Question

Which of the following statements is CORRECT?
I. $\sin \left({\cos }^{-1}\left(\tan \left(\frac{\pi }{6}\right)\right)\right)$ is equal to $\sqrt{\frac{2}{3}}$
II. Let $f\left(x\right)=\left[x\right]$ be the greatest integer function. The value of ${lim}_{x\to \infty }\frac{f\left(x\right)}{x}$ is equal to zero.
III. $f\left(x\right)=\left[x\right]+|1-x|$ is non-derivable for all integral values of $x$, where [.] denotes the greatest integer function.

• A. I and II
• B. II and III
• C. III and I
• D. I , II and III

Answer 1

Answer: (C)

III and I

$I.\sin \left({\cos }^{-1}\left(\tan \frac{\pi }{6}\right)\right)$

$=\sin \left({\cos }^{-1}\frac{1}{\sqrt{3}}\right)$

$=\sin \left({\sin }^{-1}\sqrt{\frac{2}{3}}\right)$

$=\sqrt{\frac{2}{3}}$

$II.f\left(x\right)=\left[x\right]$

$\frac{x-1}{x}\le \frac{\left[x\right]}{x}\le \frac{x}{x}$

${lim}_{x\to \infty }(1-\frac{1}{x})\le {lim}_{x\to \infty }\frac{\left[x\right]}{x}\le {lim}_{x\to \infty }1$

$1\le {lim}_{x\to \infty }\frac{\left[x\right]}{x}\le 1$

$\Rightarrow {lim}_{x\to \infty }\frac{\left[x\right]}{x}=1$

$III.f\left(x\right)=\left[x\right]+|1-x|$

$\left[x\right]$ is discontinuous

$\Rightarrow$By Algebra of continuous functions

$g\left(x\right)=\left[x\right],h\left(x\right)=|1-x|$

$f=g+h\Rightarrow f$ is also discontinuous