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Question

The number of integral values of a so that f(x)=|||x|2|+a| has exactly 7 point of non differentiablility xϵR is

A
0
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B
1
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C
3
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D
infinite
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Solution

The correct option is B 1
Clearly from the graph,
to have 7 points of non -differentiability
2+a>0a>2
and a<0
Thus only possible solution is a=1
354994_130714_ans_a97437534adf42f0b0f23b54ce09c9b0.png

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