If the point (k+1,k) lies inside the region bound by the curve `x=sqrt(25-y^2)` and the y-axis, then the integral value of k is/are

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to determine the integral values of \( k \) such that the point \( (k+1, k) \) lies inside the region bounded by the curve \( x = \sqrt{25 - y^2} \) and the y-axis.

### Step-by-step Solution:

1. **Understanding the Curve**:
The equation \( x = \sqrt{25 - y^2} \) describes a semicircle with a radius of 5 centered at the origin (0,0) in the first and second quadrants. The y-axis bounds the region on the left.

2. **Point Condition**:
The point \( (k+1, k) \) lies inside this region. For it to be inside, it must satisfy the following conditions:
- The x-coordinate must be non-negative: \( k + 1 > 0 \)
- The point must also satisfy the curve equation: \( k + 1 < \sqrt{25 - k^2} \)

3. **Finding the x-coordinate condition**:
From \( k + 1 > 0 \):
\[ k > -1 \]

4. **Finding the curve condition**:
Squaring both sides of the inequality \( k + 1 < \sqrt{25 - k^2} \):
\[ (k + 1)^2 < 25 - k^2 \]
Expanding and rearranging gives:
\[ k^2 + 2k + 1 < 25 - k^2 \]
\[ 2k^2 + 2k - 24 < 0 \]
Dividing the entire inequality by 2:
\[ k^2 + k - 12 < 0 \]

5. **Factoring the quadratic**:
We can factor \( k^2 + k - 12 \):
\[ (k + 4)(k - 3) < 0 \]

6. **Finding the intervals**:
The critical points are \( k = -4 \) and \( k = 3 \). We can test intervals:
- For \( k < -4 \): Choose \( k = -5 \) → Positive
- For \( -4 < k < 3 \): Choose \( k = 0 \) → Negative
- For \( k > 3 \): Choose \( k = 4 \) → Positive

Thus, the solution to the inequality \( (k + 4)(k - 3) < 0 \) is:
\[ -4 < k < 3 \]

7. **Combining conditions**:
Now we combine this with the condition \( k > -1 \):
\[ -1 < k < 3 \]

8. **Finding integral values**:
The integral values of \( k \) that satisfy \( -1 < k < 3 \) are:
\[ k = 0, 1, 2 \]

### Final Answer:
The integral values of \( k \) are \( 0, 1, 2 \).