Question

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f$\left(\frac{\mathrm{\pi }}{2}\right)$ = 5, find f(x)

Answer 1

$$\begin{gathered} f\text{'}\left(x\right)=a\sin x+b\cos x \\ f\text{'}\left(0\right)=4,f\left(0\right)=3 \\ f\left(\frac{\pi }{2}\right)=5 \\ f\text{'}\left(x\right)=a\sin x+b\cos x \\ \int f\text{'}\left(x\right)dx=\int \left(a\sin x+b\cos x\right)dx \\ f\left(x\right)=-a\cos x+b\sin x+\mathrm{C}...\left(i\right) \\ \mathrm{Now}\mathrm{puting}x=0\mathrm{in}\mathrm{equation}\left(i\right) \\ f\left(0\right)=-a\cos 0+b\sin 0+\mathrm{C} \\ 3=-a\times 1+b\times 0+\mathrm{C} \\ 3=-a+\mathrm{C}...\left(ii\right) \\ \mathrm{Now}\mathrm{puting}x=\frac{\pi }{2}\mathrm{in}\mathrm{equation}\left(i\right) \\ f\left(\frac{\pi }{2}\right)=-a\cos \frac{\pi }{2}+b\sin \frac{\pi }{2}+\mathrm{C} \\ 5=-a\cos \frac{\pi }{2}+b\sin \left(\frac{\pi }{2}\right)+\mathrm{C} \\ 5=-a\times 0+b\times 1+\mathrm{C} \\ 5=b+\mathrm{C}...\left(iii\right) \\ \mathrm{We}\mathrm{also}\mathrm{have}f\text{'}\left(0\right)=4 \\ f\text{'}\left(x\right)=a\sin x+b\cos x \\ f\text{'}\left(0\right)=a\sin 0+b\cos 0 \\ 4=a\times 0+b\times 1 \\ 4=b...\left(iv\right) \\ \mathrm{solving}\left(ii\right),\left(iii\right)\mathrm{and}\left(iv\right)\mathrm{we}\mathrm{get}, \\ b=4 \\ C=1 \\ a=-2 \\ \therefore f\left(x\right)=2\cos x+4\sin x+1 \end{gathered}$$