FunctionHard
Question
If q2 - 4 p r = 0, p > 0, then the domain of the function f(x) = log (p x3 + (p + q) x2 + (q + r) x + r) is:
Options
A.R - 
B.R - 
C.R - 
D.none of these
Solution
q2 - 4 p r = 0, p > 0
f(x) = log (px3 + (p + q) x2 + (q + r) x + r)
Let g(x) = px3 + (p + q) x2 + (q + r) x + r
g(x) = (x + 1) (px2 + qx + r)
Discriminant of px2 + qx + r is q2 - 4pr = 0
Domain (x + 1) (px2 + qx + r) > 0
⇒ p(x + 1)
> 2
⇒ x ≠ -
and x > - 1
∴ x ∈ R - [(-∞)] ∪
f(x) = log (px3 + (p + q) x2 + (q + r) x + r)
Let g(x) = px3 + (p + q) x2 + (q + r) x + r
g(x) = (x + 1) (px2 + qx + r)
Discriminant of px2 + qx + r is q2 - 4pr = 0
Domain (x + 1) (px2 + qx + r) > 0
⇒ p(x + 1)
⇒ x ≠ -
∴ x ∈ R - [(-∞)] ∪
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