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# Show that the function given by $f (x) = \sin x$ is strictly decreasing in $(\large {\frac{\pi}{2}},$$\pi) This is (b) part of the multi-part question q3 Can you answer this question? ## 1 Answer 0 votes Toolbox: • A function f(x) is said to be a strictly increasing function on (a,b) if x_1 < x_2\Rightarrow f(x_1) < f(x_2) for all x_1,x_2\in (a,b) • If x_1 < x_2\Rightarrow f(x_1) > f(x_2) for all x_1,x_2\in (a,b) then f(x) is said to be strictly decreasing on (a,b) • A function f(x) is said to be increasing on [a,b] if it is increasing (decreasing) on (a,b) and it is increasing (decreasing) at x=a and x=b. • The necessary sufficient condition for a differentiable function defined on (a,b) to be strictly increasing on (a,b) is that f'(x) > 0 for all x\in (a,b) • The necessary sufficient condition for a differentiable function defined on (a,b) to be strictly decreasing on (a,b) is that f'(x) < 0 for all x\in (a,b) Step 1: Given : f(x)=\sin x Differentiating w.r.t x we get, f'(x)=\cos x Step 2: Consider the interval (\large\frac{\pi}{2},$$\pi)$
The interval belongs to the second quadrant and $\cos x$ is negative in second quadrant.
$\Rightarrow f(x)$ is strictly decreasing in this interval.