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Question #40339

Difficulty: 8/10 | 🟢 Free

The Question

Let f:[k,5]→[−5,54]f: [k, 5] \to \left[-5, \frac{5}{4}\right] be a bijective function defined by f(x)=∣x2−6x+5∣−xf(x) = |x^2 - 6x + 5| - x. If g(x)g(x) is the inverse function of f(x)f(x), find the value of g(−1)+g′(−1)g(-1) + g'(-1).

Options

  • A.

    113\frac{11}{3}

  • B.

    43\frac{4}{3}

  • C.

    103\frac{10}{3}

  • D.

    83\frac{8}{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

Codomain vs Range Distinction|Bijectivity and Inverse|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to distinguish between codomain and range while handling the 'Modulus Trap'.

The absolute value function forces a folding of the parabola, creating a piecewise function with a local maximum at x=2.5x=2.5.

Students often blindly remove the modulus or ignore the domain restriction, leading to incorrect branches of the inverse function (e.g., choosing x=1x=1 instead of x=4x=4).

By enforcing bijectivity onto a specific codomain, the problem requires students to deduce the exact domain [2.5,5][2.5, 5] before applying the Inverse Function Theorem, penalizing those who perform algebraic manipulation without checking mapping properties.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptCodomain vs Range Distinction
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree