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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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Verified Answer
The correct answer is:
$3(\sin x-\cos x)+12$
$f^{\prime}(x)=k(\cos x+\sin x)$
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 $f\left(\frac{\pi}{2}\right)=k\left(\sin \frac{\pi}{2}-\cos \frac{\pi}{2}\right)+C$
$15=k+C$ ...(2)
Adding (1) $\&(2)$ we get
$2 C=24 \Rightarrow C=12 \Rightarrow k=3$
$\therefore f(x)=3(\sin x-\cos x)+12$
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 $f\left(\frac{\pi}{2}\right)=k\left(\sin \frac{\pi}{2}-\cos \frac{\pi}{2}\right)+C$
$15=k+C$ ...(2)
Adding (1) $\&(2)$ we get
$2 C=24 \Rightarrow C=12 \Rightarrow k=3$
$\therefore f(x)=3(\sin x-\cos x)+12$
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