Question

Assertion :If $\left[x\right]$ denotes the integral part of $x$, then domain of the function $f\left(x\right)=g\left(x\right)+h\left(x\right)$, where $g\left(x\right)=\frac{\sqrt{3-x}}{(x-1)(x-2)(x-3)}$ and $h\left(x\right)={\sin }^{-1}\left[\frac{3x-2}{2}\right]$ is $[0,2)-\left\{1\right\}$ Reason: Domain of $h\left(x\right)$ is $[0,2)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Domain of $g\left(x\right)$:

$g\left(x\right)$ is defined if $3-x\ge 0$ and $(x-1)(x-2)(x-2)\ne 0$
$\Rightarrow x\le 3$ and $x\ne 1,2,3$
$\therefore$ Domain of $g\left(x\right)=(\infty ,3)-1,2,3$
Domain of $h\left(x\right):$
$h\left(x\right)={\sin }^{-1}\left[\frac{3x-2}{2}\right]\Rightarrow -1\le \left[\frac{3x-2}{2}\right]\le 1$
Case I:
If $\left[\frac{3x-2}{2}\right]=-1\Rightarrow -1\le \frac{3x-2}{2}<0\Rightarrow -2\le 3x-2<0\Rightarrow 0\le x<\frac{2}{3}$ ...(1)
Case II:
$\left[\frac{3x-2}{2}\right]=0\Rightarrow 0\le \frac{3x-2}{2}<1\Rightarrow 0\le 3x-2<2$
$\Rightarrow 2\le 3x<4\Rightarrow \frac{2}{3}\le x<\frac{4}{3}$ ...(2)
Case III:
If $\left[\frac{3x-2}{2}\right]=1\Rightarrow 1\le \frac{3x-2}{2}<2\Rightarrow 2\le 3x-2<4\Rightarrow \frac{4}{3}\le x<2$ ...(3)
Thus, from (1),(2) and (3), we have
Domain of $h\left(x\right)=[0,2)$
$\therefore$ Domain of $f=[2,0)-1$