Algebra
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If x = 7 – 4√3 , then √x + 1 is equal to : √x
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x = 7 – 4√3
∴ √x = √7 – 4√3
= √7 – 2 × 2 × √3
= √4 + 3 − 2 × 2 × √3
= √(2 − √3)2 = 2 − √3∴ 1 = 1 √x 2 − √3 = 1 × 2 + √3 = 2 + √3 2 − √3 2 + √3 4 − 3
= 2 + √3∴ √x + 1 = 2 − √3 + 2 + √3 = 4 √x Correct Option: D
x = 7 – 4√3
∴ √x = √7 – 4√3
= √7 – 2 × 2 × √3
= √4 + 3 − 2 × 2 × √3
= √(2 − √3)2 = 2 − √3∴ 1 = 1 √x 2 − √3 = 1 × 2 + √3 = 2 + √3 2 − √3 2 + √3 4 − 3
= 2 + √3∴ √x + 1 = 2 − √3 + 2 + √3 = 4 √x
- If (a – 1)2 + (b + 2)2 + (c + 1)2 = 0, then the value of 2a – 3b + 7c is
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(a – 1)2 + (b + 2)2 + (c + 1)2 = 0
⇒ a – 1 = 0 ⇒ a = 1;
b + 2 = 0 ⇒ b = –2
c + 1 = 0 ⇒ c = –1
∴ 2a – 3b + 7c
= 2 – 3 (– 2) + 7 (–1)
= 2 + 6 – 7 = 1Correct Option: D
(a – 1)2 + (b + 2)2 + (c + 1)2 = 0
⇒ a – 1 = 0 ⇒ a = 1;
b + 2 = 0 ⇒ b = –2
c + 1 = 0 ⇒ c = –1
∴ 2a – 3b + 7c
= 2 – 3 (– 2) + 7 (–1)
= 2 + 6 – 7 = 1
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If 2x + 1 = 5, find the value of 5x . 3x 6x2 + 20x + 1
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2x + 1 = 5 3x
⇒ 6x2 + 1 = 15x
⇒ 6x2 + 20x + 1 = 15x + 20x = 35x⇒ 5x = 5x = 1 6x2 + 20x + 1 35x 7 Correct Option: D
2x + 1 = 5 3x
⇒ 6x2 + 1 = 15x
⇒ 6x2 + 20x + 1 = 15x + 20x = 35x⇒ 5x = 5x = 1 6x2 + 20x + 1 35x 7
- If x varies inversely as (y2 – 1) and is equal to 24 when y = 10, then the value of x when y = 5 is
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x ∝ 1 y2 − 1 ⇒ x = k y2 − 1
Where k is a constant.
When y = 10, x = 24, then∴ 24 = k ⇒ 24 = k y2 − 1 99
⇒ k = 24 × 99
When y = 5, then∴ x = k = 24 × 99 = 24 × 99 = 99 y2 − 1 52 − 1 24 Correct Option: A
x ∝ 1 y2 − 1 ⇒ x = k y2 − 1
Where k is a constant.
When y = 10, x = 24, then∴ 24 = k ⇒ 24 = k y2 − 1 99
⇒ k = 24 × 99
When y = 5, then∴ x = k = 24 × 99 = 24 × 99 = 99 y2 − 1 52 − 1 24
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If x = 1 − 1 then the value of x – x2 is : a a x
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x = 1 − 1 a a x ⇒ x = x − a a ax
⇒ x2 = x − a
⇒ x − x2 = aCorrect Option: D
x = 1 − 1 a a x ⇒ x = x − a a ax
⇒ x2 = x − a
⇒ x − x2 = a
