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Question #41135

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=x+1xf(x) = x + \frac{1}{x} and g(x)=exg(x) = e^x. Define the integrals I1=∫1/22f(g(x)) dxI_1 = \int_{1/2}^{2} f(g(x)) \, dx and I2=∫1/22g(f(x)) dxI_2 = \int_{1/2}^{2} g(f(x)) \, dx. Which of the following inequalities is correct?

Options

  • A.

    I1<e2I_1 < e^2 and I2>32e2I_2 > \frac{3}{2}e^2

  • B.

    I1>e2I_1 > e^2 and I2>32e2I_2 > \frac{3}{2}e^2

  • C.

    I1<e2I_1 < e^2 and I2<32e2I_2 < \frac{3}{2}e^2

  • D.

    I1>e2I_1 > e^2 and I2<32e2I_2 < \frac{3}{2}e^2

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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

Composition Non-Commutativity|Arithmetic Mean Inequality|Inequalities in Definite Integrals

Why This Question Is Tricky

Common Mistakes

This question tests the non-commutativity of function composition alongside the application of the AM-GM inequality within definite integrals.

A common trap is for students to incorrectly evaluate I1I_1 by neglecting the lower bound of integration (assuming ∫ex dx≈e2\int e^x \, dx \approx e^2 and adding the positive ∫e−x dx\int e^{-x} \, dx, leading to the false conclusion I1>e2I_1 > e^2).

Furthermore, students often misapply or confuse the bounds of f(g(x))f(g(x)) and g(f(x))g(f(x)), failing to recognize that the inner function dictates the domain for the outer function's inequality.

Here, AM-GM elegantly provides a strict lower bound for I2I_2, while precise algebraic evaluation bounds I1I_1 from above.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 12
ConceptComposition Non-Commutativity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree