Sample Test

Formula

Step-by-Step Derivation

1. The equations of the parabolas are $P_1: y = 4x^2$ and $P_2: y = x^2 + 27$.
First, find the points of intersection of $P_1$ and $P_2$:
4. $$ 4x^2 = x^2 + 27 \implies 3x^2 = 27 \implies x^2 = 9 \implies x = \pm 3 $$
6. The area $A_1$ enclosed by $P_1$ and $P_2$ is:
8. $$ A_1 = \int_{-3}^{3} (x^2 + 27 - 4x^2) \, dx = 2 \int_{0}^{3} (27 - 3x^2) \, dx $$
10. $$ A_1 = 2 \left[ 27x - x^3 \right]_{0}^{3} = 2(81 - 27) = 108 $$
12. Next, find the points of intersection of $P_1$ and the line $y = \alpha x$:
13. $$ 4x^2 = \alpha x \implies x = 0 \text{ or } x = \frac{\alpha}{4} $$
14. The area $A_2$ enclosed by $P_1$ and $y = \alpha x$ is:
15. $$ A_2 = \int_{0}^{\alpha/4} (\alpha x - 4x^2) \, dx = \left[ \alpha \frac{x^2}{2} - \frac{4x^3}{3} \right]_{0}^{\alpha/4} $$
16. $$ A_2 = \alpha \left( \frac{\alpha^2}{32} \right) - \frac{4}{3} \left( \frac{\alpha^3}{64} \right) = \frac{\alpha^3}{32} - \frac{\alpha^3}{48} = \frac{\alpha^3}{96} $$
17. We are given that $A_1 = 6 A_2$:
18. $$ 108 = 6 \left( \frac{\alpha^3}{96} \right) \implies 18 = \frac{\alpha^3}{96} \implies \alpha^3 = 1728 $$
19. Taking the cube root, we get $\alpha = 12$.

Final Answer

Therefore, the power of the lens is -2.0D (Option D).

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Where:

• T = Time period (s)
• l = length of string (m)
• g = acceleration due to gravity (m/s²)

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$\cos\theta = \frac{4}{5}$ (taking positive value for acute angle).

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At 25°C, $[\text{H}^+] = 10^{-7}$ M, so pH $= -\log(10^{-7}) = 7$.

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$g_{\text{moon}} \approx \frac{9.8}{6} \approx 1.63 \text{ m/s}^2 \approx 1.6 \text{ m/s}^2$.

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They perform oxidative phosphorylation through the electron transport chain located in their inner membrane.

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$f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3$.

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1 Ampere = 1 Coulomb of charge flowing per second. Volt is the unit of potential difference, Coulomb is the unit of charge, and Ohm is the unit of resistance.

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Its largest moon, Titan, is the only moon in the solar system with a dense atmosphere.