Question

The domain of definition of the function, $f(x)=[2\tan \pi x{]}^{\log [2\tan \pi x]}\left(\frac{x^2+2x-3}{4x^2-4x-3}\right)$ where [ ] denotes the greatest integer function is $\left[n+\frac{1}{4},n+\frac{1}{2}\right);n\in I$ then

• A. $n=0$
• B. $n\le -4$
• C. $n\ge 4$
• D. none

Answer 1

Answer: (A), (B), (C)

$f(x)=a^{{\log }_{a}N}N$ and $0<[2\tan \pi x]$ or $[2\tan \pi x]>1$
$\text{ and }\frac{x^2+2x-3}{4x^2-4x-3}>0\text{ i.e. }\frac{(x+3)(x-1)}{(2x-3)(2x+1)}>0$
$\Rightarrow x\in (-\infty ,-3)\cup (-1/2,1)\cup (3/2,\infty )$.....(i)
$\text{ now }0<[2\tan \pi x]<1\text{ not possible }\therefore [2\tan \pi x]>1$ $\Rightarrow 2\tan \pi x\ge 2\Rightarrow \tan \pi x\ge 1$
$\Rightarrow n\pi +\frac{\pi }{4}\le \pi x<n\pi +\frac{\pi }{2}n\in I$
$\Rightarrow n+\frac{1}{4}\le x<n+\frac{1}{2}$<....(ii)br/> Common solution of (i) and (ii) possible only if $n=0,n\ge 2$ or $n\le -4$ ]