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Question: If f<sub>1</sub>(x) = 2x, f<sub>2</sub>(x) = 3 sinx − x cosx then for x ∈\(\left( 0 , \frac { \pi } ...

If f1(x) = 2x, f2(x) = 3 sinx − x cosx then for x ∈(0,π2)\left( 0 , \frac { \pi } { 2 } \right)

A

f1(x) < f2(x)

B

f1(x) > f2(x)

C

f1(|x|) <f2(|x|)

D

None of these

Answer

f1(x) > f2(x)

Explanation

Solution

Let f(x) = f1(x) − f2(x) = 2x − 3sinx + x cosx

⇒ f '(x) = 2 − 2cosx − x sinx

⇒ f"(x) = sinx − x cosx = cosx (tanx - x)

⇒ f"(x)>0 ∀ x ∈ (0, π/2).

Thus f;(x) is increasing in (0, π/2). f'(0) = 0

⇒ f(x) > 0 ∀ x ∈ (0, π/2).

Thus f(x) is increasing in (0, π/2). f(0) = 0

⇒ f(x) > 0 ∀ x ∈ (0, π/2)

⇒ f1(x) > f2(x).