If [x] denotes the greatest integer function, | vedclass

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(C) For the function $f(x) = \sqrt{\frac{x-[x]}{\log(x^2-x)}}$ to be defined, the following conditions must be met:$1$. The expression inside the squa

Vedclass · Accepted Answer

$R \setminus \left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$

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(C) For the function $f(x) = \sqrt{\frac{x-[x]}{\log(x^2-x)}}$ to be defined, the following conditions must be met:$1$. The expression inside the square root must be non-negative: $\frac{x-[x]}{\log(x^2-x)} \geq 0$.$2$. The denominator must not be zero: $\log(x^2-x) 
eq 0 \Rightarrow x^2-x 
eq 1$.$3$. The argument of the logarithm must be positive: $x^2-x > 0$.Since $x-[x] = \{x\} \geq 0$ for all $x \in R$, the condition $\frac{x-[x]}{\log(x^2-x)} \geq 0$ simplifies to $\log(x^2-x) > 0$.This implies $x^2-x > 1$, or $x^2-x-1 > 0$.Solving $x^2-x-1 = 0$ gives $x = \frac{1 \pm \sqrt{5}}{2}$.The inequality $x^2-x-1 > 0$ holds for $x \in \left(-\infty, \frac{1-\sqrt{5}}{2}ight) \cup \left(\frac{1+\sqrt{5}}{2}, \inftyight)$.Also, we must ensure $x^2-x > 0$, which is satisfied in the intervals above.Thus, the domain is $R \setminus \left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}ight]$.

Column $I$	Column $II$
$(A)$ If $-1 < x < 1$,then $f(x)$ satisfies	$(p)$ $0 < f(x) < 1$
$(B)$ If $1 < x < 2$,then $f(x)$ satisfies	$(q)$ $f(x) < 0$
$(C)$ If $3 < x < 5$,then $f(x)$ satisfies	$(r)$ $f(x) > 0$
$(D)$ If $x > 5$,then $f(x)$ satisfies	$(s)$ $f(x) < 1$

If $[x]$ denotes the greatest integer function, then the domain of the function $f(x) = \sqrt{\frac{x-[x]}{\log(x^2-x)}}$ is

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