Question

Find the missing number in the series: \( 2, 6, 12, 20, 30, \_\_\_\_ \).

• A. \(38\)
• B. \(40\)
• C. \(42\)
• D. \(44\)

Hint:

In many number series questions, the pattern lies in the differences between consecutive terms. Always check first differences and second differences to identify the rule.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The objective is to find the area enclosed between a standard rightward-opening parabola and the vertical line passing through its focus (the latus rectum).
Step 2: Key Formula or Approach:
1. For the parabola \( y^2 = 4ax \), the latus rectum is the line \( x = a \).
2. The area is symmetric about the \( x \)-axis, so total area \( A = 2 \times \int_{0}^{a} y \, dx \).
Step 3: Detailed Explanation:
The given equation is \( y^2 = 4x \).
Comparing this with the standard form \( y^2 = 4ax \), we find \( 4a = 4 \), which implies \( a = 1 \).
The latus rectum is the line \( x = 1 \).
The required area is bounded by \( x = 0 \) to \( x = 1 \).
From \( y^2 = 4x \), we get \( y = 2\sqrt{x} \) (for the upper half).
\[ \text{Required Area} = 2 \int_{0}^{1} y \, dx \]
\[ \text{Area} = 2 \int_{0}^{1} 2\sqrt{x} \, dx \]
\[ \text{Area} = 4 \int_{0}^{1} x^{1/2} \, dx \]
\[ \text{Area} = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} \]
\[ \text{Area} = 4 \times \frac{2}{3} [x^{3/2}]_{0}^{1} \]
\[ \text{Area} = \frac{8}{3} (1 - 0) = \frac{8}{3} \text{ sq units} \]
Step 4: Final Answer:
The area bounded by the curve and its latus rectum is \( \frac{8}{3} \) sq units.