Question

If \(f(x)\) is defined by \[ f(x)= \begin{cases} \dfrac{1-\tan x}{4x-\pi}, & x\ne \dfrac{\pi}{4},\; x\in\left[0,\dfrac{\pi}{2}\right] \\[6pt] k, & x=\dfrac{\pi}{4} \end{cases} \] and \(f(x)\) is continuous in \[ \left[0,\frac{\pi}{2}\right], \] then \(k=\)

• A. \(-1\)
• B. \(-\dfrac{1}{2}\)
• C. \(\dfrac{1}{2}\)
• D. \(1\)

Hint:

For piecewise functions, continuity at the joining point requires that the function value equals the limit. When the limit gives the indeterminate form \(\frac{0}{0}\), L'Hospital's Rule is often the quickest method.

Answer 1

The Correct Option is B

Concept: For continuity at \[ x=\frac{\pi}{4}, \] we must have \[ k=\lim_{x\to\frac{\pi}{4}} \frac{1-\tan x}{4x-\pi}. \] Since both numerator and denominator approach \(0\), use L'Hospital's Rule.

Step 1: Apply the continuity condition. \[ k= \lim_{x\to\frac{\pi}{4}} \frac{1-\tan x}{4x-\pi}. \] Substituting \(x=\frac{\pi}{4}\), \[ \frac{1-\tan\frac{\pi}{4}} {4\left(\frac{\pi}{4}\right)-\pi} = \frac{0}{0}. \] Hence, L'Hospital's Rule is applicable.

Step 2: Differentiate numerator and denominator. \[ k= \lim_{x\to\frac{\pi}{4}} \frac{-\sec^2 x}{4}. \] \[ = -\frac14 \lim_{x\to\frac{\pi}{4}} \sec^2 x. \] \[ = -\frac14\sec^2\frac{\pi}{4}. \] \[ = -\frac14(2). \] \[ = -\frac12. \] Therefore, \[ \boxed{k=-\frac12} \] \[ \boxed{\text{Answer = (B)}} \]