If f^{prime}(x)=k(cos x+sin x) and f(0)=9, fl | vedclass

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(A) Given $f^{\prime}(x)=k(\cos x+\sin x)$.Integrating both sides with respect to $x$,we get:$f(x) = \int k(\cos x + \sin x) dx = k(\sin x - \cos x) +

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

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(A) Given $f^{\prime}(x)=k(\cos x+\sin x)$.Integrating both sides with respect to $x$,we get:$f(x) = \int k(\cos x + \sin x) dx = k(\sin x - \cos x) + C$.Using the condition $f(0) = 9$:$f(0) = k(\sin 0 - \cos 0) + C = k(0 - 1) + C = -k + C = 9$ ...$(1)$.Using the condition $f\left(\frac{\pi}{2}\right) = 15$:$f\left(\frac{\pi}{2}\right) = k\left(\sin \frac{\pi}{2} - \cos \frac{\pi}{2}\right) + C = k(1 - 0) + C = k + C = 15$ ...$(2)$.Adding equations $(1)$ and $(2)$:$(-k + C) + (k + C) = 9 + 15 \Rightarrow 2C = 24 \Rightarrow C = 12$.Substituting $C = 12$ into equation $(2)$:$k + 12 = 15 \Rightarrow k = 3$.Thus,$f(x) = 3(\sin x - \cos x) + 12$.

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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