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07 May 2026, 19:00
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D
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Difficulty:
35%
(medium)
Question Stats:
76% (01:09) correct
24%
(00:54)
wrong
based on 33
sessions
History
Date
Time
Result
Not Attempted Yet
If |2x – 10| < x + 1, which of the following cannot be a value of x ?
I. 0
II. 7
III. 22
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
I. 0
II. 7
III. 22
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
07 May 2026, 20:54
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Given: |2x − 10| < x + 1
First note: right side must be positive, so:
x + 1 > 0 → x > −1
Now solve by cases.
Case 1: x ≥ 5
Then |2x − 10| = 2x − 10
2x − 10 < x + 1
x < 11
So solution here: 5 ≤ x < 11
Case 2: x < 5
Then |2x − 10| = 10 − 2x
10 − 2x < x + 1
9 < 3x
x > 3
So solution here: 3 < x < 5
Combine both cases:
3 < x < 11
Now test given values:
I. 0 → not in (3, 11), also violates x > −1 condition effect → invalid
II. 7 → in range → valid
III. 22 → outside range → invalid
So values that cannot be x are I and III.
Answer: D
First note: right side must be positive, so:
x + 1 > 0 → x > −1
Now solve by cases.
Case 1: x ≥ 5
Then |2x − 10| = 2x − 10
2x − 10 < x + 1
x < 11
So solution here: 5 ≤ x < 11
Case 2: x < 5
Then |2x − 10| = 10 − 2x
10 − 2x < x + 1
9 < 3x
x > 3
So solution here: 3 < x < 5
Combine both cases:
3 < x < 11
Now test given values:
I. 0 → not in (3, 11), also violates x > −1 condition effect → invalid
II. 7 → in range → valid
III. 22 → outside range → invalid
So values that cannot be x are I and III.
Answer: D
BongBideshini
Joined: 25 May 2021
Last visit: 09 May 2026
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Location: India
Concentration: Entrepreneurship, Marketing
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08 May 2026, 07:34
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Given:
|2x − 10| < x + 1
Since absolute value is always ≥ 0 and x+1 > |2x-10|, both sides are nonnegative.
Hence, we can square both sides:
(2x − 10)^2 < (x + 1)^2
4x^2 − 40x + 100 < x^2 + 2x + 1
3x^2 − 42x + 99 < 0
x^2 − 14x + 33 < 0
(x − 11)(x − 3) < 0
3 < x < 11
Therefore, I and III cannot be values of x.
Answer: D
|2x − 10| < x + 1
Since absolute value is always ≥ 0 and x+1 > |2x-10|, both sides are nonnegative.
Hence, we can square both sides:
(2x − 10)^2 < (x + 1)^2
4x^2 − 40x + 100 < x^2 + 2x + 1
3x^2 − 42x + 99 < 0
x^2 − 14x + 33 < 0
(x − 11)(x − 3) < 0
3 < x < 11
Therefore, I and III cannot be values of x.
Answer: D
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