If f(x)=p x^3+q x^2+r x+t attains local minim | vedclass

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(C) Given $f(x) = px^3 + qx^2 + rx + t$. The derivative is $f'(x) = 3px^2 + 2qx + r$ ... $(i)$ Since $f(x)$ has local extrema at $x = -2$ and $x = 2$,

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$\frac{11}{3}$

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(C) Given $f(x) = px^3 + qx^2 + rx + t$. The derivative is $f'(x) = 3px^2 + 2qx + r$ ... $(i)$ Since $f(x)$ has local extrema at $x = -2$ and $x = 2$,$f'(x) = k(x+2)(x-2) = k(x^2 - 4)$ ... (ii) Comparing $(i)$ and (ii),we get $3p = k$,$2q = 0$,and $r = -4k$. Thus,$q = 0$ and $r = -4(3p) = -12p$. The second derivative is $f''(x) = 6px + 2q = 6px$. For local minimum at $x = -2$,$f''(-2) = -12p > 0$,which implies $p Given $p$ is a root of $9x^2 - 1 = 0$,so $x = \pm \frac{1}{3}$. Since $p Then $r = -12(-\frac{1}{3}) = 4$. Therefore,$p + q + r = -\frac{1}{3} + 0 + 4 = \frac{11}{3}$.

If $f(x)=p x^3+q x^2+r x+t$ attains local minimum and local maximum values at $x=-2$ and $x=2$ respectively and $p$ is a root of $9 x^2-1=0$,then $p+q+r=$

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