If f: R → R given by f(x) = x^3+(a + 2) x^2 + 3ax +5 is one-one, then…

The correct answer is Option (3) → (1, 4)

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 $a ∈ (1, 4)$.