If 1 < x 2, then the value of √(x - 1)^2 + √(3 - x)^2?
- 4
- 5
- 2
- None of these
Explanation
√(x - 1)^2 + √(3 - x)^2
Since 1 < x < 2, we can simplify the expressions inside the square roots:
(x - 1)^2 = (x - 1)(x - 1) = x^2 - 2x + 1
(3 - x)^2 = (3 - x)(3 - x) = 9 - 6x + x^2
Now, add the two expressions:
x^2 - 2x + 1 + 9 - 6x + x^2
= 2x^2 - 8x + 10
Take the square root of the result:
√(2x^2 - 8x + 10)
Since 1 < x < 2, we can plug in x = 1 and x = 2 to find the maximum and minimum values:
At x = 1: √(2 - 8 + 10) = √4 = 2
At x = 2: √(8 - 16 + 10) = √2
So, the minimum value of the expression is 2, which occurs when x = 1.
Therefore, the correct answer is: 2
Related MCQs
- (t² + 4)(t - 2)(t - 2)
- (t² - 4)(t² - 4)
- (t² + 4)(t + 2)(t - 2)
- None of these
اس سوال کو وضاحت کے ساتھ پڑھیں
- +4
- -4
- +16
- None of these
اس سوال کو وضاحت کے ساتھ پڑھیں
- (x − 5)(x + 1)
- (x − √5)(x + √5)
- (x − 2)(x + 3)
- (x − 1)(x + 5)
اس سوال کو وضاحت کے ساتھ پڑھیں
- a = 1/2, b = 1/3
- a = 7/16, b = 3/4
- a = 3/4, b = 7/16
- a = 2, b = 1
اس سوال کو وضاحت کے ساتھ پڑھیں
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