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Question #41148

Difficulty: 8/10 | 🟢 Free

The Question

Let $a, b \in \mathbb{R}$ and $k \in \mathbb{Z}$ . Consider the function $f(x) = 2x^3 + a x^2 + b x + (k^2 - 4)\text{sgn}(x)$ defined on the interval $[-2, 2]$ . It is given that Lagrange's Mean Value Theorem is applicable to $f(x)$ on $[-2, 2]$ . Let $S$ be the set of all points $c \in (-2, 2)$ satisfying the conclusion of the theorem for $f$ on $[-2, 2]$ . If $S = \{c_1, c_2\}$ and $c_1 + c_2 = 1$ , find the maximum possible value of $a + k$ .

Options

A.
-1
B.
0
C.
1
D.
-5

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

Domain Restrictions|Mean Value Theorem Applicability|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

This question tests the applicability conditions of Lagrange's Mean Value Theorem (LMVT), specifically the strict requirement of continuity on the closed interval.

The trap lies in the signum function, which introduces a jump discontinuity at the origin.

Students often compute the derivative algebraically (ignoring the discontinuity at x=0) and solve $f'(c) = \frac{f(b)-f(a)}{b-a}$ .

Doing so yields a parameter-dependent quadratic $3c^2 + ac - (4 + \frac{k^2-4}{4}) = 0$ .

The sum of the roots remains $-a/3 = 1 \implies a = -3$ .

However, checking the root location condition for $c \in (-2, 2)$ seemingly allows larger integer values for $k$ (up to $k=3$ ), leading to the incorrect maximum sum of 0.

Recognizing that LMVT strictly requires continuity forces $k^2 - 4 = 0$ , correctly restricting $k$ to $\pm 2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Domain Restrictions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free