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Question #41900

Difficulty: 8/10 | 🟢 Free

The Question

Let S=(−1,1)S = (-1, 1). A binary operation ∗* is defined on SS by x∗y=f−1(f(x)+f(y))x * y = f^{-1}(f(x) + f(y)), where f(x)=∫0xa−∣t∣1−t2dtf(x) = \int_0^x \frac{a - |t|}{1 - t^2} dt and aa is a real parameter. The set of all values of aa for which ∗* is a well-defined binary operation on SS is:

Options

  • A.

    (−∞,0]∪(1,∞)(-\infty, 0] \cup (1, \infty)

  • B.

    (−∞,0]∪[1,∞)(-\infty, 0] \cup [1, \infty)

  • C.

    (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)

  • D.

    (1,∞)(1, \infty)

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Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Invertibility in Binary Operations|Newton-Leibniz Formula with variable limits|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

The problem defines a binary operation via an integral function, requiring students to deduce that closure implies the function's range must be R\mathbb{R}.

Applying the Newton-Leibniz formula gives the derivative, where the modulus function forces a piecewise monotonicity analysis, yielding a≤0a \le 0 or a≥1a \ge 1.

The 'lethal collision' occurs when students assume strict monotonicity is sufficient for invertibility on R\mathbb{R}, failing to check the improper integral at the boundary a=1a=1.

At a=1a=1, the singularity cancels, restricting the range to (−ln⁡2,ln⁡2)(-\ln 2, \ln 2) and destroying the binary operation's closure.

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Question Information

SubjectMathematics
ChapterRelations and Functions - Class 12
ConceptInvertibility in Binary Operations
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree