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Question #41398

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a function defined by f(x)={x3+ax2+bx+c,x<1max⁡{2−∣x−2∣,x2−4x+4},x≥1f(x) = \begin{cases} x^3 + ax^2 + bx + c, & x < 1 \\ \max\{2 - |x-2|, x^2 - 4x + 4\}, & x \ge 1 \end{cases} where a,b,c∈Ra, b, c \in \mathbb{R}. If f(x)f(x) is differentiable at x=1x = 1 and has a local extremum at x=−1x = -1, then the sum of all values of xx where f(x)f(x) attains a local extremum is:

Options

  • A.

    296\frac{29}{6}

  • B.

    356\frac{35}{6}

  • C.

    4

  • D.

    −16-\frac{1}{6}

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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

Non-Differentiable Extrema|Polynomial Continuity/Differentiability Link|Limit Existence at Discontinuity

Why This Question Is Tricky

Common Mistakes

This question deploys the 'Lethal Collision' architecture by splicing a polynomial with a non-differentiable piecewise function.

The primary trap tests the boundary of the 'Critical Point' definition.

Students often incorrectly assume that local extrema only occur where f′(x)=0f'(x) = 0, completely missing the points where f′(x)f'(x) does not exist (such as x=2x=2 and x=3x=3 generated by the absolute value and max functions).

Furthermore, the junction point x=1x=1 acts as a decoy; students frequently flag piecewise boundaries as extrema by default, but the forced differentiability constraint (f′(1)=1f'(1)=1) ensures the function is strictly increasing there, disqualifying it.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Derivatives - Class 12
ConceptNon-Differentiable Extrema
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree