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Question #40795

Difficulty: 8/10 | 🟢 Free

The Question

Let $\Delta_1(t) = \begin{vmatrix} 1 & t & t^2 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{vmatrix}$ and $\Delta_2(t) = \begin{vmatrix} 1 & t & t^2 \\ 1 & 3 & 9 \\ 1 & 4 & 16 \end{vmatrix}$ . Let $S$ be the set of all integers $t \in [0, 10]$ that satisfy the inequality $\frac{\Delta_1(t)}{\Delta_2(t)} \ge 1$ . Let $h = n(S)$ (the number of elements in $S$ ) and $k = \sum_{t \in S} t$ . A parabola $P$ has its vertex at $(h, k)$ and passes through the point $(h+2, k+12)$ . The $y$ -intercept of the parabola $P$ is:

Options

A.
153
B.
244
C.
30
D.
108

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

Parabola Shifted Vertex|Factor Theorem in Determinants|Determinant Scaling Property|Cross Multiplication Trap

Why This Question Is Tricky

Common Mistakes

This question tests the integration of algebraic determinants, rational inequalities, and coordinate geometry.

The primary trap lies in the rational inequality $\frac{\Delta_1(t)}{\Delta_2(t)} \ge 1$ .

Students often incorrectly cross-multiply the denominator, assuming $(t-3)(t-4) > 0$ , which reduces the inequality to $4t - 10 \ge 0 \implies t \ge 2.5$ .

This 'Cross Multiplication Trap' leads them to erroneously include $t=3$ and $t=4$ in the solution set $S$ , resulting in an incorrect vertex $(8, 52)$ and a wrong $y$ -intercept of 244.

Proper resolution requires subtracting 1 and using the wavy curve method to respect the undefined points at the roots of the denominator.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Conic Sections - Class 11
Concept	Parabola Shifted Vertex
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free