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Question: Domain of f(x) = ln (ax<sup>3</sup> + (a + b)x<sup>2</sup> + (b + c)x + c), where a \> 0, b<sup>2</s...

Domain of f(x) = ln (ax3 + (a + b)x2 + (b + c)x + c), where a > 0, b2 - 4ac = 0, is

A

(-1, ∞) ~ {−b2a}\left\{ - \frac{b}{2a} \right\}

B

(1, ∞) ~ {−b2a}\left\{ - \frac{b}{2a} \right\}

C

(−1, 1) ~ {−b2a}\left\{ - \frac{b}{2a} \right\}

D

None of these

Answer

(-1, ∞) ~ {−b2a}\left\{ - \frac{b}{2a} \right\}

Explanation

Solution

We must have

ax3 + (a + b)x2 + (b + c)x + c > 0

⇒ ax2(x + 1) + bx(x + 1) + c(x + 1) > 0

⇒ (x + 1) (ax2 + bx + c) > 0

⇒ a(x + 1) (ax2 + bx + c) > 0

⇒ a(x+ 1) (x+b2a)2\left( x + \frac { b } { 2 a } \right) ^ { 2 } > 0 as b2 = 4ac

⇒ x > −1 and ≠ −b2a- \frac { b } { 2 a }