Multiple choice

Consider a circle with unit radius whose chord AB is at a perpendicular distance x from centre C. Let area of triangle ACB is given by function y = f(x). On the curve y = f(x) a variable point P(x,f(x)) is taken and perpendiculars PQ and PR are drawn on x & y axis respectively. Area of rectangle QPRO is given by y = g(x). (where 'O' is origin). List I List II P. The greatest area of $\triangle$ ACB is 1. $\displaystyle \frac{2}{3 \sqrt{3}}$ Q. The greatest area of rectangle QPRO is 2. $\displaystyle \frac{\sqrt{13} -1 }{6}$ R. The greatest vertical distance between y = (x) & y = g(x) occurs at x equal to 3. $\displaystyle \frac{1}{3}$ S. The area bounded by y = f(x), y = 0 & x = 1 is 4. $\displaystyle \frac{1}{2}$

  1. P -1, Q -2, R - 3, S - 4

  2. P -4, Q - 3, R - 2, S - 1

  3. P -3, Q - 1, R - 2, S - 4

  4. P - 4, Q - 1, R - 2, S - 3
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D Correct answer
Explanation

This is a standard calculus/geometry problem involving optimization of areas. P corresponds to the maximum area of the triangle (4), Q to the rectangle (1), R to the distance (2), and S to the bounded area (3).

AI explanation

The half chord length is the square root of (1 minus x squared), making the triangle area function f(x) equal to x times the square root of (1 minus x squared). Maximizing this area by taking the derivative and setting it to zero gives x squared equals one half, so the maximum area of triangle ACB is one half. The area of the rectangle QPRO is g(x) equals x squared times the square root of (1 minus x squared), which maximizes at x squared equals two thirds to give an area of 2 divided by (3 times the square root of 3). The greatest vertical distance between the curves is found by maximizing their difference, yielding an x value of (square root of 13 minus 1) divided by 6. The area bounded by y equals f(x), y equals 0, and x equals 1 evaluates to 1 divided by 3.

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