Algebra
-
If a (2 + √3) = b (2 – √3) = 1, then the value of 1 + 1 is a2 + 1 b2 + 1
-
View Hint View Answer Discuss in Forum
a (2 + √3) = b (2 – √3) = 1
⇒ a = 1 = 2 - √3 = 2 - √3 = 2 - √3 2 + √3 (2 + √3)(2 - √3) 4 - 3 ⇒ b = 1 = 2 + √3 = 2 + √3 = 2 + √3 2 - √3 (2 - √3)(2 + √3) 4 - 3
∴ a² + 1 = (2 – 3 )² + 1
= 4 + 3 – 4√3 + 1 = 8 – 4√3 b² + 1 = (2 + 3 )² + 1
= 4 + 3 + 4√3 + 1 = 8 + 4√3∴ 1 + 1 a² + 1 b² + 1 = 8 + 4√3 + 8 - 4√3 (8 - 4√3)(8 + 4√3) = 16 = 16 = 16 = 1 64 - 16 × 3 64 - 48 16 Correct Option: B
a (2 + √3) = b (2 – √3) = 1
⇒ a = 1 = 2 - √3 = 2 - √3 = 2 - √3 2 + √3 (2 + √3)(2 - √3) 4 - 3 ⇒ b = 1 = 2 + √3 = 2 + √3 = 2 + √3 2 - √3 (2 - √3)(2 + √3) 4 - 3
∴ a² + 1 = (2 – 3 )² + 1
= 4 + 3 – 4√3 + 1 = 8 – 4√3 b² + 1 = (2 + 3 )² + 1
= 4 + 3 + 4√3 + 1 = 8 + 4√3∴ 1 + 1 a² + 1 b² + 1 = 8 + 4√3 + 8 - 4√3 (8 - 4√3)(8 + 4√3) = 16 = 16 = 16 = 1 64 - 16 × 3 64 - 48 16
- If a2 + b2 + c2 = 2a - 2b - 2,then the value of 3a - 2b + c is
-
View Hint View Answer Discuss in Forum
a2 + b2 + c2 = 2a – 2b – 2
⇒ a2 + b2 + c2 – 2a + 2b + 2 = 0
⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 = 0
⇒ (a – 1)2 + (b + 1)2 + c2 = 0
⇒ a – 1 = 0 ⇒ a = 1;
⇒ b + 1 = 0 ⇒ b = –1;
and c = 0
∴ 3a – 2b + c = 3 × 1 – 2 (–1) + 0
= 3 + 2 = 5Correct Option: C
a2 + b2 + c2 = 2a – 2b – 2
⇒ a2 + b2 + c2 – 2a + 2b + 2 = 0
⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 = 0
⇒ (a – 1)2 + (b + 1)2 + c2 = 0
⇒ a – 1 = 0 ⇒ a = 1;
⇒ b + 1 = 0 ⇒ b = –1;
and c = 0
∴ 3a – 2b + c = 3 × 1 – 2 (–1) + 0
= 3 + 2 = 5
- If a = √6 + √5 , b = √6 − √5 then 2a2 – 5 ab + 2b2 = ?
-
View Hint View Answer Discuss in Forum
a = √6 + √5 , b = √6 − √5
a − b = √6 + √5 − √6 + √5 = 2√5
ab = (√6 + √5)(√6 − √5)
= 6 – 5 = 1
∴ 2a2 – 5ab + 2b2= 2 
a2 − 5 ab + b2 
2 = 2 a2 − 2ab + b2 − 1 ab 2
= 2 (a2 – 2ab + b2) – ab
= 2 (a – b)2 – ab
= 2 × (2√5)2 − 1
= 2 × 4 × 5 – 1 = 40 – 1 = 39Correct Option: B
a = √6 + √5 , b = √6 − √5
a − b = √6 + √5 − √6 + √5 = 2√5
ab = (√6 + √5)(√6 − √5)
= 6 – 5 = 1
∴ 2a2 – 5ab + 2b2= 2 a2 − 5 ab + b2 2 = 2 a2 − 2ab + b2 − 1 ab 2
= 2 (a2 – 2ab + b2) – ab
= 2 (a – b)2 – ab
= 2 × (2√5)2 − 1
= 2 × 4 × 5 – 1 = 40 – 1 = 39
- If a1/3 + b1/3 + c1/3 = 0, then a relation among a, b , c is
-
View Hint View Answer Discuss in Forum
If a + b + c = 0
a3 + b3 + c3 = 3abc
∴ If a1/3 + b1/3 + c1/3 = 0
⇒ (a1/3)3 + (b1/3)3 + (c1/3)3
= 3.a1/3.b1/3.c1/3
⇒ a + b + c = 3.a1/3.b1/3.c1/3
⇒ (a + b + c)3
= 33.(a1/3.b1/3.c1/3)3 = 27abcCorrect Option: B
If a + b + c = 0
a3 + b3 + c3 = 3abc
∴ If a1/3 + b1/3 + c1/3 = 0
⇒ (a1/3)3 + (b1/3)3 + (c1/3)3
= 3.a1/3.b1/3.c1/3
⇒ a + b + c = 3.a1/3.b1/3.c1/3
⇒ (a + b + c)3
= 33.(a1/3.b1/3.c1/3)3 = 27abc
- If a + b + c + d = 4, then find the value of
1 + 1 + 1 + 1 . (1 − a)(1 − b)(1 − c) (1 − b)(1 − c)(1 − d) (1 − c)(1 − d)(1 − a) (1 − d)(1 − a)(1 − b)
-
View Hint View Answer Discuss in Forum
a + b + c + d = 4 (Given)
Expression = 1 + 1 + 1 + 1 . (1 − a)(1 − b)(1 − c) (1 − b)(1 − c)(1 − d) (1 − c)(1 − d)(1 − a) (1 − d)(1 − a)(1 − b) = 1 − d + 1 − a + 1 − b + 1 − c (1 − a)(1 − b)(1 − c)(1 − d) = 4 − (a + b + c + d) (1 − a)(1 − b)(1 − c)(1 − d) = 4 − 4 = 0 (1 − a)(1 − b)(1 − c)(1 − d) Correct Option: A
a + b + c + d = 4 (Given)
Expression = 1 + 1 + 1 + 1 . (1 − a)(1 − b)(1 − c) (1 − b)(1 − c)(1 − d) (1 − c)(1 − d)(1 − a) (1 − d)(1 − a)(1 − b) = 1 − d + 1 − a + 1 − b + 1 − c (1 − a)(1 − b)(1 − c)(1 − d) = 4 − (a + b + c + d) (1 − a)(1 − b)(1 − c)(1 − d) = 4 − 4 = 0 (1 − a)(1 − b)(1 − c)(1 − d)
