If: R → R given by $f(x) = x^3 + (a + 2) x^2 + 3ax +5$ is one-one, then a belongs to the interval
Answer & explanation
Correct answer: option 3
Since f: R → R is one-one. Therefore, f (x) is either strictly increasing or strictly decreasing.
$⇒f'(x)>0$ or $f'(x) <0$ for all x
$⇒ 3x^2 + 2x (a + 2) + 3a > 0$ for all $x ∈ R$
or, $3x^2 + 2x (a + 2) + 3a <0$ for all $x ∈ R$
$⇒3x^2 + 2x (a + 2) + 3a > 0$ for all x
$⇒4 (a + 2)^2 - 36a <0$ [$∵ ax^2+ bx + c>0$ for all $x ⇒ Disc <0$]
$⇒4 (a^2+4a+4-9a) <0$
$⇒(a^2-5a+4) <0⇒ (a-1) (a-4) <0⇒ 1<a<4$
Hence, f(x) is one-one if $a ∈ (1, 4)$.