Calculus

Application of Integrals

Hard

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.

Figure for CUET Mathematics 2022 6 Aug Shift 2 question 43 (Calculus)

Based on the above information answer the Questions

The maximum height (in meters) achieved in the first jump is

$\frac{2}{3}$

$1$

$\frac{1}{2}$

$\frac{1}{6}$

✅ Correct Option: 2

Maximum occurs at $x = \frac{1}{2}$ .

$y(\frac{1}{2}) = \frac{1}{2} + \sqrt{1 - \frac{3}{4}} = \frac{1}{2} + \frac{1}{2} = 1$ .

Question 41

The value of $x$ for which $\frac{dy}{dx} = 0$ , is

Question 42

$y(x)$ is strictly increasing in the interval

Question 44

Evaluate $\int_0^{\frac{1}{3}} y \, dx$

Question 45

The value of $\frac{dy}{dx}$ when $x = \frac{1}{6}$ , is

The maximum height (in meters) achieved in the first jump is

Solution

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