Let S= [a,b] where `a lt b`. Suppose `f:S to [2,28]` defined by `f(x) = 5 sin x + 12 cos x + 15`. If f is one-to-one and onto, then the least value of b-a is

Options

A. `pi/2`
B. `pi`
C. `3/2pi`
D. `2pi`

Answer

To solve the problem, we need to analyze the function \( f(x) = 5 \sin x + 12 \cos x + 15 \) and determine the conditions under which it is one-to-one and onto from the interval \( S = [a, b] \) to the interval \( [2, 28] \).

### Step-by-step Solution:

1. **Understanding the Function**:
The function is given as:
\[ f(x) = 5 \sin x + 12 \cos x + 15 \]
We need to find the range of \( f(x) \) to ensure it covers the codomain \( [2, 28] \).

2. **Finding the Range of \( 5 \sin x + 12 \cos x \)**:
The expression \( 5 \sin x + 12 \cos x \) can be rewritten in a simpler form using the amplitude formula:
\[ R = \sqrt{a^2 + b^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]
Hence, the range of \( 5 \sin x + 12 \cos x \) is:
\[ [-13, 13] \]

3. **Adjusting for the Constant**:
Adding 15 to the range:
\[ f(x) = 5 \sin x + 12 \cos x + 15 \implies \text{Range of } f(x) = [-13 + 15, 13 + 15] = [2, 28] \]
Thus, the function \( f(x) \) maps \( S \) onto \( [2, 28] \).

4. **Condition for One-to-One**:
The function \( f(x) \) must be either strictly increasing or strictly decreasing on the interval \( S \). The derivative \( f'(x) \) is:
\[ f'(x) = 5 \cos x - 12 \sin x \]
For \( f(x) \) to be one-to-one, \( f'(x) \) must not change sign in the interval \( S \).

5. **Finding Critical Points**:
Set the derivative to zero to find critical points:
\[ 5 \cos x - 12 \sin x = 0 \implies \tan x = \frac{5}{12} \]
This gives us a critical point where the function changes from increasing to decreasing or vice versa.

6. **Determining the Period**:
The period of \( f(x) \) is \( 2\pi \). To ensure the function is one-to-one, we can restrict the interval \( S \) to half the period, which is \( \pi \).

7. **Calculating the Minimum Length of Interval**:
Since the function is periodic with period \( 2\pi \), the least value of \( b - a \) that ensures \( f(x) \) is one-to-one and onto is:
\[ b - a = \pi \]

### Conclusion:
The least value of \( b - a \) is \( \pi \).