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Question #41500

Difficulty: 8/10 | 🟢 Free

The Question

Let f:[0,3]→[12,6110]f: [0, 3] \rightarrow \left[\frac{1}{2}, \frac{61}{10}\right] be a bijective function defined by f(x)=2x+∣x−1∣x2+3x+2f(x) = 2x + \frac{|x - 1|}{x^2 + 3x + 2} Find the value of the integral ∫1/261/10f−1(x)dx\int_{1/2}^{61/10} f^{-1}(x) dx.

Options

  • A.

    9310−3ln⁡(109)\frac{93}{10} - 3 \ln\left(\frac{10}{9}\right)

  • B.

    9310+3ln⁡(109)\frac{93}{10} + 3 \ln\left(\frac{10}{9}\right)

  • C.

    18310−3ln⁡(109)\frac{183}{10} - 3 \ln\left(\frac{10}{9}\right)

  • D.

    9310−3ln⁡(59)\frac{93}{10} - 3 \ln\left(\frac{5}{9}\right)

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

Partial Fractions Decomposition|Inverse Function Integration|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

The question tests the integration of an inverse function using the identity ∫abf−1(x)dx=bf−1(b)−af−1(a)−∫f−1(a)f−1(b)f(x)dx\int_a^b f^{-1}(x) dx = b f^{-1}(b) - a f^{-1}(a) - \int_{f^{-1}(a)}^{f^{-1}(b)} f(x) dx.

The primary trap lies in the modulus sign dependency within the resulting integral.

Students often blindly apply partial fraction decomposition without splitting the integral at the critical point x=1x=1 where the modulus argument changes sign.

Failing to partition the domain leads to incorrect logarithmic coefficients.

Furthermore, the problem requires careful algebraic bookkeeping during the definite integration of the rational function to avoid sign errors.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptPartial Fractions Decomposition
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree