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Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsFunctions2 Marks
MCQ
The domain of the function
$f(x)=\sin ^{-1}\left(\frac{8.3^{x-2}}{1-3^{2 (x-1)}}\right)$ is
  • A
    $(-\infty, 0]$
  • B
    $[2, \infty)$
  • ✓
    $(-\infty, 0) \cup[2, \infty)$
  • D
    $(-\infty,-1) \cup[1, \infty)$
✓

Answer

Correct option: C.
$(-\infty, 0) \cup[2, \infty)$
(C)
$f (x)$ is defined for
$-1 \leq \frac{8.3^{x-2}}{1-3^{2(x-1)}} \leq 1$
$\Rightarrow-1 \leq \frac{\left(3^2-1\right)\left(3^{x-2}\right)}{1-3^{2 x-2}} \leq 1$
$\Rightarrow-1 \leq \frac{3^x-3^{x-2}}{1-3^{2 x-2}} \leq 1$
$\Rightarrow \frac{3^x-3^{x-2}}{1-3^{2 x-2}}+1 \geq 0$ and $\frac{3^x-3^{x-2}}{1-3^{2 x-2}}-1 \leq 0$
$\Rightarrow \frac{1+3^x-3^{x-2}-3^{2 x-2}}{1-3^{2 x-2}} \geq 0$ and
$\frac{3^x-3^{x-2}-1+3^{2 x-2}}{1-3^{2 x-2}} \leq 0$
$\Rightarrow \frac{\left(3^x+1\right)\left(3^{x-2}-1\right)}{\left(3^x \cdot 3^{x-2}-1\right)} \geq 0$ and $\frac{\left(3^x-1\right)\left(3^{x-2}+1\right)}{\left(3^{2 x-2}-1\right)} \geq 0$
$\Rightarrow \frac{\left(3^{x-2}-1\right)}{\left(3^x \cdot 3^{x-2}-1\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^{2 x-2}-1\right)} \geq 0$
$\Rightarrow \frac{\left(3^x-3^2\right)}{\left(3^{2 x}-3^2\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^{2 x}-3^2\right)} \geq 0$
$\Rightarrow \frac{\left(3^x-3^2\right)}{\left(3^x-3\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^x-3\right)} \geq 0$
$\Rightarrow x \in(-\infty, 1] \cup[2, \infty)$ and $x \in(-\infty, 0] \cup(1, \infty)$
$\Rightarrow x \in(-\infty, 0] \cup[2, \infty)$

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