Question

If $x = f(t)$ changes the interval $ \alpha \le t \le \alpha + 2C $ to $ \beta \le x \le \beta + 2\pi $, then $\frac{f(t)}{\beta} = $

• A. t
• B. $\frac{t}{\alpha}$
• C. $ \alpha t$
• D. $t - \alpha$

Answer 1

The Correct Option is B

We are given that a function \( x = f(t) \) transforms the interval \( \alpha \le t \le \alpha + 2C \) to the interval \( \beta \le x \le \beta + 2\pi \).

We are asked to find the value of \( \frac{f(t)}{\beta} \).

Suppose the transformation is linear, and let us assume that \( f(t) = \frac{\beta}{\alpha} t \). Let's verify if this assumption satisfies the given mapping condition.

When \( t = \alpha \), we have:

\[ f(\alpha) = \frac{\beta}{\alpha} \cdot \alpha = \beta \]

When \( t = \alpha + 2C \), we get:

\[ f(\alpha + 2C) = \frac{\beta}{\alpha} (\alpha + 2C) = \beta + \frac{2C \beta}{\alpha} \]

We are told this should be equal to \( \beta + 2\pi \). Therefore, we equate:

\[ \beta + \frac{2C \beta}{\alpha} = \beta + 2\pi \]

Subtracting \( \beta \) from both sides:

\[ \frac{2C \beta}{\alpha} = 2\pi \Rightarrow \frac{C \beta}{\alpha} = \pi \Rightarrow \beta = \frac{\pi \alpha}{C} \]

So the assumption \( f(t) = \frac{\beta}{\alpha} t \) is valid under the condition \( \beta = \frac{\pi \alpha}{C} \).

Now compute the required expression:

\[ \frac{f(t)}{\beta} = \frac{\frac{\beta}{\alpha} t}{\beta} = \frac{t}{\alpha} \]

Final Answer: \( \boxed{\frac{f(t)}{\beta} = \frac{t}{\alpha}} \)