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Question #40761

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha$ and $\beta$ ( $\alpha < \beta$ ) be the critical points of the function $f(x) = \begin{vmatrix} x & 1 & 0 \\ 0 & x & 1 \\ 72x & 21x & 2x \end{vmatrix}$ . Consider two lines $L_1: \beta x - \alpha y + 10 = 0$ and $L_2: (\alpha \beta)x + (\beta^2 - \alpha^2 - 2)y + 26 = 0$ . Let $m$ be the slope of the acute angle bisector of $L_1$ and $L_2$ . Let $M = \begin{pmatrix} 8m & 2 \\ -2 & 8m \end{pmatrix}$ . If the characteristic equation of $M$ is $\lambda^2 - p\lambda + q = 0$ , find the value of $p+q$ .

Options

A.
4228
B.
3
C.
3972
D.
7

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Angle Bisector Logic|Differentiation of Determinants|Characteristic Equation

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of calculus, coordinate geometry, and linear algebra.

The primary trap lies in the angle bisector logic (the '227' trap).

Students often blindly apply the '+' sign for the bisector without checking the $a_1a_2 + b_1b_2$ condition.

Here, $a_1a_2 + b_1b_2 = 33 > 0$ , meaning the origin is in the obtuse angle, so the '+' sign yields the obtuse bisector (slope $m = -1/8$ ).

Using this incorrect slope in the characteristic equation matrix yields $p+q = 3$ , a highly plausible distractor.

The correct path requires identifying the '-' sign for the acute bisector (slope $m = 8$ ), leading to $p+q = 4228$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Angle Bisector Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free