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`        given that a,b,c are real numbers such that b^2=4ac and a>0 the maximam possible set d on which the function is defined isf(x)=log{ax^3+(a+b)x^2+(b+c)x+c`
10 months ago

```							obviously, the expression inside the log should be greater than zero.or ax^3+(a+b)x^2+(b+c)x+c should be greater than zero.note that ax^3+(a+b)x^2+(b+c)x+c is factorisable as (x+1)(ax^2+bx+c)so (x+1)(ax^2+bx+c) should be greater than zero.(x+1)(ax^2+bx+b^2/4a) should be greater than zero.a(x+1)(x+b/2a)^2 should be greater than zero.(x+1)(x+b/2a)^2 should be greater than zero (as a is greater than zero)hence, we conclude that x belongs to the set where x is greater than – 1 and does not equal – b/2a (in case – b/2a is greater than – 1)
```
10 months ago
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