If $f^{\prime}(x)=k(\cos x+\sin x)$ and $f(0)=9, f\left(\frac{\pi}{2}\right)=15$, then $f(x)=$

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Aicharya · Accepted Answer

$f^{\prime}(x)=k(\cos x+\sin x)$ On integrating both sides, we get $f(x)=k(\sin x-\cos x)+C$ $f(0)=k(0-1)+C$ $f(0)=-k+C \Rightarrow-k+C=9$ ...(1) Also $f\left(\frac{\pi}{2}ight)=k\left(\sin \frac{\pi}{2}-\cos \frac{\pi}{2}ight)+C$ $15=k+C$ ...(2) Adding (1) $\&(2)$ we get $2 C=24 \Rightarrow C=12 \Rightarrow k=3$ $	herefore f(x)=3(\sin x-\cos x)+12$

If $f^{\prime}(x)=k(\cos x+\sin x)$ and $f(0)=9, f\left(\frac{\pi}{2}\right)=15$, then $f(x)=$

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