MCQ
If ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}},$then $x =$
- A$1$
- B$3$
- ✓$4$
- D$0$
Clearly $x + 2 = 2x - 2$ ==> $x = 4$
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$(A)$ Range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(B)$ Range of $f \circ g$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(C)$ $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{\pi}{6}$
$(D)$ There is an $x \in R$ such that $( g \circ f )(x)=1$