If the function f(x)=2 cot x+(2 a+1) ln | operatornamecosec x|+(2-a) x is strictly decreasing in (0, (π/2)) then range of ' a ' is [ mathrmm, ∞), find the value of mathrmm.

Question

If the function f(x)=2 cot x+(2 a+1) ln | operatornamecosec x|+(2-a) x is strictly decreasing in (0, (π/2)) then range of ' a ' is [ mathrmm, ∞), find the value of mathrmm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

⇒ mathrmf prime( mathrmx)=-2 operatornamecosec2 mathrmx-(2 mathrma+1) cot mathrmx+(2- mathrma)=-2 cot 2 mathrmx-(2 mathrma+1) cot mathrmx- mathrma ⇒ mathrmf prime( mathrmx)=( cot mathrmx+ mathrma)(-2 cot mathrmx-1) ≤ 0 in (0, (π/2)) ∴ cot mathrmx+ mathrma ≥ 0 in (0, (π/2)) Hence a ≥ 0

Q. If the function $f (x) = 2 cot x + (2 a + 1) ln ∣ cosec x ∣ + (2 - a) x$ is strictly decreasing in $(0, \frac{π}{2})$ then range of ' $a$ ' is $[m, \infty)$ , find the value of $m$ .

$\Rightarrow f^{'} (x) = - 2 cosec^{2} x - (2 a + 1) cot x + (2 - a) = - 2 cot^{2} x - (2 a + 1) cot x - a$
$\Rightarrow f^{'} (x) = (cot x + a) (- 2 cot x - 1) \leq 0$ in $(0, \frac{π}{2})$
$∴ cot x + a \geq 0$ in $(0, \frac{π}{2})$
Hence $a \geq 0$

Q. If the function f(x)=2cotx+(2a+1)ln∣cosecx∣+(2−a)x is strictly decreasing in (0,2π​) then range of ' a ' is [m,∞), find the value of m.

⇒f′(x)=−2cosec2x−(2a+1)cotx+(2−a)=−2cot2x−(2a+1)cotx−a ⇒f′(x)=(cotx+a)(−2cotx−1)≤0 in (0,2π​) ∴cotx+a≥0 in (0,2π​) Hence a≥0

Q. If the function $f (x) = 2 cot x + (2 a + 1) ln ∣ cosec x ∣ + (2 - a) x$ is strictly decreasing in $(0, \frac{π}{2})$ then range of ' $a$ ' is $[m, \infty)$ , find the value of $m$ .

$\Rightarrow f^{'} (x) = - 2 cosec^{2} x - (2 a + 1) cot x + (2 - a) = - 2 cot^{2} x - (2 a + 1) cot x - a$
$\Rightarrow f^{'} (x) = (cot x + a) (- 2 cot x - 1) \leq 0$ in $(0, \frac{π}{2})$
$∴ cot x + a \geq 0$ in $(0, \frac{π}{2})$
Hence $a \geq 0$