**Correct option (a) (–3, –1) ∪ (3, 5)**

**Explanation**

D > 0

ƒ(–2)ƒ(4) < 0

(–2m)^{2 } –4.1.(m^{2}–1) > 0

(4 + 4m + m^{2 }– 1) (16 – 8m + m^{2 }– 1) < 0

4 > 0 (m^{2} +4m + 3) (m^{2 }– 8m + 3)

*⇒ *m ∈ R ......(1)

(m + 1) (m + 3) (m – 3) (m – 5) < 0

*⇒ *m ∈(–3, –1) ∪ (3, 5) ......(2)

From (1) and (2) , m ∈ (–3, –1) ∪ (3, 5)