At the point `P(a , a^n)` on the graph of `y=x^n ,(n in N),` in the first quadrant, a normal is drawn. The normal intersects the `y-a xi s` at the point `(0, b)dot` If `("lim")_(avec0)b=1/2,` then `n` equals 1 (b) 3 (c) 2 (d) 4

Answer

To solve the problem, we need to find the value of \( n \) given the conditions of the normal line to the curve \( y = x^n \) at the point \( P(a, a^n) \) and the limit condition involving \( b \).

### Step 1: Find the derivative of the function
The function is given by:
\[ y = x^n \]
To find the slope of the tangent line at the point \( P(a, a^n) \), we compute the derivative:
\[ \frac{dy}{dx} = n x^{n-1} \]
At the point \( P(a, a^n) \), the slope of the tangent line is:
\[ \frac{dy}{dx} \bigg|_{x=a} = n a^{n-1} \]

### Step 2: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{n a^{n-1}} \]

### Step 3: Write the equation of the normal line
Using the point-slope form of the line equation, the equation of the normal line at point \( P(a, a^n) \) can be written as:
\[ b - a^n = -\frac{1}{n a^{n-1}}(0 - a) \]
Simplifying this, we have:
\[ b - a^n = \frac{a}{n a^{n-1}} \]
\[ b - a^n = \frac{1}{n} a^{2-n} \]
Thus, we can express \( b \) as:
\[ b = a^n + \frac{1}{n} a^{2-n} \]

### Step 4: Apply the limit as \( a \to 0 \)
We need to find the limit:
\[ \lim_{a \to 0} b = \lim_{a \to 0} \left( a^n + \frac{1}{n} a^{2-n} \right) \]
We analyze the two terms separately:
1. \( \lim_{a \to 0} a^n = 0 \) for \( n > 0 \)
2. \( \lim_{a \to 0} \frac{1}{n} a^{2-n} \)

The behavior of the second term depends on the value of \( n \):
- If \( n < 2 \), \( a^{2-n} \to \infty \) as \( a \to 0 \).
- If \( n = 2 \), \( \frac{1}{n} a^{2-n} = \frac{1}{2} \).
- If \( n > 2 \), \( a^{2-n} \to 0 \).

### Step 5: Set the limit equal to \( \frac{1}{2} \)
From the problem statement, we have:
\[ \lim_{a \to 0} b = \frac{1}{2} \]
This condition is satisfied only when \( n = 2 \).

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{2} \]