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: Understanding the Concept:
Continuity at \(x=\pi/4\) means \(f(\pi/4)=\lim_{x\to\pi/4}f(x)\). At \(x=\pi/4\) both numerator and denominator are zero, so we simplify first.

Step 2: Substitution:
Let \(u=\cos x+\sin x\). Then \(u^2=1+\sin2x\), so \(1-\sin2x=2-u^2\). Also \(2\sqrt2=(\sqrt2)^3\). As \(x\to\pi/4\), \(u\to\sqrt2\).

Step 3: Simplify:
\[ f(x)=\frac{(\sqrt2)^3-u^3}{2-u^2}=\frac{(\sqrt2-u)(2+\sqrt2u+u^2)}{(\sqrt2-u)(\sqrt2+u)}=\frac{2+\sqrt2u+u^2}{\sqrt2+u} \]

Step 4: Take the Limit:
\[ \lim_{u\to\sqrt2}\frac{2+\sqrt2u+u^2}{\sqrt2+u}=\frac{2+2+2}{2\sqrt2}=\frac{6}{2\sqrt2}=\frac{3}{\sqrt2}=\frac{3\sqrt2}{2} \]
So \(f(\pi/4)=\dfrac{3\sqrt2}{2}\).

Final Answer:
The value is \(\dfrac{3\sqrt2}{2}\), option (A). \[ \boxed{\text{(A) } \frac{3\sqrt{2}}{2}} \]