Application of Derivatives 2 Question 26
26. Prove that $\sin x+2 x \geq \frac{3 x(x+1)}{\pi}, \forall x \in 0, \frac{\pi}{2}$ (Justify the inequality, if any used).
(2004, 4M)
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Answer:
Correct Answer: 26. (b)
Solution:
- Let $f(x)=\sin x+2 x-\frac{3 x(x+1)}{\pi}$
On differentiating w.r.t. $x$, we get
$\Rightarrow \quad f^{\prime}(x)=\cos x+2-\frac{(6 x+3)}{\pi}$
$\Rightarrow \quad f^{\prime \prime}(x)=-\sin x-\frac{6}{\pi}<0, \forall x \in 0, \frac{\pi}{2}$
$\therefore \quad f^{\prime}(x)$ is decreasing for all $x \in 0, \frac{\pi}{2}$.
$$ \begin{array}{lll} \Rightarrow & f^{\prime}(x)>0 & {[\because x<\pi / 2]} \\ \Rightarrow & f^{\prime}(x)>f^{\prime}(\pi / 2) & \end{array} $$
$\therefore f(x)$ is increasing.
Thus, when $x \geq 0, f(x) \geq f(0)$
$$ \begin{aligned} & \Rightarrow \quad \sin x+2 x-\frac{3 x(x+1)}{\pi} \geq 0 \\ & \Rightarrow \quad \sin x+2 x \geq \frac{3 x(x+1)}{\pi} \end{aligned} $$
