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If $f(x)=\left|\begin{array}{ccc}x^3+x & x+1 & x-2 \\ 2 x^3+3 x-1 & 3 x & 3 x-3 \\ x^3+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|$, then $\frac{d}{d x}(f(x))$ is equal to
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The correct answer is:
24
Given, $f(x)=\left|\begin{array}{ccc}x^3+x & x+1 & x-2 \\ 2 x^3+3 x-1 & 3 x & 3 x-3 \\ x^3+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|$
Applying $R_1 \rightarrow R_1+R_3-R_2$, we get
$$
\begin{aligned}
f(x) & =\left|\begin{array}{ccc}
4 & 0 & 0 \\
2 x^3+3 x-1 & 3 x & 3 x-3 \\
x^3+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right| \\
\Rightarrow \quad f(x) & =4[(3 x)(2 x-1)-(3 x-3)(2 x-1)] \\
& =4\left(6 x^2-3 x-6 x^2+3 x+6 x-3\right) \\
& =4(6 x-3) \\
\therefore \quad \frac{d f(x)}{d x} & =4(6 \times 1-0)=24
\end{aligned}
$$
Applying $R_1 \rightarrow R_1+R_3-R_2$, we get
$$
\begin{aligned}
f(x) & =\left|\begin{array}{ccc}
4 & 0 & 0 \\
2 x^3+3 x-1 & 3 x & 3 x-3 \\
x^3+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right| \\
\Rightarrow \quad f(x) & =4[(3 x)(2 x-1)-(3 x-3)(2 x-1)] \\
& =4\left(6 x^2-3 x-6 x^2+3 x+6 x-3\right) \\
& =4(6 x-3) \\
\therefore \quad \frac{d f(x)}{d x} & =4(6 \times 1-0)=24
\end{aligned}
$$
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