If $[\sqrt{n^2 +λ}=[\sqrt{n^2 +1}]+2$, where $λ,\, n∈N$ then λ can assume

$(2n + 4)$ different values $(2n + 3)$ different values $(2n+5)$ different values $(2n + 6)$ different values

$(2n+5)$ different values

We have, $[\sqrt{n^2 +1}]=n$ $∴[\sqrt{n^2 +λ}=[\sqrt{n^2 +1}]+2$ $⇒[\sqrt{n^2 +λ}]=n+2$ $⇒n+2≤\sqrt{n^2 +λ} <n+3$ $⇒(n+ 2)^2 ≤n^2 +λ<(n+3)^2$ $⇒4n+4≤λ<6n+9$ $⇒λ=4n+4, 4n+5,...., 6n+ 8$. Hence, λ can assume $(2n+5)$ distinct values.