Question

y is the sum of three numbers, one of which is a constant, the 2nd varies as x and the 3rd varies inversely as x. The values of y at x = 1, -1 and 3 are respectively 6, -4 and 8. Then, y is equal to

• A. \(1 + x - \frac{1}{x}\)
• B. \(1 + 2x + \frac{3}{x}\)
• C. \(2 + x + \frac{1}{x}\)
• D. \(2 - x + \frac{1}{x}\)

Hint:

Form equations carefully and eliminate variables systematically.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to model the relationship based on the given conditions. Constant means a fixed value, varies as \( x \) means \( kx \), and varies inversely as \( x \) means \( p/x \).
: Key Formula or Approach:
The general form is \( y = a + bx + c/x \). We will plug in the given data points to find constants \( a, b, \) and \( c \).
Step 2: Detailed Explanation:
Given \( y = a + bx + c/x \).
1. At \( x = 1, y = 6 \): \[ a + b + c = 6 \quad \dots \text{(Eq. 1)} \] 2. At \( x = -1, y = -4 \): \[ a - b - c = -4 \quad \dots \text{(Eq. 2)} \] 3. At \( x = 3, y = 8 \): \[ a + 3b + c/3 = 8 \quad \dots \text{(Eq. 3)} \] Solve Eq. 1 and Eq. 2: Adding (Eq. 1) and (Eq. 2): \( (a+b+c) + (a-b-c) = 6 - 4 \implies 2a = 2 \implies a = 1 \).
Subtracting (Eq. 2) from (Eq. 1): \( (a+b+c) - (a-b-c) = 6 - (-4) \implies 2b + 2c = 10 \implies b + c = 5 \).
Substitute \( a = 1 \) into Eq. 3: \[ 1 + 3b + c/3 = 8 \implies 3b + c/3 = 7 \implies 9b + c = 21 \quad \dots \text{(Eq. 4)} \] From \( b + c = 5 \), we get \( c = 5 - b \). Substitute in Eq. 4: \[ 9b + (5 - b) = 21 \implies 8b = 16 \implies b = 2 \] Then \( c = 5 - 2 = 3 \).
So, \( y = 1 + 2x + 3/x \).
Step 3: Final Answer:
The required equation is \( y = 1 + 2x + 3/x \).