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Question #40789

Difficulty: 8/10 | 🟢 Free

The Question

Consider the family of parabolas $P_t : y = x^2 - 4tx + 5t^2 + 2t$ in the $xy$ -plane, where $t \in \mathbb{R}$ . Let the origin be shifted to a point $V_t(h(t), k(t))$ such that the equation of $P_t$ in the new coordinate system $(X, Y)$ takes the form $Y = X^2$ . A function $f(t)$ is defined as $f(t) = [k(t)]$ , where $[\cdot]$ denotes the greatest integer function. Evaluate the integral $I = \int_{0}^{3} f(t) dt$ .

Options

A.
$61 - \sum_{r=1}^{16} \sqrt{r}$
B.
$64 - \sum_{r=1}^{16} \sqrt{r}$
C.
$61 - \sum_{r=1}^{15} \sqrt{r}$
D.
$58 - \sum_{r=1}^{16} \sqrt{r}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Shifting Origin|Completing the Square in Quadratics|Greatest Integer Function (GIF)

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of coordinate transformations and piecewise integration.

The primary trap lies in the shifting of the origin: students must correctly complete the square to identify the dynamic vertex $(h(t), k(t))$ that reduces the parabola to $Y=X^2$ .

The secondary, more lethal trap is the integration of the GIF $[t^2 + 2t]$ .

Students often attempt to split the integral at roots of $t^2+2t=k$ , leading to a messy, error-prone summation.

The elegant bypass requires recognizing the algebraic structure to complete the square inside the GIF, substituting $u = t+1$ to transform the integrand into $[u^2 - 1]$ .

This allows the use of the standard telescoping area formula for $\int [x^2] dx$ , testing the student's ability to simplify discrete step-functions before brute-force integration.

Quick Facts About This Question

Question Information

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