If f^{prime}(x)=a sin x+b cos x,f^{prime}(0)= | vedclass

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(B) Given,$f^{\prime}(x)=a \sin x+b \cos x$ and $f^{\prime}(0)=4$.
Substituting $x=0$ in the derivative: $f^{\prime}(0)=a \sin(0)+b \cos(0) = b = 4$.

Vedclass · Accepted Answer

$2 \cos x+4 \sin x+1$

Vedclass · Answer

(B) Given,$f^{\prime}(x)=a \sin x+b \cos x$ and $f^{\prime}(0)=4$.
Substituting $x=0$ in the derivative: $f^{\prime}(0)=a \sin(0)+b \cos(0) = b = 4$.
Now,integrate $f^{\prime}(x)$ to find $f(x)$:
$f(x) = \int (a \sin x + 4 \cos x) dx = -a \cos x + 4 \sin x + C$.
Using $f(0)=3$: $f(0) = -a \cos(0) + 4 \sin(0) + C = -a + C = 3 \Rightarrow C = a+3$.
Using $f\left(\frac{\pi}{2}\right)=5$: $f\left(\frac{\pi}{2}\right) = -a \cos\left(\frac{\pi}{2}\right) + 4 \sin\left(\frac{\pi}{2}\right) + C = 0 + 4(1) + C = 4+C = 5 \Rightarrow C = 1$.
Substituting $C=1$ into $C=a+3$: $1 = a+3 \Rightarrow a = -2$.
Thus,$f(x) = -(-2) \cos x + 4 \sin x + 1 = 2 \cos x + 4 \sin x + 1$.

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

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