If f:[1, 2] rightarrow R defined by f(x) = x^ | vedclass

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(B) Given $f(x) = x^2 + 2kx + k < 0$ for all $x \in [1, 2]$.Since the coefficient of $x^2$ is positive,the parabola opens upward.For a quadratic $f(x)

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$(-\infty, -4/5)$

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(B) Given $f(x) = x^2 + 2kx + k Since the coefficient of $x^2$ is positive,the parabola opens upward.For a quadratic $f(x)$ to be negative on $[1, 2]$,the maximum value of $f(x)$ on the interval $[1, 2]$ must be less than $0$.Since the parabola opens upward,the maximum value on $[1, 2]$ occurs at the endpoints $x=1$ or $x=2$.$f(1) = 1^2 + 2k(1) + k = 1 + 3k $f(2) = 2^2 + 2k(2) + k = 4 + 5k For $f(x) Thus,$k Therefore,the interval for $k$ is $(-\infty, -4/5)$.

If $f:[1, 2] \rightarrow R$ defined by $f(x) = x^2 + 2kx + k$ is always negative for all $x \in [1, 2]$,then the interval in which $k$ lies is:

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