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

Given,

$f^{\text{'}}\left(x\right)=a\sin x+b\cos x ...\left(i\right)$

and $f^{\text{'}}\left(0\right)=4$

$\Rightarrow f^{\text{'}}\left(0\right)=a \sin 0+b\cos 0$

$\Rightarrow b=4$

Integrating both the sides of equation $\left(i\right)$ with respect to $x$, 

$\int f^{\text{'}}\left(x\right) dx=\int \left(a\sin x+b\cos x\right)dx$

$f\left(x\right)=-a\cos x+b\sin x+c$

$f\left(x\right)=-a\cos x+4\sin x+c$

$\because f\left(0\right)=3, f\left(\frac{\mathrm{\pi }}{2}\right)=5$

$\Rightarrow f\left(\frac{\mathrm{\pi }}{2}\right)=-a\cos \frac{\mathrm{\pi }}{2}+4\sin \frac{\mathrm{\pi }}{2}+c=4+c$

$\Rightarrow 5=4+c$

$\Rightarrow c=1$

$\Rightarrow f\left(0\right)=-a\cos 0+4\sin 0+1=-a+1$

$\Rightarrow a=-2$

$\therefore f\left(x\right)=2\cos x+4\sin x+1$.