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Question #41363

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=∫0x(t2−3t+2) dtf(x) = \int_0^x (t^2 - 3t + 2) \,dt and g(x)=x2−6x+8g(x) = x^2 - 6x + 8. If F(x)=f(g(x))F(x) = f(g(x)), then the set of all values of xx for which F(x)F(x) is strictly increasing is:

Options

  • A.

    (3−3,3−2)∪(3,3+2)∪(3+3,∞)(3-\sqrt{3}, 3-\sqrt{2}) \cup (3, 3+\sqrt{2}) \cup (3+\sqrt{3}, \infty)

  • B.

    (−∞,3−3)∪(3−2,3)∪(3+2,3+3)(-\infty, 3-\sqrt{3}) \cup (3-\sqrt{2}, 3) \cup (3+\sqrt{2}, 3+\sqrt{3})

  • C.

    (3−3,3−2)∪(3+2,3+3)(3-\sqrt{3}, 3-\sqrt{2}) \cup (3+\sqrt{2}, 3+\sqrt{3})

  • D.

    (−∞,3−2)∪(3,3+2)∪(3+3,∞)(-\infty, 3-\sqrt{2}) \cup (3, 3+\sqrt{2}) \cup (3+\sqrt{3}, \infty)

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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

Increasing/Decreasing Strictness|Maximum/Minimum of Quadratic Expressions|Compound Function Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze the monotonicity of a composite function F(x)=f(g(x))F(x) = f(g(x)) using the chain rule F′(x)=f′(g(x))g′(x)F'(x) = f'(g(x))g'(x).

A common trap is analyzing only f′(g(x))>0f'(g(x)) > 0 and ignoring the sign change of the inner function's derivative g′(x)g'(x) at its vertex.

By nesting a quadratic inside a function with multiple roots, the problem creates compound boundaries where the global behavior of the quadratic interacts with the local roots of f′(x)f'(x).

Students must perform a rigorous multi-interval sign analysis (wavy curve method) across five critical points, correctly identifying that the monotonicity of F(x)F(x) inverts whenever g(x)g(x) is decreasing.

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Question Information

SubjectMathematics
ChapterApplication of Derivatives - Class 12
ConceptIncreasing/Decreasing Strictness
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree