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 41 (Calculus)

Based on the above information answer the Questions

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

$\frac{1}{2}$

$\frac{1}{7}$

$\frac{2}{3}$

$\frac{1}{3}$

✅ Correct Option: 1

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

Setting this to zero: $\sqrt{2x - 3x^2} = 3x - 1$ (need $x \geq 1/3$ ).

Squaring: $2x - 3x^2 = 9x^2 - 6x + 1$ , giving $12x^2 - 8x + 1 = 0$ .

$x = \frac{8 \pm 4}{24} = \frac{1}{2}$ or $\frac{1}{6}$ . Since $x \geq 1/3$ , $x = \frac{1}{2}$ .

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

Question 45

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

The value of xxx for which dydx=0\frac{dy}{dx} = 0dxdy​=0, is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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