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. **Identify the Curve**:
The equation \( x = \sqrt{25 - y^2} \) represents the right half of a circle centered at the origin with a radius of 5. The full equation of the circle is \( x^2 + y^2 = 25 \).

2. **Determine the Constraints**:
Since the point \( (k+1, k) \) lies inside the region, we need to ensure:
- \( k + 1 \geq 0 \) (since \( x \) must be non-negative)
- The point must satisfy the inequality derived from the circle:
\[ (k + 1)^2 + k^2 < 25 \]

3. **Set Up the Inequality**:
Expanding the inequality:
\[ (k + 1)^2 + k^2 < 25 \]
\[ (k^2 + 2k + 1) + k^2 < 25 \]
\[ 2k^2 + 2k + 1 < 25 \]
\[ 2k^2 + 2k - 24 < 0 \]
Dividing everything by 2 gives:
\[ k^2 + k - 12 < 0 \]

4. **Factor the Quadratic**:
We can factor the quadratic:
\[ (k - 3)(k + 4) < 0 \]

5. **Determine the Intervals**:
The critical points are \( k = -4 \) and \( k = 3 \). We need to test the intervals:
- For \( k < -4 \): Choose \( k = -5 \) → \( (-5 - 3)(-5 + 4) = (-8)(-1) > 0 \) (not valid)
- For \( -4 < k < 3 \): Choose \( k = 0 \) → \( (0 - 3)(0 + 4) = (-3)(4) < 0 \) (valid)
- For \( k > 3 \): Choose \( k = 4 \) → \( (4 - 3)(4 + 4) = (1)(8) > 0 \) (not valid)

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

6. **Combine with Other Constraints**:
We also have the constraint from the non-negativity of \( k + 1 \):
\[ k + 1 \geq 0 \implies k \geq -1 \]

Combining \( -4 < k < 3 \) with \( k \geq -1 \):
\[ -1 \leq k < 3 \]

7. **Identify Integral Values**:
The integral values of \( k \) in the interval \( -1 \leq k < 3 \) are:
- \( k = -1, 0, 1, 2 \)

### Final Answer:
The integral values of \( k \) are:
\[ \{-1, 0, 1, 2\} \]