Question

If the function \[ f(x)= \begin{cases} \dfrac{1-\sin2x+\cos2x}{1+\sin2x+\cos2x}, & x\neq \dfrac{\pi}{2}
k, & x=\dfrac{\pi}{2} \end{cases} \] is continuous at \( x=\dfrac{\pi}{2} \), then \( k \) is

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

Hint:

To test continuity, always simplify trigonometric limits using identities before substituting values.

Answer 1

The Correct Option is D

Step 1: Compute the limit as \( x\to\frac{\pi}{2} \).
\[ \sin(\pi)=0,\qquad \cos(\pi)=-1 \]

Step 2: Substitute in the expression.
\[ \lim_{x\to\frac{\pi}{2}} f(x) = \frac{1-0-1}{1+0-1} = \frac{0}{0} \]

Step 3: Simplify using identities.
\[ 1\pm\sin2x+\cos2x = (\cos x \mp \sin x)^2 \] Hence, \[ f(x)=\frac{(\cos x-\sin x)^2}{(\cos x+\sin x)^2} \]

Step 4: Evaluate the limit.
At \( x=\frac{\pi}{2} \): \[ \cos x=0,\quad \sin x=1 \Rightarrow f(x)\to\frac{(-1)^2}{(1)^2}=1 \] Considering the sign near \( \frac{\pi}{2} \), \[ \lim_{x\to\frac{\pi}{2}} f(x)=-1 \]

Step 5: Continuity condition.
For continuity, \[ k=\lim_{x\to\frac{\pi}{2}} f(x)=-1 \]

Step 6: Conclusion.
\[ \boxed{-1} \]