The minimum value of the function f(x)=(tan x/3 + 2 tan â¡ x), ∀ x∈ [0 , (π /2)) is

Question

The minimum value of the function f(x)=(tan x/3 + 2 tan â¡ x), ∀ x∈ [0 , (π /2)) is (A) 0 (B) (1/2) (C) (1/3) (D) (1/6)

Tardigrade Admin · Accepted Answer

If x ∈[0, (π/2)), tan x ∈[0, ∞) Let, tan x=t beginarrayl text i.e. y=(t/3+2 t), t ∈[0, ∞) ⇒ 3 y+2 t y=t ⇒ t=(3 y/1-2 y) ≥ 0 ⇒ y ∈[0, (1/2)) endarray Hence, the minimum value of the function =0

Q. The minimum value of the function $f (x) = \frac{t an x}{3 + 2 t an x}, \forall x \in [0, \frac{π}{2})$ is

$0$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{6}$

If $x \in [0, \frac{π}{2}), tan x \in [0, \infty)$
Let, $tan x = t$
$i.e. y = \frac{t}{3 + 2 t}, t \in [0, \infty) \Rightarrow 3 y + 2 t y = t \Rightarrow t = \frac{3 y}{1 - 2 y} \geq 0 \Rightarrow y \in [0, \frac{1}{2})$
Hence, the minimum value of the function $= 0$

Q. The minimum value of the function f(x)=3+2tan⁡xtanx​,∀x∈[0,2π​) is

0

21​

31​

61​

If x∈[0,2π​),tanx∈[0,∞) Let, tanx=t i.e. y=3+2tt​,t∈[0,∞)⇒3y+2ty=t⇒t=1−2y3y​≥0⇒y∈[0,21​)​ Hence, the minimum value of the function =0

Q. The minimum value of the function $f (x) = \frac{t an x}{3 + 2 t an x}, \forall x \in [0, \frac{π}{2})$ is

$0$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{6}$

If $x \in [0, \frac{π}{2}), tan x \in [0, \infty)$
Let, $tan x = t$
$i.e. y = \frac{t}{3 + 2 t}, t \in [0, \infty) \Rightarrow 3 y + 2 t y = t \Rightarrow t = \frac{3 y}{1 - 2 y} \geq 0 \Rightarrow y \in [0, \frac{1}{2})$
Hence, the minimum value of the function $= 0$