Question

If the function \(f(x) = \frac{2\sqrt{2}-(cosx+sinx)^3}{1-sin2x}\) is continuous at \(x = \frac{π}{4}\), then the value of \(f(\frac{π}{4})\) is ...

• A. \(\frac{3\sqrt{2}}{2}\)
• B. \(\frac{5\sqrt{2}}{2}\)
• C. \(0\)
• D. \(\sqrt{2}\)

Hint:

For continuity, f(pi/4) equals the limit; factor the numerator as a difference of cubes.

Answer 1

The Correct Option is A

Step 1: Shift the Variable:
Put $t=x-\pi/4$, so $t\to0$. Then $\cos x+\sin x=\sqrt2\cos t$ and $\sin2x=\sin(2t+\pi/2)=\cos2t$.

Step 2: Rewrite:
$1-\sin2x=1-\cos2t=2\sin^2t$. The numerator is $2\sqrt2-2\sqrt2\cos^3t=2\sqrt2(1-\cos t)(1+\cos t+\cos^2t)$.

Step 3: Limit:
\[ f=\frac{2\sqrt2(1-\cos t)(1+\cos t+\cos^2t)}{2\sin^2t} \]
As $t\to0$: $\dfrac{1-\cos t}{\sin^2t}\to\dfrac12$ and $1+\cos t+\cos^2t\to3$. So the limit is $\dfrac{2\sqrt2\cdot3\cdot\tfrac12}{2}=\dfrac{3\sqrt2}{2}$. Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A) } \frac{3\sqrt{2}}{2}} \]