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If $f(x)$ is differentiable on $\mathbb{R}, f(x) f^{\prime}(-x)-f(-x) f^{\prime}(x)=0$, $f(x) f^{\prime}(-x)=0, f(0)=3$ and $f(3)=9$, then $(1+f(-3))^3+1=$
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Verified Answer
The correct answer is:
$9$
Given, $f(x) f^{\prime}(-x)-f(-x) f^{\prime}(x)=0$
$$
\Rightarrow f(-x) f^{\prime}(x)-f^{\prime}(-x) f(x)=0
$$
$$
\Rightarrow \frac{d}{d x}[f(-x) f(x)]=0
$$
$\Rightarrow f(-x) \cdot f(x)=c$ where $c=$ Arbitrary constant
when $x=0$, then
$$
f(0) \cdot f(0)=c \Rightarrow c=9
$$
when $x=3$, then
$$
\begin{aligned}
& f(-3) f(3)=c \Rightarrow f(-3) .(9)=9 \\
& \Rightarrow f(-3)=1
\end{aligned}
$$
Now $[1+f(-3)]^3+1=[1+1]^3+1=9$
$$
\Rightarrow f(-x) f^{\prime}(x)-f^{\prime}(-x) f(x)=0
$$
$$
\Rightarrow \frac{d}{d x}[f(-x) f(x)]=0
$$
$\Rightarrow f(-x) \cdot f(x)=c$ where $c=$ Arbitrary constant
when $x=0$, then
$$
f(0) \cdot f(0)=c \Rightarrow c=9
$$
when $x=3$, then
$$
\begin{aligned}
& f(-3) f(3)=c \Rightarrow f(-3) .(9)=9 \\
& \Rightarrow f(-3)=1
\end{aligned}
$$
Now $[1+f(-3)]^3+1=[1+1]^3+1=9$
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