Consider a function $f (x) = 1 + 12 | x | - 3 x^{2}$ defined on $[- 2, 5] .$ The absolute difference between the global maximum and global minimum values of $f (x)$ is

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Solution

$f (x) = 1 + 12 | x | - 3 x^{2}$
$f (x)$ is an even function.
$∴$ We take only $x \in [0, 5]$
For $x \in [0, 5],$
$f (x) = 1 + 12 x - 3 x^{2}$
Since the coefficient of $x^{2}$ is negative, we get downward parabola whose maximum value is $- \frac{D}{4 a} = \frac{156}{4 \times 3} = 13$
As the parabola is downward, we get its minimum value in $[- 2, 5]$ at $x = 5$
$f (5) = - 14$
Required difference is $| 13 - (- 14) | = 27$

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