Question

Let f(x) = 4x + 8 cosx - ln(cosx(1 + sinx)) + tanx − 2secx − 6.

If f(x) > 0 ∀ x ∈ (0, a) then

• A. a = $\frac { \pi } { 6 }$
• B. a = $\frac { \pi } { 3 }$
• C. a = $\frac { \pi } { 2 }$
• D. ) None of these

Answer 1

Answer: (A)

a = $\frac { \pi } { 6 }$

Answer 2

f'(x) = 4 − 8 sinx −$\frac { 4 \left( - \sin x + \cos ^ { 2 } x - \sin ^ { 2 } x \right) } { \cos x ( 1 + \sin x ) }$ +

sec²x - secx. tanx

= 4 (1 − 2 sinx) + sec²x (1 – 2 sinx) − 4 secx(1 - 2 sinx)

⇒ (secx − 2)² (1 − 2 sinx)

f(0 + 0) = 0. If f(x) > 0 ∀ x ∈ (0, a) then f(x) must be increasing in (0, a)

⇒ f '(x) > 0 ∀ x ∈ (0, a)

⇒