If f(x) = left| begin{array}{ccc} x^3+x & x+1 | vedclass

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(A) Given,$f(x) = \left| \begin{array}{ccc} x^3+x & x+1 & x-2 \ 2x^3+3x-1 & 3x & 3x-3 \ x^3+2x+3 & 2x-1 & 2x-1 \end{array} ight|$.Applying the row

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$24$

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(A) Given,$f(x) = \left| \begin{array}{ccc} x^3+x & x+1 & x-2 \ 2x^3+3x-1 & 3x & 3x-3 \ x^3+2x+3 & 2x-1 & 2x-1 \end{array} ight|$.Applying the row operation $R_1 ightarrow R_1 + R_3 - R_2$,we get:$f(x) = \left| \begin{array}{ccc} (x^3+x) + (x^3+2x+3) - (2x^3+3x-1) & (x+1) + (2x-1) - 3x & (x-2) + (2x-1) - (3x-3) \ 2x^3+3x-1 & 3x & 3x-3 \ x^3+2x+3 & 2x-1 & 2x-1 \end{array} ight|$$f(x) = \left| \begin{array}{ccc} 4 & 0 & 0 \ 2x^3+3x-1 & 3x & 3x-3 \ x^3+2x+3 & 2x-1 & 2x-1 \end{array} ight|$Expanding along the first row:$f(x) = 4 [ (3x)(2x-1) - (3x-3)(2x-1) ]$$f(x) = 4 [ (2x-1)(3x - (3x-3)) ]$$f(x) = 4 [ (2x-1)(3) ] = 12(2x-1) = 24x - 12$Now,differentiating with respect to $x$:$\frac{d}{dx}(f(x)) = \frac{d}{dx}(24x - 12) = 24$.

If $f(x) = \left| \begin{array}{ccc} x^3+x & x+1 & x-2 \\ 2x^3+3x-1 & 3x & 3x-3 \\ x^3+2x+3 & 2x-1 & 2x-1 \end{array} \right|$,then $\frac{d}{dx}(f(x))$ is equal to

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