If largest subset of $(0, p)$ , in which the function $f (x) = 3 {cos}^{4} x + 10 {cos}^{3} x + 6 {cos}^{2} x - 3$ is strictly decreasing, is $(0, \frac{π}{p}) \cup (\frac{2 π}{r}, π)$ , then find the value of $(p + r)$ .

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Solution

$f^{'} (x) = 12 {cos}^{3} x (- sin x) + 30 {cos}^{2} x (- sin x) + 12 cos x (= sin x)$
$= - 3 sin 2 x (2 c o s^{2} x + 5 cos x + 2)$
$= - 3 sin 2 x (2 cos x + 1) (cos x + 2)$
$f^{'} (x) < 0$ for strictly decreasing function.
$\Rightarrow (sin 2 x) (2 cos x + 1) > 0$
This will be possible in $(0, \frac{π}{2}) \cup (\frac{2 π}{3}, π)$
Hence, $p = 2$ and $r = 3$

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(0, π)

f (x) = 3 {cos}^{4} x + 10 {cos}^{3} x + 6 {cos}^{2} x - 3

(0, \frac{π}{p}) \cup (\frac{2 π}{r}, π)

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(0, π)

f (x) = 3 {cos}^{4} x + 10 {cos}^{3} x + 6 {cos}^{2} x - 3

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Prove that function f given by $f (x) = l o g (c o s x)$ is strictly decreasing on $(0, \frac{π}{2})$ and strictly increasing on $(\frac{π}{2} π)$ .

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1.It is periodic with period $2 π$ .

2.Domain of the function is R and the range is [-1, 1]

3.F(x) decreases strictly from 1 to -1 as x increases from 0 to $π$ . [For eg. If $x_{2} > x_{1}, F (x_{1}) > f (x_{2}), x ϵ [0, π]$

4.F(x) increases strictly from -1 to 1 as x increases from $π t o 2 π$ . (foreg. If $x_{2} > x_{1}, f (x_{2}) > f (x_{1}), x ϵ [π, 2 π]$

Show that the function given by $f (x) = s i n x$ is
strictly increasing in $(0, \frac{π}{2})$

strictly decreasing in $(\frac{π}{2}, π)$

neither increasing nor decreasing in $(0, π)$

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