Answer is: option2
\( \sqrt{2} \)Solution:
First, find the derivative of \( y \) with respect to \( x \):
\[ y = \frac{\cos x - m}{\sin x} \]
Using the quotient rule:
\[ y' = \frac{(-\sin x)(\sin x) - (\cos x - m)(\cos x)}{\sin^2 x} \]
Simplify the numerator:
\[ y' = \frac{-\sin^2 x - \cos^2 x + m \cos x}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x) + m \cos x}{\sin^2 x} \]
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ y' = \frac{-1 + m \cos x}{\sin^2 x} \]
For the function to have a maximum at \( x = \frac{\pi}{4} \), the derivative must be zero at that point:
\[ y'\left(\frac{\pi}{4}\right) = \frac{-1 + m \cos\left(\frac{\pi}{4}\right)}{\sin^2\left(\frac{\pi}{4}\right)} = 0 \]
Compute \( \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
Substitute these values:
\[ \frac{-1 + m \cdot \frac{\sqrt{2}}{2}}{\left(\frac{\sqrt{2}}{2}\right)^2} = 0 \Rightarrow \frac{-1 + m \cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 0 \]
Multiply numerator and denominator by 2 to eliminate the fraction:
\[ -2 + m \sqrt{2} = 0 \Rightarrow m \sqrt{2} = 2 \Rightarrow m = \frac{2}{\sqrt{2}} = \sqrt{2} \]
The value of \( m \) is \( \sqrt{2} \).
Answer: (B)