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Solution:

Given $f^{\prime}(x)=a \sin x+b \cos x$ and $f^{\prime}(0)=4$

On substituting $x=0$ in $f^{\prime}(x)$, we get

$f^{\prime}(0)=a \sin 0+b \cos 0$

$\Rightarrow 4=a \times 0+b \times 1$

$\Rightarrow 4=0+b$

$\therefore b=4$

Hence, $f^{\prime}(x)=a \sin x+4 \cos x$

On integrating this equation, we have

$\int \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}=\int(a \sin \mathrm{x}+4 \cos \mathrm{x}) \mathrm{dx}$

We know $\int \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}=\mathrm{f}(\mathrm{x})$

$\Rightarrow \mathrm{f}(\mathrm{x})=\int(a \sin \mathrm{x}+4 \cos \mathrm{x}) \mathrm{dx}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\int a \sin \mathrm{x} \mathrm{dx}+\int 4 \cos \mathrm{x} \mathrm{dx}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{a} \int \sin \mathrm{x} \mathrm{d} \mathrm{x}+4 \int \cos \mathrm{x} \mathrm{dx}$

Recall $\int \sin x d x=-\cos x+c$ and $\int \cos x d x=\sin x+c$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{a}(-\cos \mathrm{x})+4(\sin \mathrm{x})+\mathrm{c}$

$\Rightarrow \mathrm{f}(\mathrm{x})=-\mathrm{a} \cos \mathrm{x}+4 \sin \mathrm{x}+\mathrm{c}$

On substituting $x=0$ in $f(x)$, we get

$f(0)=-a \cos 0+4 \sin 0+c$

$\Rightarrow 3=-a \times 1+4 \times 0+c$

$\Rightarrow 3=-a+c$

$\Rightarrow c-a=3$           ......(1)

On substituting $x=\frac{\pi}{2}$ in $f(x)$, we get

$f(0)=-a \cos 0+4 \sin 0+c$

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

$\Rightarrow 5=-a \times 0+4 \times 1+c$

$\Rightarrow 5=0+4+c$

$\Rightarrow 5=4+c$

$\therefore c=1$

On substituting $c=1$ in equation (1), we get

$1-a=3$

$\Rightarrow a=1-3$

$\therefore a=-2$

On substituting the values of $c$ and $a$ in $f(x)$, we get

$f(x)=-(-2) \cos x+4 \sin x+1$

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

Thus, $f(x)=2 \cos x+4 \sin x+1$