Suppose `lim_(xrarr0)(int_(0)^(x)(t^(2) dt)/((a+t^(r))^(1//p)))/(bx- sinx)=l`,
`p in N, p ge 2,a gt gt 0,rgt 0 and b ne 0`
If `p=3 and l=1,` then the value of 'a' is equal to

Options

A. 8
B. 3
C. 6
D. `3//2`

Answer

To solve the given limit problem step by step, we start with the limit expression:

\[ \lim_{x \to 0} \frac{\int_{0}^{x} t^{2} \, dt}{(a + t^{r})^{1/p}} \cdot \frac{1}{bx - \sin x} = l \]

Given that \( p = 3 \) and \( l = 1 \), we need to find the value of \( a \).

### Step 1: Evaluate the integral in the numerator

The integral \(\int_{0}^{x} t^{2} \, dt\) can be calculated as follows:

\[ \int_{0}^{x} t^{2} \, dt = \left[ \frac{t^{3}}{3} \right]_{0}^{x} = \frac{x^{3}}{3} \]

### Step 2: Substitute the integral back into the limit

Now, substituting the integral back into the limit gives us:

\[ \lim_{x \to 0} \frac{\frac{x^{3}}{3}}{(a + x^{r})^{1/3}} \cdot \frac{1}{bx - \sin x} = 1 \]

### Step 3: Analyze the denominator \(bx - \sin x\)

As \(x \to 0\), we can use the Taylor series expansion for \(\sin x\):

\[ \sin x \approx x - \frac{x^{3}}{6} \]

Thus, we have:

\[ bx - \sin x \approx bx - \left(x - \frac{x^{3}}{6}\right) = (b - 1)x + \frac{x^{3}}{6} \]

### Step 4: Rewrite the limit expression

Now the limit expression can be rewritten as:

\[ \lim_{x \to 0} \frac{\frac{x^{3}}{3}}{(a + x^{r})^{1/3} \cdot ((b - 1)x + \frac{x^{3}}{6})} = 1 \]

### Step 5: Apply L'Hôpital's Rule

Since we have a \(0/0\) form, we can apply L'Hôpital's Rule. We differentiate the numerator and denominator with respect to \(x\):

- The derivative of the numerator \(\frac{x^{3}}{3}\) is \(x^{2}\).
- The derivative of the denominator \((a + x^{r})^{1/3} \cdot ((b - 1)x + \frac{x^{3}}{6})\) requires the product rule.

Let \(u = (a + x^{r})^{1/3}\) and \(v = (b - 1)x + \frac{x^{3}}{6}\).

Using the product rule:

\[ \frac{d}{dx}(uv) = u'v + uv' \]

### Step 6: Evaluate the derivatives

1. **For \(u\)**:
\[ u' = \frac{1}{3}(a + x^{r})^{-2/3} \cdot r x^{r-1} \]

2. **For \(v\)**:
\[ v' = b - 1 + \frac{x^{2}}{2} \]

### Step 7: Substitute back into the limit

Now we substitute back into the limit and evaluate as \(x \to 0\):

\[ \lim_{x \to 0} \frac{x^{2}}{u'v + uv'} = 1 \]

### Step 8: Set up the equation

At \(x = 0\):

\[ u(0) = (a)^{1/3}, \quad v(0) = 0 \]

This means we need to ensure that:

\[ (b - 1) = 0 \implies b = 1 \]

### Step 9: Solve for \(a\)

Now we have:

\[ \lim_{x \to 0} \frac{x^{2}}{(a)^{1/3} \cdot (b - 1)} = 1 \]

Substituting \(b = 1\) gives us:

\[ \lim_{x \to 0} \frac{x^{2}}{(a)^{1/3} \cdot 0} \text{ is undefined, so we need to differentiate again.} \]

After further analysis, we find that:

\[ \frac{2}{(a)^{1/3}} = 1 \implies a = 8 \]

### Final Answer

Thus, the value of \(a\) is:

\[ \boxed{8} \]