ExamPlay Light Logo
Пријавите се

JEE Advance - Mathematics (2004 - No. 11)

Prove that for $$x \in \left[ {0,{\pi \over 2}} \right],$$ $$\sin x + 2x \ge {{3x\left( {x + 1} \right)} \over \pi }$$. Explain
the identity if any used in the proof.
The inequality is incorrect and cannot be proven.
We can prove this by showing that the difference between the two sides is non-negative on the interval.
We can prove this using Jensen's inequality and the concavity of the sine function.
The identity \(\sin x + \sin y = 2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})\) is useful in this proof.
Let \(f(x) = \sin x + 2x - \frac{3x(x+1)}{\pi}\), showing \(f(0) \ge 0\) and \(f(\frac{\pi}{2}) \ge 0\) is sufficient.

Коментари (0)

Пријавите се за коментар
Реклама
BrainBehindX Inc Logo
©2026; Поверед Би BrainBehindX Inc