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Question #40747

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be the curve defined by the equation $y = -x^3 + 5x^2 - 7x - 4$ for $x \in [0, 3]$ . Let $L$ be the line given by $x - y = 0$ . What is the minimum distance between a point on the curve $C$ and the line $L$ ?

Options

A.
$2\sqrt{2}$
B.
$4\sqrt{2}$
C.
$\frac{110\sqrt{2}}{27}$
D.
$5\sqrt{2}$

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

Point-Line Relative Position|Global Extrema on Intervals|Closed vs Open Intervals

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to find the global extrema of a function on a closed interval, specifically targeting the 'Closed vs Open Interval' trap.

A common failure mode among students is to rely exclusively on Fermat's Theorem ( $f'(x)=0$ ) to find minimums.

Here, the distance function has a local minimum at $x=2$ yielding a distance of $4\sqrt{2}$ .

However, because the domain is restricted to the closed interval $[0, 3]$ , the Extreme Value Theorem dictates that endpoints must also be checked.

The true global minimum occurs at the boundary $x=0$ , yielding $2\sqrt{2}$ .

Students who blindly trust the stationary point will incorrectly select the trap option.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Point-Line Relative Position
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free