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Question

Consider the function f(x)=x2+bx+c, where D=b24c>0. If b<0, c>0 then the number of points of non-differentiability of g(x)=|f(|x|)| is


Your Answer
A
1
Your Answer
B
2
Your Answer
C
3
Correct Answer
D
5

Solution

The correct option is D 5
f(x)=x2+bx+c
D>0 the function has real zeros.
b<0, c>0α+β=b>0 & αβ=c>0
both roots are positive
Graph of f(x):
Graph of f(|x|):
Graph of g(x)=|f(|x|)|:

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