For the function f ( x ) = sin x + 3 x - 2 ( x 2 + x ) , where, consider the following two statements :(I) f is increasing in. (II) f ' is decreasing in. Between the above two statements, [2024]

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Solution

Q.
For the function $f (x) = \sin x + 3 x - \frac{2}{π} (x^{2} + x), where x \in [0, \frac{π}{2}],$ consider the following two statements :

(I) $f$ is increasing in $(0, \frac{π}{2}) .$

(II) $f'$ is decreasing in $(0, \frac{π}{2})$ .

Between the above two statements, [2024]

1 neither (I) nor (II) is true.

2 only (II) is true.

3 both (I) and (II) are true.

4 only (I) is true.

Ans.
(3)

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

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

$f' (x) > 0 for x \in [0, \frac{π}{4}]$

For $x \in [π / 4, π / 2],$ we have $π / 4 \leq x < π / 2$

$\Rightarrow \frac{π}{2} + 1 \leq 2 x + 1 < π + 1$

$\Rightarrow 1 + \frac{2}{π} \leq \frac{2}{π} (2 x + 1) < 2 + \frac{2}{π}$

Since, $\frac{2}{π} (2 x + 1) < 3$

So, $f' (x) > 0, x \in (0, π / 2)$

So $f (x)$ is increasing in $(0, π / 2)$

Now, $f'' (x) = - \sin x - \frac{4}{π} < 0, x \in (0, π / 2)$

$\Rightarrow f' (x) is decreasing in (0, π / 2) .$

Hence, both statements are true.

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