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 42 (Calculus)

Based on the above information answer the Questions

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

$(0, \frac{1}{6})$

$(\frac{1}{6}, \frac{2}{3})$

$(0, \frac{1}{2})$

$(\frac{1}{2}, \frac{2}{3})$

✅ Correct Option: 3

Domain: $0 \leq x \leq \frac{2}{3}$ . Critical point at $x = \frac{1}{2}$ (where $\frac{dy}{dx} = 0$ ).

At $x = 1/6$ : $\frac{dy}{dx} = 1 + \frac{1/2}{1/2} = 2 > 0$ (increasing).

At $x = 0.6$ : $\frac{dy}{dx} < 0$ (decreasing).

So $y(x)$ strictly increases on $(0, \frac{1}{2})$ .

Question 41

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

Question 43

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

Question 44

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

Question 45

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

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

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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