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 45 (Calculus)

Based on the above information answer the Questions

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

$2$

$0$

$3$

$1$

✅ Correct Option: 1

$\frac{dy}{dx} = 1 + \frac{1 - 3x}{\sqrt{2x - 3x^2}}$ .

At $x = \frac{1}{6}$ : $2x - 3x^2 = \frac{1}{3} - \frac{1}{12} = \frac{1}{4}$ , so $\sqrt{2x - 3x^2} = \frac{1}{2}$ .

$1 - 3x = \frac{1}{2}$ .

$\frac{dy}{dx} = 1 + \frac{1/2}{1/2} = 2$ .

Question 41

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

Question 42

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

Question 43

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

Question 44

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

The value of dydx\frac{dy}{dx}dxdy​ when x=16x = \frac{1}{6}x=61​, is

Solution

More from this set:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

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