A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by \(y = y(x) = x + \sqrt{2x - 3x^2}\) where ‘x’ (in meters) is the horizontal distance moved and ‘y’ (in meters) is the corresponding height. Assuming that his initial point is at origin.
Based on the above information answer the Questions
The maximum height (in meters) achieved in the first jump is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
From
\( \displaystyle \frac{dy}{dx} = 1 + \frac{1 - 3x}{\sqrt{2x - 3x^2}} \)
dy/dx = 0 only at \(x = \frac{1}{2}\) (the root \(x = \frac{1}{6}\) of the squared equation does not satisfy it). y increases on \(\left(0, \frac{1}{2}\right)\) and decreases on \(\left(\frac{1}{2}, \frac{2}{3}\right)\), so the maximum height is at \(x = \frac{1}{2}\):
\( \displaystyle y\left(\frac{1}{2}\right) = \frac{1}{2} + \sqrt{1 - \frac{3}{4}} \)
\( \displaystyle = \frac{1}{2} + \frac{1}{2} = 1 \)
At the ends of the path, y(0) = 0 and \(y\left(\frac{2}{3}\right) = \frac{2}{3}\), both less than 1. So the maximum height is 1 m.