Question

Let f '(sinx) < 0 and f'(sinx) > 0 ∀ x ∈ $\left( 0 , \frac { \pi } { 2 } \right)$ and g(x) =

f(sinx) + f(cosx), then g(x) is decreasing in

• A. $\left( \frac { \pi } { 4 } , \frac { \pi } { 2 } \right)$
• B. $\left( 0 , \frac { \pi } { 4 } \right)$
• C. $\left( 0 , \frac { \pi } { 2 } \right)$
• D. $\left( \frac { \pi } { 6 } , \frac { \pi } { 2 } \right)$

Answer 1

Answer: (B)

$\left( 0 , \frac { \pi } { 4 } \right)$

Answer 2

g'(x) = f(sinx).cosx − f '(cosx).sinx

⇒ g"(x) = −f '(cosx).sinx + cos²x. f ''(sinx) +

f "(cosx).sin²x − f '(cosx).cosx > 0 ∀ x ∈ (0, π/2)

⇒ g'(x) is increasing in (0, π/2). Also g'(π/4) = 0

⇒ g'(x) > 0 and g'(x) < 0

∀ x ∈ (0, π/4)