Let f(x)=5 x tan x+8 sin ( tan x)+5 ln ( cos x), x ∈((-π/4), 0) then

Question

Let f(x)=5 x tan x+8 sin ( tan x)+5 ln ( cos x), x ∈((-π/4), 0) then (A) f(x) is strictly increasing in ((-π/4), 0). (B) f ( x ) has a point of local maximum. (C) the equation f ( x )=0 has a root in ((-π/4), 0) (D)  f(x)<0 for all x ∈((-π/4), 0).

Tardigrade Admin · Accepted Answer

: f(x)=5 x tan x+8 sin ( tan x)+5 ln ( cos x), x ∈((-π/4), 0) ⇒ f prime(x)= sec 2 x(5 x+8 cos ( tan x)) For x ∈((-π/4), 0), 5 x ∈(-3.9,0) and 8 cos ( tan x)>4 Hence, f prime(x)>0 for .x ∈((-π/4), 0) ]

Q. Let $f (x) = 5 x tan x + 8 sin (tan x) + 5 ln (cos x), x \in (\frac{- π}{4}, 0)$ then

$f (x)$ is strictly increasing in $(\frac{- π}{4}, 0)$ .

$f (x)$ has a point of local maximum.

the equation $f (x) = 0$ has a root in $(\frac{- π}{4}, 0)$

$f (x) < 0$ for all $x \in (\frac{- π}{4}, 0)$ .

$: f (x) = 5 x tan x + 8 sin (tan x) + 5 ln (cos x), x \in (\frac{- π}{4}, 0)$
$\Rightarrow f^{'} (x) = sec^{2} x (5 x + 8 cos (tan x))$
For $x \in (\frac{- π}{4}, 0), 5 x \in (- 3.9, 0)$ and $8 cos (tan x) > 4$
Hence, $f^{'} (x) > 0$ for $x \in (\frac{- π}{4}, 0)]$

Q. Let f(x)=5xtanx+8sin(tanx)+5ln(cosx),x∈(4−π​,0) then

f(x) is strictly increasing in (4−π​,0).

f(x) has a point of local maximum.

the equation f(x)=0 has a root in (4−π​,0)

f(x)<0 for all x∈(4−π​,0).