Difference between maximum and minimum values of `(60sin alpha+p cos alpha)` is 122 then p can be

Options

A. 61
B. 11
C. `-61`
D. `-11`

Explanation

To solve the problem, we need to find the value of \( p \) such that the difference between the maximum and minimum values of the expression \( 60 \sin \alpha + p \cos \alpha \) is 122.

### Step-by-Step Solution:

1. **Identify the Maximum and Minimum Values**:
The expression \( a \sin \alpha + b \cos \alpha \) has a maximum value of \( \sqrt{a^2 + b^2} \) and a minimum value of \( -\sqrt{a^2 + b^2} \).
Here, \( a = 60 \) and \( b = p \).

2. **Calculate Maximum Value**:
\[ \text{Maximum value} = \sqrt{60^2 + p^2} = \sqrt{3600 + p^2} \]

3. **Calculate Minimum Value**:
\[ \text{Minimum value} = -\sqrt{60^2 + p^2} = -\sqrt{3600 + p^2} \]

4. **Set Up the Equation for the Difference**:
The difference between the maximum and minimum values is given by:
\[ \text{Difference} = \sqrt{3600 + p^2} - (-\sqrt{3600 + p^2}) = 2\sqrt{3600 + p^2} \]
According to the problem, this difference equals 122:
\[ 2\sqrt{3600 + p^2} = 122 \]

5. **Solve for \( \sqrt{3600 + p^2} \)**:
Divide both sides by 2:
\[ \sqrt{3600 + p^2} = 61 \]

6. **Square Both Sides**:
\[ 3600 + p^2 = 61^2 \]
Calculate \( 61^2 \):
\[ 61^2 = 3721 \]
So, we have:
\[ 3600 + p^2 = 3721 \]

7. **Isolate \( p^2 \)**:
\[ p^2 = 3721 - 3600 \]
\[ p^2 = 121 \]

8. **Find \( p \)**:
Taking the square root of both sides:
\[ p = \pm 11 \]

### Conclusion:
The possible values for \( p \) are \( 11 \) and \( -11 \).