Let the functions f and g be f: [0, frac{pi}{ | vedclass

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(A) For $f(x) = \sin x$ on $[0, \frac{\pi}{2}]$,the function is strictly increasing,so it is one-one.For $g(x) = \cos x$ on $[0, \frac{\pi}{2}]$,the f

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Statement $(I)$ is true,statement $(II)$ is false

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(A) For $f(x) = \sin x$ on $[0, \frac{\pi}{2}]$,the function is strictly increasing,so it is one-one.For $g(x) = \cos x$ on $[0, \frac{\pi}{2}]$,the function is strictly decreasing,so it is one-one.Thus,Statement $(I)$ is true.Now,consider $(f+g)(x) = \sin x + \cos x$.We evaluate the function at the endpoints of the interval:$(f+g)(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.$(f+g)(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1$.Since $(f+g)(0) = (f+g)(\frac{\pi}{2})$ but $0 
eq \frac{\pi}{2}$,the function $f+g$ is not one-one.Thus,Statement $(II)$ is false.

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