Let f: R arrow((-π/2), 0] be a function defined by f ( x )= tan -1(2 x - x 2+λ). If f is onto, then λ lies in the interval

Question

Let f: R arrow((-π/2), 0] be a function defined by f ( x )= tan -1(2 x - x 2+λ). If f is onto, then λ lies in the interval (A) (-2,0) (B) (0,2) (C) (-1,1) (D) none

Tardigrade Admin · Accepted Answer

∵ f is onto ∴ Range of tan -1(2 x - x 2+λ) should be ((-π/2), 0] ⇒ Range of 2 x-x2+λ should be (-∞, 0]; hence D=0 ⇒ 4+4 λ=0, hence λ=-1

Q. Let $f : R \to (\frac{- π}{2}, 0]$ be a function defined by $f (x) = tan^{- 1} (2 x - x^{2} + λ)$ . If $f$ is onto, then $λ$ lies in the interval

$(- 2, 0)$

$(0, 2)$

$(- 1, 1)$

none

$∵ f$ is onto
$∴$ Range of $tan^{- 1} (2 x - x^{2} + λ)$ should be $(\frac{- π}{2}, 0]$
$\Rightarrow$ Range of $2 x - x^{2} + λ$ should be $(- \infty, 0]$ ; hence $D = 0 \Rightarrow 4 + 4 λ = 0$ , hence $λ = - 1$

Q. Let f:R→(2−π​,0] be a function defined by f(x)=tan−1(2x−x2+λ). If f is onto, then λ lies in the interval

(−2,0)

(0,2)

(−1,1)