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Question #40906

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all critical points of the function $f(x) = \int_0^x (3t-4)(3t-8)[t] \, dt$ in the interval $(1, 4)$ , where $[t]$ denotes the greatest integer less than or equal to $t$ .

A point $P$ divides the line segment joining $A(2, 1, 3)$ and $B(5, 2, 1)$ internally in the ratio $k:1$ , where $k \in S$ . A line $L$ passing through $P$ with direction ratios $(k, [k], k^2 - 5k + 6)$ lies entirely in the plane $\Pi: x - y + z = c$ for some real constant $c$ .

Let $k_{max}$ be the maximum value of $k \in S$ for which this is possible. Find the value of $k_{max}^2 + c^2$ .

Options

A.
25
B.
20
C.
$\frac{208}{9}$
D.
41

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

Ratio of Division (k:1)|Line lying in a plane|Critical Point vs Stationary Point|Greatest Integer Function (GIF)

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between stationary points (where $f'(x)=0$ ) and critical points (where $f'(x)=0$ or does not exist).

Students often incorrectly assume that critical points only occur where the derivative is zero, thereby missing the non-differentiable points at integers caused by the Greatest Integer Function (GIF) jump discontinuities.

The 3D geometry condition $\{k\} + (k-2)(k-3) = 0$ is specifically designed to reject the stationary points ( $4/3, 8/3$ ) and only accept the non-differentiable integer critical points ( $2, 3$ ).

Missing this trap leaves the student with no valid $k$ or forces an incorrect guess.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Ratio of Division (k:1)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free