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Q.
The function $f\left(x\right)=cosx-2px$ is monotonically decreasing for
NTA AbhyasNTA Abhyas 2022
Solution:
Given $f\left(x\right)=cosx-2px$ .
Differentiating $f\left(x\right)$ with respect to $x$ , we get
$f^{'}\left(x\right)=-sinx-2p$
For $f\left(x\right)$ to be monotonically decreasing, $f^{'}\left(x\right) < 0$ .
$\Rightarrow -sinx-2p < 0$
$\Rightarrow p>-\frac{1}{2}sinx..........\left(i\right)$
Since, $-1\leq sinx\leq 1$
$\Rightarrow -\frac{1}{2}\leq -\frac{1}{2}sinx\leq \frac{1}{2}..........\left(i i\right)$
From equations $\left(i\right)\&\left(i i\right)$ , we get
$p>\frac{1}{2}$