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Question #41323

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be a curve defined parametrically by $x = t + 1$ and $y = t^4 - 15t^2 - 12$ for $t \in \left[-\frac{3}{2}, \frac{5}{2}\right]$ . The origin of the coordinate system is shifted to the point $(1, 0)$ without rotating the axes, establishing a new $(X, Y)$ coordinate system. Let $S$ be the sum of the slopes of all lines passing through the new origin $(0,0)$ that are tangent to $C$ . Find the value of $S$ .

Options

A.
$-28$
B.
$0$
C.
$28$
D.
$-54$

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

Parametric Derivative Endpoint Trap|Condition of Tangency|Shifting Origin

Why This Question Is Tricky

Common Mistakes

This question tests the 'Parametric Domain Endpoint Trap' combined with a coordinate shift.

Students typically set up the tangency condition algebraically by equating the slope of the secant from the origin to the derivative, yielding a polynomial in parameter $t$ .

Solving this gives four roots: $t \in \{-2, -1, 1, 2\}$ .

The trap lies in blindly accepting all roots and summing their corresponding slopes (which incorrectly yields 0).

The coordinate shift obscures the original parametric domain $t \in [-1.5, 2.5]$ .

Candidates must verify that the tangency points actually exist on the restricted curve segment, leading to the necessary rejection of the extraneous root $t = -2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Parametric Derivative Endpoint Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free