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Question #41554

Difficulty: 8/10 | 🟢 Free

The Question

Let f:(−∞,3]→(−∞,5]f: (-\infty, 3] \to (-\infty, 5] be defined by f(x)=−x2+6x−4f(x) = -x^2 + 6x - 4. If x0x_0 is the unique real root of the equation f(x)=f−1(x)f(x) = f^{-1}(x), evaluate the integral I=∫−4x0f−1(x)dxI = \int_{-4}^{x_0} f^{-1}(x) dx

Options

  • A.

    73\frac{7}{3}

  • B.

    203\frac{20}{3}

  • C.

    43\frac{4}{3}

  • D.

    113\frac{11}{3}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Inverse Function Integration|Substitution (Change of Variable)|Intersection Point Accuracy|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle the intersection of a function and its inverse on a restricted domain, combined with inverse function integration.

While students correctly identify that f(x)=f−1(x)f(x) = f^{-1}(x) simplifies to f(x)=xf(x) = x for strictly increasing functions, they frequently fail to cross-reference the resulting roots with the function's domain.

Here, solving −x2+6x−4=x-x^2 + 6x - 4 = x yields x=1x=1 and x=4x=4.

The primary trap is the extraneous root x=4x=4, which lies outside the domain (−∞,3](-\infty, 3].

Using x=4x=4 as the upper limit leads directly to the incorrect distractor option 20/3.

Furthermore, the problem rigorously tests the application of the inverse integration formula ∫f−1(x)dx=xf−1(x)−∫f(y)dy\int f^{-1}(x)dx = xf^{-1}(x) - \int f(y)dy, requiring precise mapping of the boundary conditions to avoid sign errors.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Integrals - Class 12
ConceptInverse Function Integration
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree