SATHEE: Application of Derivatives 2 Question 26

Application of Derivatives 2 Question 26

####26. Prove that $\sin x + 2 x \geq \frac{3 x (x + 1)}{π}, \forall x \in 0, \frac{π}{2}$ (Justify the inequality, if any used).

(2004, 4M)

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Answer:

Correct Answer: 26. (b)

Solution:

Let $f (x) = \sin x + 2 x - \frac{3 x (x + 1)}{π}$

On differentiating w.r.t. $x$ , we get

$\Rightarrow f^{'} (x) = \cos x + 2 - \frac{(6 x + 3)}{π}$

$\Rightarrow f^{''} (x) = - \sin x - \frac{6}{π} < 0, \forall x \in 0, \frac{π}{2}$

$∴ f^{'} (x)$ is decreasing for all $x \in 0, \frac{π}{2}$ .

$\begin{array}{lll} \Rightarrow & f^{'} (x) > 0 & [∵ x < π / 2] \\ \Rightarrow & f^{'} (x) > f^{'} (π / 2) \end{array}$

$∴ f (x)$ is increasing.

Thus, when $x \geq 0, f (x) \geq f (0)$

$\begin{aligned} \Rightarrow \sin x + 2 x - \frac{3 x (x + 1)}{π} \geq 0 \\ \Rightarrow \sin x + 2 x \geq \frac{3 x (x + 1)}{π} \end{aligned}$

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Application of Derivatives 2 Question 26

Answer:

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