Algebra
- If x2 + ax + b is a perfect square, then which one of the following relations between a and b is
true ?
-
View Hint View Answer Discuss in Forum
ax2 + bx + c will be a perfect square, if b2 = 4ac
∴ x2 + ax + b will be a perfect square if a2 = 4b
Look : x2 + 2√b x + b
= x2 + 2.x.√b + (√b)2
= (x + √b)2Correct Option: B
ax2 + bx + c will be a perfect square, if b2 = 4ac
∴ x2 + ax + b will be a perfect square if a2 = 4b
Look : x2 + 2√b x + b
= x2 + 2.x.√b + (√b)2
= (x + √b)2
-
If x = √3 − 1 and y = √3 + 1 , then the value of x2 + y2 is √3 √3 y x
-
View Hint View Answer Discuss in Forum
x = √3 − 1 √3 y = √3 + 1 √3 x + y = √3 − 1 + √3 + 1 = 2√3 √3 √3 xy = 
√3 − 1 
√3 + 1 √3 √3 = 3 – 1 = 9 − 1 = 8 3 3 3 ∴ x2 + y2 = x3 + y3 y x xy = (x + y)3 − 3xy(x + y) xy = (2√3)3 − 3 × 8 (2√3) 3 8 3 = 24√3 − 16√3 8/3 = 8√3 × 3 = 3√3 8 Correct Option: B
x = √3 − 1 √3 y = √3 + 1 √3 x + y = √3 − 1 + √3 + 1 = 2√3 √3 √3 xy = √3 − 1 √3 + 1 √3 √3 = 3 – 1 = 9 − 1 = 8 3 3 3 ∴ x2 + y2 = x3 + y3 y x xy = (x + y)3 − 3xy(x + y) xy = (2√3)3 − 3 × 8 (2√3) 3 8 3 = 24√3 − 16√3 8/3 = 8√3 × 3 = 3√3 8
- If a + b = 12, ab = 22, then (a2 + b2) is equal to
-
View Hint View Answer Discuss in Forum
a + b = 12, ab = 22
∴ a2 + b2 = (a + b)2 – 2ab
= (12)2 – 2 × 22
= 144 – 44 = 100Correct Option: D
a + b = 12, ab = 22
∴ a2 + b2 = (a + b)2 – 2ab
= (12)2 – 2 × 22
= 144 – 44 = 100
-
If x = 4 , then the value of 4 + 2y − x is y 5 7 2y + x
-
View Hint View Answer Discuss in Forum
x = 4 (Given) y 5 Expression = 4 + 2y − x 7 2y + x = (4/7) + 2y − x y y 2y + x y y = (4/7) + 2 − x y 2 + x y = (4/7) + 2 − 4 5 2 + 4 5 = (4/7) + 10 − 4 5 10 + 4 5 = 4 + 16 7 14 = 4 + 3 = 7 = 1 7 7 7 Correct Option: C
x = 4 (Given) y 5 Expression = 4 + 2y − x 7 2y + x = (4/7) + 2y − x y y 2y + x y y = (4/7) + 2 − x y 2 + x y = (4/7) + 2 − 4 5 2 + 4 5 = (4/7) + 10 − 4 5 10 + 4 5 = 4 + 16 7 14 = 4 + 3 = 7 = 1 7 7 7
-
If a + 1 = b + 1 = c + 1 (a ≠ b ≠ c), then the value of abc is b c a
-
View Hint View Answer Discuss in Forum
a + 1 = b + 1 = c + 1 = ± 1 b c a ⇒ a + 1 = 1 b
⇒ ab + 1 = b ⇒ ab = b – 1b + 1 = 1, ⇒ 1 = 1 – b c c ⇒ c = 1 1 – b ∴ abc = b – 1 = – 1 1 – b Again, a + 1 = – 1 b
⇒ ab + 1 = – b ⇒ ab = – b – 1b + 1 = – 1 ⇒ 1 = = – 1 – b c c ⇒ c = 1 – 1 – b
∴ abc = 1
∴ abc = ± 1Correct Option: A
a + 1 = b + 1 = c + 1 = ± 1 b c a ⇒ a + 1 = 1 b
⇒ ab + 1 = b ⇒ ab = b – 1b + 1 = 1, ⇒ 1 = 1 – b c c ⇒ c = 1 1 – b ∴ abc = b – 1 = – 1 1 – b Again, a + 1 = – 1 b
⇒ ab + 1 = – b ⇒ ab = – b – 1b + 1 = – 1 ⇒ 1 = = – 1 – b c c ⇒ c = 1 – 1 – b
∴ abc = 1
∴ abc = ± 1
