If f(x)=k x^3-9 x^2+9 x+3 (k0) is increasing | vedclass

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(B) Given $f(x)=k x^3-9 x^2+9 x+3$ $(k>0)$ is increasing for all $x$. Since $f(x)$ is increasing,its derivative $f^{\prime}(x) \geq 0$ for all $x$. $f

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$k \geq 3$

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(B) Given $f(x)=k x^3-9 x^2+9 x+3$ $(k>0)$ is increasing for all $x$. Since $f(x)$ is increasing,its derivative $f^{\prime}(x) \geq 0$ for all $x$. $f^{\prime}(x) = 3 k x^2-18 x+9$. Setting $f^{\prime}(x) \geq 0$,we get $3 k x^2-18 x+9 \geq 0$,which simplifies to $k x^2-6 x+3 \geq 0$. For a quadratic expression $a x^2+b x+c$ to be non-negative for all $x$ where $a>0$,the discriminant $D$ must be less than or equal to $0$. Here $a=k$,$b=-6$,and $c=3$. $D = b^2-4 a c = (-6)^2-4(k)(3) = 36-12 k$. Setting $D \leq 0$,we have $36-12 k \leq 0$,which implies $12 k \geq 36$,so $k \geq 3$. Thus,the correct condition is $k \geq 3$.

If $f(x)=k x^3-9 x^2+9 x+3$ $(k>0)$ is increasing for all $x$,then

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