If f^{prime}(x)=k(cos x-sin x),f^{prime}(0)=3 | vedclass

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(A) Given $f^{\prime}(x)=k(\cos x-\sin x)$.Substitute $x=0$ into the derivative: $f^{\prime}(0)=k(\cos 0-\sin 0)=k(1-0)=k$.Since $f^{\prime}(0)=3$,we

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$3(\sin x+\cos x)+12$

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(A) Given $f^{\prime}(x)=k(\cos x-\sin x)$.Substitute $x=0$ into the derivative: $f^{\prime}(0)=k(\cos 0-\sin 0)=k(1-0)=k$.Since $f^{\prime}(0)=3$,we have $k=3$.Now,integrate $f^{\prime}(x)$ to find $f(x)$:$f(x)=\int 3(\cos x-\sin x) \, dx = 3(\sin x+\cos x)+C$.Use the condition $f\left(\frac{\pi}{2}\right)=15$ to find $C$:$f\left(\frac{\pi}{2}\right)=3(\sin \frac{\pi}{2}+\cos \frac{\pi}{2})+C = 3(1+0)+C = 3+C$.Setting $3+C=15$,we get $C=12$.Therefore,$f(x)=3(\sin x+\cos x)+12$.

If $f^{\prime}(x)=k(\cos x-\sin x)$,$f^{\prime}(0)=3$,and $f\left(\frac{\pi}{2}\right)=15$,then $f(x)=$

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