Question

If $f\left(x\right)=(1+\sin \frac{\pi }{8x})(1-\sin \frac{(4x-1)\pi }{8x})(1+\sin \frac{(4x+1)\pi }{8x})(1-\sin \frac{(8x-1)\pi }{8x})$ then

• A. $\underset{x\to 0}{lim}{x}^{2}f\left(x\right)=\frac{{\pi }^2}{16}$
• B. For $f\left(x\right)=0$, the largest positive solution of $x$ is $\frac{1}{4}$
• C. $f\left(2\right)=\frac{1}{8}(2-\sqrt{2})$
• D. $f\left(x\right)$ is a even function

Answer 1

Answer: (B), (D)

The correct options are
B For $f\left(x\right)=0$, the largest positive solution of $x$ is $\frac{1}{4}$
D $f\left(x\right)$ is a even function

$f\left(x\right)=(1+\sin \frac{\pi }{8x})(1-\sin (\frac{\pi }{2}-\frac{\pi }{8x}))(1+\sin (\frac{\pi }{2}+\frac{\pi }{8x}))(1-\sin (\pi -\frac{\pi }{8x}))$

$=(1+\sin \frac{\pi }{8x})(1-\sin \frac{\pi }{8x})(1-\cos \frac{\pi }{8x})(1+\cos \frac{\pi }{8x})$

$=(1-{\sin }^2\frac{\pi }{8x})(1-{\cos }^2\frac{\pi }{8x})={\sin }^2\frac{\pi }{8x}{\cos }^2\frac{\pi }{8x}f\left(x\right)=\frac{1}{4}{\sin }^2\frac{\pi }{4x}$

$\underset{x\to 0}{lim}{x}^{2}f\left(x\right)=\underset{x\to 0}{lim}\frac{{\left(\frac{\pi }{4}\right)}^2{\sin }^2\frac{\pi }{4x}}{4{\left(\frac{\pi }{4x}\right)}^2}=\frac{{\pi }^2}{64}$

$f\left(x\right)=0\Rightarrow \frac{\pi }{4x}=n\pi ,n\in Nx=\frac{1}{4n}$
Largest value of $x$ is $\frac{1}{4}$


$f\left(2\right)=\frac{1}{8}(1-\cos \frac{\pi }{4})=\frac{1}{8}(1-\frac{1}{\sqrt{2}})=\frac{(2-\sqrt{2})}{16}$

$f(-x)=f\left(x\right)\Rightarrow f\left(x\right)$ is an even function