The set of values of a such that x^2 - 2ax + | vedclass

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Question

(B) Let $f(x) = x^2 - 2ax + a^2 - 6a$. For $f(x) \leqslant 0$ to hold for all $x \in [1, 2]$,the values of the function at the endpoints must satisfy

Vedclass · Accepted Answer

$[5 - \sqrt{21}, 4 + \sqrt{15}]$

Vedclass · Answer

(B) Let $f(x) = x^2 - 2ax + a^2 - 6a$. For $f(x) \leqslant 0$ to hold for all $x \in [1, 2]$,the values of the function at the endpoints must satisfy $f(1) \leqslant 0$ and $f(2) \leqslant 0$.Step $1$: Solve $f(1) \leqslant 0$.$f(1) = 1 - 2a + a^2 - 6a = a^2 - 8a + 1 \leqslant 0$.The roots of $a^2 - 8a + 1 = 0$ are $a = \frac{8 \pm \sqrt{64 - 4}}{2} = 4 \pm \sqrt{15}$.Thus,$a \in [4 - \sqrt{15}, 4 + \sqrt{15}]$ ... $(i)$.Step $2$: Solve $f(2) \leqslant 0$.$f(2) = 4 - 4a + a^2 - 6a = a^2 - 10a + 4 \leqslant 0$.The roots of $a^2 - 10a + 4 = 0$ are $a = \frac{10 \pm \sqrt{100 - 16}}{2} = 5 \pm \sqrt{21}$.Thus,$a \in [5 - \sqrt{21}, 5 + \sqrt{21}]$ ... $(ii)$.Step $3$: Find the intersection of $(i)$ and $(ii)$.Since $5 - \sqrt{21} \approx 5 - 4.58 = 0.42$ and $4 - \sqrt{15} \approx 4 - 3.87 = 0.13$,the lower bound is $5 - \sqrt{21}$.Since $4 + \sqrt{15} \approx 4 + 3.87 = 7.87$ and $5 + \sqrt{21} \approx 5 + 4.58 = 9.58$,the upper bound is $4 + \sqrt{15}$.Therefore,$a \in [5 - \sqrt{21}, 4 + \sqrt{15}]$.

The set of values of $a$ such that $x^2 - 2ax + a^2 - 6a \leqslant 0$ for all $x \in [1, 2]$ is:

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