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

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Aicharya · Accepted Answer

$f^{\prime}(x)=k(\cos x-\sin x)$ $f^{\prime}(0)=3 \quad f(\pi / 2)=15$ $k=3$ then $f(x)=8$ Integrate $f^{\prime}(x)$ $f(x)=k \sin x+k \cos x+c$ $f(x)=3 \sin x+3 \cos x+c$ $f(\pi / 2)=15$ $c+3=15$ $c=12$ $f(x)=3 \sin x+3 \cos x+12$

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

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