Question

Find the value of \( k \) if the function \( f(x) = \dfrac{k\cos x}{\pi - 2x} \) is continuous at \( x = \dfrac{\pi}{2} \).

• A. \(1\)
• B. \(2\)
• C. \(-1\)
• D. \(0\)

Hint:

When a trigonometric limit gives \( \frac{0}{0} \), use small-angle approximations such as \( \sin h \approx h \) and \( \cos\left(\frac{\pi}{2}+h\right)\approx -h \).

Answer 1

The Correct Option is B

Concept: A function is continuous at a point \(x=a\) if \[ \lim_{x \to a} f(x) = f(a). \] If the function gives an indeterminate form such as \( \frac{0}{0} \), we evaluate the limit using standard limit techniques.

Step 1: Check the value of the function at \( x = \frac{\pi}{2} \). \[ f\left(\frac{\pi}{2}\right) = \frac{k\cos\left(\frac{\pi}{2}\right)}{\pi-2\left(\frac{\pi}{2}\right)} = \frac{k\cdot0}{\pi-\pi} = \frac{0}{0}. \] Thus, the expression gives an indeterminate form.

Step 2: Evaluate the limit \( \lim_{x\to\frac{\pi}{2}} f(x) \). \[ \lim_{x\to\frac{\pi}{2}} \frac{k\cos x}{\pi-2x} = k\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{\pi-2x}. \] Let \( h = x-\frac{\pi}{2} \). As \(x\to\frac{\pi}{2}\), \(h\to0\). \[ \cos x = \cos\left(\frac{\pi}{2}+h\right) = -\sin h \approx -h \] \[ \pi-2x = \pi-2\left(\frac{\pi}{2}+h\right) = -2h \] Therefore, \[ \lim_{x\to\frac{\pi}{2}}\frac{\cos x}{\pi-2x} = \frac{-h}{-2h} = \frac{1}{2}. \]

Step 3: Apply the continuity condition. \[ \lim_{x\to\frac{\pi}{2}} f(x) = k\left(\frac{1}{2}\right). \] For continuity, the limit must be finite and well-defined, giving \[ k\left(\frac{1}{2}\right)=1 \] \[ k=2. \] Thus, \[ \boxed{k=2} \]