Vikas has Rs `(x^3 + 2ax + b)`, with this money he can buy exactly (x - 1) jeans or (x + 1) shirts with no money left. How much money does Vikas have, if x = 4 ?

Options

A. Rs 80
B. Rs 120
C. Rs 30
D. Rs 60

Answer

To solve the problem step by step, we need to find the values of \( a \) and \( b \) from the equation \( x^3 + 2ax + b \) given that Vikas can buy \( (x - 1) \) jeans or \( (x + 1) \) shirts with the same amount of money.

### Step 1: Set up the equations
Given that Vikas can buy \( (x - 1) \) jeans, we can say that the cost of one jean is:
\[ \text{Cost of one jean} = \frac{x^3 + 2ax + b}{x - 1} \]
Since he can buy exactly \( (x - 1) \) jeans, we can set up the first equation:
\[ x^3 + 2ax + b = k(x - 1) \quad \text{for some integer } k \]

Similarly, for \( (x + 1) \) shirts:
\[ \text{Cost of one shirt} = \frac{x^3 + 2ax + b}{x + 1} \]
Setting up the second equation:
\[ x^3 + 2ax + b = m(x + 1) \quad \text{for some integer } m \]

### Step 2: Substitute values for \( x \)
To find \( a \) and \( b \), we can use specific values of \( x \).

1. For \( x = 1 \):
\[ 1^3 + 2a(1) + b = k(1 - 1) \implies 1 + 2a + b = 0 \implies 2a + b = -1 \quad \text{(Equation 1)} \]

2. For \( x = -1 \):
\[ (-1)^3 + 2a(-1) + b = m(-1 + 1) \implies -1 - 2a + b = 0 \implies -2a + b = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( 2a + b = -1 \)
2. \( -2a + b = 1 \)

We can subtract the first equation from the second:
\[ (-2a + b) - (2a + b) = 1 - (-1) \]
This simplifies to:
\[ -4a = 2 \implies a = -\frac{1}{2} \]

Now substitute \( a \) back into Equation 1:
\[ 2(-\frac{1}{2}) + b = -1 \implies -1 + b = -1 \implies b = 0 \]

### Step 4: Substitute \( a \) and \( b \) back into the original equation
Now we have \( a = -\frac{1}{2} \) and \( b = 0 \). We can find the total money Vikas has when \( x = 4 \):
\[ x^3 + 2ax + b = 4^3 + 2(-\frac{1}{2})(4) + 0 \]
Calculating this:
\[ = 64 - 4 + 0 = 60 \]

### Final Answer
Thus, the total amount of money Vikas has is **Rs 60**.

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