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If $\int f(x) \cos x d x=\frac{1}{2}[f(x))^2+C$ and $f(0)=0$, then $f^{\prime}(0)=$
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1
We have,
$$
\int f(x) \cos x d x=\frac{1}{2}\langle f(x))^2+C
$$
On differentiating w.r.t ' $x$ ', we get
$$
\begin{aligned}
& f(x) \cos x=\frac{1}{2} \times 2 f(x) \cdot f^{\prime}(x) \\
& \Rightarrow \quad f(x) \cos x=f(x) \cdot f^{\prime}(x) \\
& \Rightarrow \quad f^{\prime}(x)=\cos x \\
& \Rightarrow \quad f^{\prime}(0)=\cos 0 \\
& \Rightarrow \quad f^{\prime}(0)=1 \\
& {[\because \cos 0=1]} \\
&
\end{aligned}
$$
$$
\int f(x) \cos x d x=\frac{1}{2}\langle f(x))^2+C
$$
On differentiating w.r.t ' $x$ ', we get
$$
\begin{aligned}
& f(x) \cos x=\frac{1}{2} \times 2 f(x) \cdot f^{\prime}(x) \\
& \Rightarrow \quad f(x) \cos x=f(x) \cdot f^{\prime}(x) \\
& \Rightarrow \quad f^{\prime}(x)=\cos x \\
& \Rightarrow \quad f^{\prime}(0)=\cos 0 \\
& \Rightarrow \quad f^{\prime}(0)=1 \\
& {[\because \cos 0=1]} \\
&
\end{aligned}
$$
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