AMU 2026 — Mathematics PYQ
AMU | Mathematics | 2026The function is:
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The function sin−1(2x1+x2) is:
Continuous but not differentiable at x=1
Differentiable at x=1
Neither continuous nor differentiable at x=1
(Correct Answer)Continuous everywhere
Neither continuous nor differentiable at x=1
For any inverse sine function y=sin−1(θ), the inner argument θ is strictly constrained by the standard trigonometric domain boundaries:
−1≤θ≤1⟹∣θ∣≤1
Therefore, for our given function to be mathematically defined and exist in the real number system, its argument must satisfy:
2x1+x2≤1
Let's evaluate the behavior of the expression 2x1+x2 using the Arithmetic Mean-Geometric Mean (AM-GM) Inequality for real positive values of x:
2x+x1≥x⋅x1
2xx2+1≥1
Similarly, if x is negative, applying the inequality yields:
2x1+x2≤−1
Combining both scenarios, we discover an essential algebraic truth for this expression:
2x1+x2≥1for all x=0
Now, let's look at the intersection between our required domain condition and the algebraic reality of the expression:
Required Condition: 2x1+x2≤1
Algebraic Reality: 2x1+x2≥1
The only way both conditions can be satisfied simultaneously is when the absolute value is precisely equal to 1:
2x1+x2=1
This equality occurs exclusively at two isolated points on the entire real number line:
When x=1⟹2(1)1+12=22=1
When x=−1⟹2(−1)1+(−1)2=−22=−1
Thus, the domain of the function consists solely of a set containing two isolated singleton numbers:
Domain={−1,1}
Continuity: By definition, a function cannot be continuous in a neighborhood around a point if it does not exist in that neighborhood. Since the function is completely undefined on both the immediate left-hand side (1−δ) and the right-hand side (1+δ) of x=1, we cannot evaluate a standard limit. Therefore, it is not continuous at x=1.
Differentiability: Differentiability fundamentally requires a well-defined continuous functional curve to calculate a local derivative slope (limh→0hf(x+h)−f(x)). Since the function does not exist around x=1, it is not differentiable at x=1.
Conclusively, the function is neither continuous nor differentiable at x=1.
The correct option is (c) Neither continuous nor differentiable at x=1.
For any inverse sine function y=sin−1(θ), the inner argument θ is strictly constrained by the standard trigonometric domain boundaries:
−1≤θ≤1⟹∣θ∣≤1
Therefore, for our given function to be mathematically defined and exist in the real number system, its argument must satisfy:
2x1+x2≤1
Let's evaluate the behavior of the expression 2x1+x2 using the Arithmetic Mean-Geometric Mean (AM-GM) Inequality for real positive values of x:
2x+x1≥x⋅x1
2xx2+1≥1
Similarly, if x is negative, applying the inequality yields:
2x1+x2≤−1
Combining both scenarios, we discover an essential algebraic truth for this expression:
2x1+x2≥1for all x=0
Now, let's look at the intersection between our required domain condition and the algebraic reality of the expression:
Required Condition: 2x1+x2≤1
Algebraic Reality: 2x1+x2≥1
The only way both conditions can be satisfied simultaneously is when the absolute value is precisely equal to 1:
2x1+x2=1
This equality occurs exclusively at two isolated points on the entire real number line:
When x=1⟹2(1)1+12=22=1
When x=−1⟹2(−1)1+(−1)2=−22=−1
Thus, the domain of the function consists solely of a set containing two isolated singleton numbers:
Domain={−1,1}
Continuity: By definition, a function cannot be continuous in a neighborhood around a point if it does not exist in that neighborhood. Since the function is completely undefined on both the immediate left-hand side (1−δ) and the right-hand side (1+δ) of x=1, we cannot evaluate a standard limit. Therefore, it is not continuous at x=1.
Differentiability: Differentiability fundamentally requires a well-defined continuous functional curve to calculate a local derivative slope (limh→0hf(x+h)−f(x)). Since the function does not exist around x=1, it is not differentiable at x=1.
Conclusively, the function is neither continuous nor differentiable at x=1.
The correct option is (c) Neither continuous nor differentiable at x=1.