If f(x) = operatorname{sgn}((x^2 - kx + 6)(si | vedclass

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(C) The function $f(x) = \operatorname{sgn}(g(x))$ is discontinuous where $g(x) = 0$. Let $g(x) = (x^2 - kx + 6)(\sin x - 1/2)$. The points of discont

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$6$

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(C) The function $f(x) = \operatorname{sgn}(g(x))$ is discontinuous where $g(x) = 0$. Let $g(x) = (x^2 - kx + 6)(\sin x - 1/2)$. The points of discontinuity occur where $x^2 - kx + 6 = 0$ or $\sin x = 1/2$. For $\sin x = 1/2$ in $(0, 6)$,the solutions are $x = \pi/6 \approx 0.52$ and $x = 5\pi/6 \approx 2.62$. These provide $2$ points of discontinuity. For $f(x)$ to have exactly $4$ points of discontinuity,the quadratic $x^2 - kx + 6 = 0$ must have $2$ distinct roots in $(0, 6)$ that are not equal to $\pi/6$ or $5\pi/6$. For the quadratic $x^2 - kx + 6 = 0$ to have $2$ distinct roots in $(0, 6)$: $1) D = k^2 - 24 > 0 \Rightarrow k > \sqrt{24} \approx 4.89$. $2) 0 $3) f(0) = 6 > 0$ (always true). $4) f(6) = 36 - 6k + 6 > 0 \Rightarrow 42 > 6k \Rightarrow k Combining these,$k \in (4.89, 7)$. The integral values for $k$ are $5$ and $6$. Since we need the maximum integral value,$k = 6$.

If $f(x) = \operatorname{sgn}((x^2 - kx + 6)(\sin x - 1/2))$ (where $k > 0$) has exactly $4$ points of discontinuity in $(0, 6)$,then the maximum integral value of $k$ is:

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