The least positive integral value of x for which `""^(10)C_(x-1)gt2(""^(10)C_(x))` is

Options

A. 5
B. 8
C. 9
D. 6

Answer

To solve the inequality \( \binom{10}{x-1} > 2 \cdot \binom{10}{x} \), we will follow these steps:

### Step 1: Write the binomial coefficients in terms of factorials
The binomial coefficient \( \binom{n}{r} \) is defined as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Thus, we can express \( \binom{10}{x-1} \) and \( \binom{10}{x} \) as:
\[ \binom{10}{x-1} = \frac{10!}{(x-1)!(10-(x-1))!} = \frac{10!}{(x-1)!(11-x)!} \]
\[ \binom{10}{x} = \frac{10!}{x!(10-x)!} \]

### Step 2: Substitute the binomial coefficients into the inequality
Substituting these into the inequality gives:
\[ \frac{10!}{(x-1)!(11-x)!} > 2 \cdot \frac{10!}{x!(10-x)!} \]

### Step 3: Simplify the inequality
We can cancel \( 10! \) from both sides:
\[ \frac{1}{(x-1)!(11-x)!} > 2 \cdot \frac{1}{x!(10-x)!} \]
This simplifies to:
\[ \frac{1}{(x-1)!(11-x)!} > \frac{2}{x!(10-x)!} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ x!(10-x)! > 2(x-1)!(11-x)! \]

### Step 5: Rewrite \( x! \) and \( (10-x)! \)
Recall that \( x! = x \cdot (x-1)! \) and \( (10-x)! = (10-x)(9-x)! \):
\[ x \cdot (x-1)! \cdot (10-x)(9-x)! > 2 \cdot (x-1)! \cdot (11-x)(10-x)! \]
We can cancel \( (x-1)! \) from both sides:
\[ x(10-x)(9-x)! > 2(11-x)(10-x)! \]

### Step 6: Further simplify
Since \( (10-x)! = (10-x)(9-x)! \), we can substitute:
\[ x(10-x)(9-x)! > 2(11-x)(10-x)(9-x)! \]
Now cancel \( (9-x)! \):
\[ x(10-x) > 2(11-x) \]

### Step 7: Expand and rearrange the inequality
Expanding gives:
\[ 10x - x^2 > 22 - 2x \]
Rearranging yields:
\[ x^2 - 12x + 22 > 0 \]

### Step 8: Solve the quadratic inequality
To find the roots of the equation \( x^2 - 12x + 22 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{144 - 88}}{2} = \frac{12 \pm \sqrt{56}}{2} = \frac{12 \pm 2\sqrt{14}}{2} = 6 \pm \sqrt{14} \]
Calculating \( \sqrt{14} \approx 3.74 \), we find:
\[ x \approx 6 + 3.74 \quad \text{and} \quad x \approx 6 - 3.74 \]
This gives us:
\[ x \approx 9.74 \quad \text{and} \quad x \approx 2.26 \]

### Step 9: Determine the intervals
The quadratic opens upwards, so the inequality \( x^2 - 12x + 22 > 0 \) holds outside the roots:
\[ x < 2.26 \quad \text{or} \quad x > 9.74 \]

### Step 10: Find the least positive integral value of \( x \)
The least positive integral value of \( x \) that satisfies the inequality is:
\[ x = 10 \]

### Final Answer
Thus, the least positive integral value of \( x \) for which \( \binom{10}{x-1} > 2 \cdot \binom{10}{x} \) is \( \boxed{10} \).