Let S be set of domain of the function f ( x )=√(π/2)- tan -1 √- x 2+5 x -6. If λ=α+(1/α) where α ∈ S and λ is an integer then find the value of (λ2).

Question

Let S be set of domain of the function f ( x )=√(π/2)- tan -1 √- x 2+5 x -6. If λ=α+(1/α) where α ∈ S and λ is an integer then find the value of (λ2). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For domain of function, -x2+5 x+6 ≥ 0 ⇒ x2-5 x+6 ≤ 0 ⇒ (x-2)(x-3) ≤ 0 ⇒ x ∈[2,3] text Now, λ=α+(1/α) ⇒ λ ∈[(5/2), (10/3)] So, integral value of λ is 3 . Hence, λ2=9.

Q. Let $S$ be set of domain of the function $f (x) = \frac{π}{2} - tan^{- 1} - x^{2} + 5 x - 6$ . If $λ = α + \frac{1}{α}$ where $α \in S$ and $λ$ is an integer then find the value of $(λ^{2})$ .

For domain of function,
$- x^{2} + 5 x + 6 \geq 0 \Rightarrow x^{2} - 5 x + 6 \leq 0$
$\Rightarrow (x - 2) (x - 3) \leq 0$
$\Rightarrow x \in [2, 3]$
$Now, λ = α + \frac{1}{α} \Rightarrow λ \in [\frac{5}{2}, \frac{10}{3}]$
So, integral value of $λ$ is 3 .
Hence, $λ^{2} = 9$ .

Q. Let S be set of domain of the function f(x)=2π​−tan−1−x2+5x−6​​. If λ=α+α1​ where α∈S and λ is an integer then find the value of (λ2).

For domain of function, −x2+5x+6≥0⇒x2−5x+6≤0 ⇒(x−2)(x−3)≤0 ⇒x∈[2,3] Now, λ=α+α1​⇒λ∈[25​,310​] So, integral value of λ is 3 . Hence, λ2=9.

Q. Let $S$ be set of domain of the function $f (x) = \frac{π}{2} - tan^{- 1} - x^{2} + 5 x - 6$ . If $λ = α + \frac{1}{α}$ where $α \in S$ and $λ$ is an integer then find the value of $(λ^{2})$ .

For domain of function,
$- x^{2} + 5 x + 6 \geq 0 \Rightarrow x^{2} - 5 x + 6 \leq 0$
$\Rightarrow (x - 2) (x - 3) \leq 0$
$\Rightarrow x \in [2, 3]$
$Now, λ = α + \frac{1}{α} \Rightarrow λ \in [\frac{5}{2}, \frac{10}{3}]$
So, integral value of $λ$ is 3 .
Hence, $λ^{2} = 9$ .