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(C) Given the function $f(x) = \frac{5+[x]}{\sqrt{11+[x]-6 \sqrt{2+[x]}}}$.For the function to be discontinuous,the denominator must be zero or the ex

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$[7, 8)$

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(C) Given the function $f(x) = \frac{5+[x]}{\sqrt{11+[x]-6 \sqrt{2+[x]}}}$.For the function to be discontinuous,the denominator must be zero or the expression inside the square root must be negative.Let $n = [x]$. The denominator is zero when $11+n - 6\sqrt{2+n} = 0$.Let $u = \sqrt{2+n}$,then $n = u^2 - 2$.Substituting this into the equation: $11 + (u^2 - 2) - 6u = 0$.$u^2 - 6u + 9 = 0$.$(u-3)^2 = 0$,which gives $u = 3$.Since $u = \sqrt{2+n} = 3$,we have $2+n = 9$,so $n = 7$.Thus,$[x] = 7$,which implies $x \in [7, 8)$.Also,for the square root to be defined,$11+[x]-6\sqrt{2+[x]} \ge 0$. Since $([x]-7)^2 \ge 0$,the expression is always non-negative for $n \ge -2$. The discontinuity occurs exactly when the denominator is zero,which is at $[x] = 7$.

Let $[x]$ represent the greatest integer not more than $x$. The discontinuous points of the function $f(x) = \frac{5+[x]}{\sqrt{11+[x]-6 \sqrt{2+[x]}}}$ lie in the interval

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