If the rth term in the expansion of `(x/3-2/x^(2))^(10)` contains `x^(4)`, then r is equal to

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the value of \( r \) such that the \( r \)th term in the expansion of \( \left( \frac{x}{3} - \frac{2}{x^2} \right)^{10} \) contains \( x^4 \), we will follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
In our case, \( a = \frac{x}{3} \), \( b = -\frac{2}{x^2} \), and \( n = 10 \).

### Step 2: Write the expression for the \( r \)th term
Substituting the values, we have:
\[ T_r = \binom{10}{r-1} \left(\frac{x}{3}\right)^{10-(r-1)} \left(-\frac{2}{x^2}\right)^{r-1} \]
This simplifies to:
\[ T_r = \binom{10}{r-1} \left(\frac{x}{3}\right)^{11-r} \left(-\frac{2}{x^2}\right)^{r-1} \]

### Step 3: Simplify the term
Now, we can simplify \( T_r \):
\[ T_r = \binom{10}{r-1} \left(\frac{x^{11-r}}{3^{11-r}}\right) \left(-\frac{2^{r-1}}{x^{2(r-1)}}\right) \]
This gives:
\[ T_r = \binom{10}{r-1} \left(-\frac{2^{r-1}}{3^{11-r}}\right) x^{11-r - 2(r-1)} \]
Simplifying the exponent of \( x \):
\[ T_r = \binom{10}{r-1} \left(-\frac{2^{r-1}}{3^{11-r}}\right) x^{11 - r - 2r + 2} \]
\[ T_r = \binom{10}{r-1} \left(-\frac{2^{r-1}}{3^{11-r}}\right) x^{13 - 3r} \]

### Step 4: Set the exponent of \( x \) equal to 4
We need the term to contain \( x^4 \):
\[ 13 - 3r = 4 \]

### Step 5: Solve for \( r \)
Rearranging the equation:
\[ 13 - 4 = 3r \]
\[ 9 = 3r \]
\[ r = 3 \]

Thus, the value of \( r \) is \( 3 \).

### Summary
The value of \( r \) such that the \( r \)th term in the expansion contains \( x^4 \) is \( r = 3 \).

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