MAH-CET 2026 — Mathematics PYQ
MAH-CET | Mathematics | 2026The solution set of is:

The solution set of ∣3−x∣=a is:
{3−a,3+a} for a≥0 and ϕ for a < 0
(Correct Answer){3−a,3+a}
ϕ
insufficient information
{3−a,3+a} for a≥0 and ϕ for a < 0
Step 1: Understand the nature of the modulus function
The modulus function ∣3−x∣ represents an absolute value, which denotes distance on a number line. By definition, the absolute value of any real expression is always non-negative. That means:
∣3−x∣≥0
Step 2: Analyze based on the value of a
Case 1: When a < 0 (Negative value)
An absolute value can never be equal to a negative number. For example, ∣3−x∣=−5 has no real solution.
Therefore, if a < 0, there is no solution, which is represented by the empty set:
ϕ
Case 2: When a≥0 (Non-negative value)
If a is greater than or equal to zero, we can remove the modulus sign by considering both positive and negative possibilities:
3−x=±a
Let's solve for x in both scenarios:
Taking the positive sign (+a):
3−x=a⟹x=3−a
Taking the negative sign (−a):
3−x=−a⟹x=3+a
Thus, when a≥0, the solution set contains the elements:
{3−a,3+a}
Step 3: Combine both cases
Bringing both cases together, the complete solution set is {3−a,3+a} for a≥0 and ϕ for a < 0.
(a) {3−a,3+a} for a≥0 and ϕ for a < 0
Step 1: Understand the nature of the modulus function
The modulus function ∣3−x∣ represents an absolute value, which denotes distance on a number line. By definition, the absolute value of any real expression is always non-negative. That means:
∣3−x∣≥0
Step 2: Analyze based on the value of a
Case 1: When a < 0 (Negative value)
An absolute value can never be equal to a negative number. For example, ∣3−x∣=−5 has no real solution.
Therefore, if a < 0, there is no solution, which is represented by the empty set:
ϕ
Case 2: When a≥0 (Non-negative value)
If a is greater than or equal to zero, we can remove the modulus sign by considering both positive and negative possibilities:
3−x=±a
Let's solve for x in both scenarios:
Taking the positive sign (+a):
3−x=a⟹x=3−a
Taking the negative sign (−a):
3−x=−a⟹x=3+a
Thus, when a≥0, the solution set contains the elements:
{3−a,3+a}
Step 3: Combine both cases
Bringing both cases together, the complete solution set is {3−a,3+a} for a≥0 and ϕ for a < 0.
(a) {3−a,3+a} for a≥0 and ϕ for a < 0