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Question #41574

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \int_0^1 \left[ \frac{4}{\pi} \tan^{-1}(xt) \right] dt$ and $g(x) = \int_0^1 \left[ \frac{6}{\pi} \cos^{-1}(xt) \right] dt$ , where $[ \cdot ]$ denotes the greatest integer function. An ellipse is given by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , where $a = 10 f'(\sqrt{2})$ and $b = -4 g'\left(\frac{1}{\sqrt{2}}\right)$ . The area of the ellipse is:

Options

A.
$20\pi$
B.
$10\pi$
C.
$40\pi$
D.
$15\pi$

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

Standard Ellipse/Circle Area|Leibniz Rule for Differentiation under Integral|Greatest Integer Function (GIF)|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle parametric differentiation under the integral sign when the integrand contains a Greatest Integer Function (GIF).

The specific trap is the blind application of the Leibniz rule, which would incorrectly yield a derivative of zero because the derivative of a constant integer is zero.

To avoid this 'Lethal Collision', students must recognize that the GIF creates jump discontinuities that shift with the parameter $x$ .

By analyzing the principal value branches of the inverse trigonometric functions, students must partition the integral into sub-intervals where the integrand is constant, thereby expressing the integral as a piecewise rational function of $x$ before differentiating.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Standard Ellipse/Circle Area
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free