Before You Begin

Question #66590

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha, \beta$ ( $\alpha < \beta$ ) be the roots of the equation $x^2 - 4x + 3 = 0$ . Let $\{t_n\}$ be an arithmetic progression with first term $t_1 = \alpha$ and common difference $d = \beta - \alpha$ . Let $p, q \in \mathbb{Z}$ . Define the function $f: [t_1, t_3] \to \mathbb{R}$ as: $f(x) = \begin{cases} x^3 - t_2 x^2 + (t_4 - 1)x + p & \text{if } t_1 \le x < t_2 \\ x^2 - t_1 x + q & \text{if } t_2 \le x \le t_3 \end{cases}$ If $f(x)$ is strictly increasing on $[t_1, t_3]$ and has a point of discontinuity at $x = t_2$ , find the minimum value of $q - p$ .

Options

A.
13
B.
12
C.
11
D.
0

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

Increasing/Decreasing Strictness|Quadratic Roots in Sequences|Transitivity of Identity

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the 'Transitivity of Identity' trap in the context of piecewise monotonic functions.

Students often incorrectly assume that if a function is strictly increasing on $(a, b)$ and $(b, c)$ , it is strictly increasing on $(a, c)$ without verifying the boundary condition $\lim_{x \to b^-} f(x) \le f(b)$ .

Furthermore, those who do check the boundary frequently conflate the monotonicity condition with the continuity condition (assuming they must be equal).

By explicitly stating the function is discontinuous, the problem forces the strict inequality $q - p > 12$ , trapping students who default to the continuity equation and incorrectly answer 12.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Increasing/Decreasing Strictness
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free