If f ( x )=3 tan x +(3 a +1) ln | sec x |+( a -3) x is increasing in (0, (π/2)) then minimum integral value of

Question

If f ( x )=3 tan x +(3 a +1) ln | sec x |+( a -3) x is increasing in (0, (π/2)) then minimum integral value of (A) 0 (B) 1 (C) 2 (D) -1

Tardigrade Admin · Accepted Answer

f prime(x)=3 sec 2 x+(3 a+1) tan x+(a-3) =3 tan 2 x+3 a tan x+ tan x+a =(3 tan x+1)( tan x+a) f prime( x )>0 ∀ x ∈(0, (π/2)) ⇒ tan x+a>0 ∀ x ∈(0, (π/2)) ∴ a ≥ 0 ∴ text minimum integral value of a=0

Q. If $f (x) = 3 tan x + (3 a + 1) ln ∣ sec x ∣ + (a - 3) x$ is increasing in $(0, \frac{π}{2})$ then minimum integral value of

0

1

2

-1

$f^{'} (x) = 3 sec^{2} x + (3 a + 1) tan x + (a - 3)$
$= 3 tan^{2} x + 3 a tan x + tan x + a$
$= (3 tan x + 1) (tan x + a)$
$f^{'} (x) > 0\forall x \in (0, \frac{π}{2})$
$\Rightarrow tan x + a > 0\forall x \in (0, \frac{π}{2})$
$∴ a \geq 0$
$∴ minimum integral value of a = 0$

Q. If f(x)=3tanx+(3a+1)ln∣secx∣+(a−3)x is increasing in (0,2π​) then minimum integral value of

0

1

2

-1

f′(x)=3sec2x+(3a+1)tanx+(a−3) =3tan2x+3atanx+tanx+a =(3tanx+1)(tanx+a) f′(x)>0∀x∈(0,2π​) ⇒tanx+a>0∀x∈(0,2π​) ∴a≥0 ∴ minimum integral value of a=0

Q. If $f (x) = 3 tan x + (3 a + 1) ln ∣ sec x ∣ + (a - 3) x$ is increasing in $(0, \frac{π}{2})$ then minimum integral value of

$f^{'} (x) = 3 sec^{2} x + (3 a + 1) tan x + (a - 3)$
$= 3 tan^{2} x + 3 a tan x + tan x + a$
$= (3 tan x + 1) (tan x + a)$
$f^{'} (x) > 0\forall x \in (0, \frac{π}{2})$
$\Rightarrow tan x + a > 0\forall x \in (0, \frac{π}{2})$
$∴ a \geq 0$
$∴ minimum integral value of a = 0$