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For all integral values of x,
|x - 4| x< 5
- a)-1 ≤x≤5
- b)1 ≤x≤5
- c)-1 ≤ x ≤ 1
- d)x<5
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
For all integral values of x,|x- 4| x< 5a)-1 ≤x≤5 b)1 ≤x&l...
At x = 0 inequality is satisfied, option (b) is rejected.
At x = 2, inequality is satisfied, option (c) is rejected.
At x = 5, LHS = RHS.
Thus, option (d) is correct.
Most Upvoted Answer
For all integral values of x,|x- 4| x< 5a)-1 ≤x≤5 b)1 ≤x&l...
If x is greater than or equal to 4, then |x-4| = x-4 and the expression becomes (x-4)x = x^2 - 4x.
If x is less than 4, then |x-4| = -(x-4) = 4-x and the expression becomes (4-x)x = 4x - x^2.
So the overall expression is:
x^2 - 4x if x >= 4
4x - x^2 if x < />
We can graph these two equations separately and then combine them into one graph:
When x is greater than or equal to 4, the graph is a parabola opening upwards, with its vertex at (2, -4).
When x is less than 4, the graph is a parabola opening downwards, with its vertex at (2, 4).
We can see that the two parabolas intersect at x=2, where the value of the expression is -4.
Therefore, the values of the expression for all integral values of x are:
-20, -12, -6, -4, -4, -6, -12, -20... (with x=2 being the minimum value).
If x is less than 4, then |x-4| = -(x-4) = 4-x and the expression becomes (4-x)x = 4x - x^2.
So the overall expression is:
x^2 - 4x if x >= 4
4x - x^2 if x < />
We can graph these two equations separately and then combine them into one graph:
When x is greater than or equal to 4, the graph is a parabola opening upwards, with its vertex at (2, -4).
When x is less than 4, the graph is a parabola opening downwards, with its vertex at (2, 4).
We can see that the two parabolas intersect at x=2, where the value of the expression is -4.
Therefore, the values of the expression for all integral values of x are:
-20, -12, -6, -4, -4, -6, -12, -20... (with x=2 being the minimum value).
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Community Answer
For all integral values of x,|x- 4| x< 5a)-1 ≤x≤5 b)1 ≤x&l...
At x = 0 inequality is satisfied, option (b) is rejected.
At x = 2, inequality is satisfied, option (c) is rejected.
At x = 5, LHS = RHS.
Thus, option (d) is correct.
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For all integral values of x,|x- 4| x< 5a)-1 ≤x≤5 b)1 ≤x≤5c)-1 ≤ x ≤1 d)x<5Correct answer is option 'D'. Can you explain this answer?
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