c)² – 2 (ab + bc + ca)
= 25 – 2 × 7 = 11.
Clearly,
a = b = 1, c = 3
a = c = 1, b = 3
b = c = 1, a = 3∴ a + b + b + c + c + a b a c b a c = (1 + 1) + 1 + 3 + 3 + 1 3 3 = 8 + 1 + 1 = 8 2 3 3 3
- p and q are positive numbers satisfying 3p + 2pq = 4 and 5q + pq = 3. Find the value of p.
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3p + 2pq = 4
→ p (3 + 2q) = 4⇒ p = 4 .........(i) 3 + 2q
Now, putting the value of p in
5q + pq = 3,
we get5q + 4 (q) = 3 3 + 2q ⇒ 15q + 10q² + 4q = 3 3 + 2q
⇒ 19q + 10q² = 9 + 6q
⇒ 10q² + 13q – 9 = 0
⇒ 10q² + 18q – 5q – 9 = 0
⇒ 2q (5q + 9) – 1 (5q + 9) = 0
⇒ (2q – 1) (5q + 9) = 0⇒ q = 1 or - 9 2 2
Putting q = 1/2 in (i),⇒ p = 4 = 1 3 + 2 × (1/2) putting q = - 9 5 = p 4 3 + 2 - 9 5 = 4 × 5 15 - 18 = - 20 3 Correct Option: C
3p + 2pq = 4
→ p (3 + 2q) = 4⇒ p = 4 .........(i) 3 + 2q
Now, putting the value of p in
5q + pq = 3,
we get5q + 4 (q) = 3 3 + 2q ⇒ 15q + 10q² + 4q = 3 3 + 2q
⇒ 19q + 10q² = 9 + 6q
⇒ 10q² + 13q – 9 = 0
⇒ 10q² + 18q – 5q – 9 = 0
⇒ 2q (5q + 9) – 1 (5q + 9) = 0
⇒ (2q – 1) (5q + 9) = 0⇒ q = 1 or - 9 2 2
Putting q = 1/2 in (i),⇒ p = 4 = 1 3 + 2 × (1/2) putting q = - 9 5 = p 4 3 + 2 - 9 5 = 4 × 5 15 - 18 = - 20 3
