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 squar

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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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