Mathematics
Assertion (A) : x < -2 and x ≥ 1.
⇒ Solution set S = {x | -2 < x ≤ 1, x ∈ R}
Reason (R) : Two inequations can be written in a combined expression.
Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.
Answer
Given, x < -2 and x ≥ 1
Means, x is less than -2 and x is greater than or equal to 1.
There is no real number that can satisfy both conditions at the same time.
So, assertion (A) is false.
In mathematics, it's common to combine multiple inequalities into a single expression using logical connectors.
For example, x > 3 and x ≤ 5 can be combined as 3 < x ≤ 5.
So, reason (R) is true.
∴ A is false, but R is true.
Hence, option 4 is the correct option.
Related Questions
Assertion (A) : The solution set for :
x + 5 ≤ 10, if the replacement set is {x | x ≤ 5, x ∈ W} is {0, 1, 2, 3, 4, 5}.
Reason (R) : The set of elements of the replacement which satisfy the given inequation is called the solution set.
Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.
Assertion (A) : The solution set for :
x + 3 ≥ 15 is Φ, if the replacement set is {x | x < 10, x ∈ N}.
Reason (R) : If we change over the sides of an equality, we must change the sign from < to > or > to < or ≥ to ≤ or ≤ to ≥.
Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.
Assertion (A) : The solution set in the system of negative integers for : 0 > -4 - p is {-3, -2. -1}.
Reason (R) : If the same quantity is subtracted from both sides of an equation, the symbol of inequality is reversed.
Both A and R are correct, and R is the correct explanation for A.
Both A and R are correct, and R is not the correct explanation for A.
A is true, but R is false.
A is false, but R is true.
If the replacement set is the set of natural numbers, solve:
(i) x - 5 < 0
(ii) x + 1 ≤ 7
(iii) 3x - 4 > 6
(iv) 4x + 1 ≥ 17