Start Learning Free

Before You Begin

Question #40337

Difficulty: 8/10 | 🟢 Free

The Question

Let f:[−1,1]→Rf: [-1, 1] \to \mathbb{R} be defined by f(x)=∫0xln⁡(t+t2+1)dtf(x) = \int_0^x \ln\left(t + \sqrt{t^2+1}\right) dt Which of the following represents the value of f−1(1−2+ln⁡(1+2))f^{-1}\left( 1 - \sqrt{2} + \ln(1 + \sqrt{2}) \right)?

Options

  • A.

    1

  • B.

    -1

  • C.

    0

  • D.

    \text{Does not exist}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Function Definition (The 'One-to-One' Constraint)|Bijectivity and Inverse|Even and Odd Function Symmetry

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental condition for the existence of an inverse function: bijectivity.

The 'Lethal Collision' occurs because students often rush to algebraically evaluate the integral and solve f(x)=yf(x) = y to find f−1(y)f^{-1}(y), completely ignoring the domain.

By defining the function as the integral of an odd function, f(x)f(x) becomes an even function on a symmetric interval [−1,1][-1, 1].

Even functions inherently violate the one-to-one (injectivity) constraint.

Thus, despite x=1x=1 yielding the target value, the lack of global injectivity renders the inverse function non-existent.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptFunction Definition (The 'One-to-One' Constraint)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree