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The graph of $f (x) = | | x - 2 | - a | - 3$ has exactly 3-intercepts on x-axis then sum of all values of $a \in [- 10, 10]$ and a is an integer.

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answer is D.

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Detailed Solution

$a = | x - 2 | - 3 o r a = | x - 2 | + 3$

$a > 3 \Rightarrow a = 4, 5, 6, 7, 8, 9, 10$

$\sum a = 4 + 5 + 6 + 7 + 8 + 9 + 10$

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