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

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(B) Let $f(x) = x^2 - 2ax + a^2 - 6a$. For $f(x) \leqslant 0$ to hold for all $x \in [1, 2]$,we must have $f(1) \leqslant 0$ and $f(2) \leqslant 0$.St

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

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(B) Let $f(x) = x^2 - 2ax + a^2 - 6a$. For $f(x) \leqslant 0$ to hold for all $x \in [1, 2]$,we must have $f(1) \leqslant 0$ 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 $a = \frac{8 \pm \sqrt{64 - 4}}{2} = 4 \pm \sqrt{15}$.So,$a \in [4 - \sqrt{15}, 4 + \sqrt{15}]$  $(i)$.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 $a = \frac{10 \pm \sqrt{100 - 16}}{2} = 5 \pm \sqrt{21}$.So,$a \in [5 - \sqrt{21}, 5 + \sqrt{21}]$  $(ii)$.Step $3$: Find the intersection of $(i)$ and $(ii)$.$a \in [4 - \sqrt{15}, 4 + \sqrt{15}] \cap [5 - \sqrt{21}, 5 + \sqrt{21}] = [5 - \sqrt{21}, 4 + \sqrt{15}]$.

Column-$I$	Column-$II$
$A$. Imaginary roots	$P$. $a \in (-1, 4)$
$B$. One root less than $3$ and other greater than $3$	$Q$. $a \in (-\infty, -1)$
$C$. One root less than $1$ and other greater than $3$	$R$. $a \in (-\infty, -4/3)$

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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