(A) Given $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}$.Expanding along the first column:$f(x) = x(12x^2 - 6x^2)

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(A) Given $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}$.Expanding along the first column:$f(x) = x(12x^2 - 6x^2) - 1(6x^3 - 2x^3) + 0$$f(x) = x(6x^2) - 1(4x^3) = 6x^3 - 4x^3 = 2x^3$.Now,find the derivatives:$f'(x) = \frac{d}{dx}(2x^3) = 6x^2$.$f''(x) = \frac{d}{dx}(6x^2) = 12x$.Therefore,the ratio is $\frac{f''(x)}{f'(x)} = \frac{12x}{6x^2} = \frac{2}{x}$ or $2 : x$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Rank of Matrices , Some special determinants, differentiation and integration of determinants

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If $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}$,then the ratio $f^{\prime \prime}(x) : f^{\prime}(x) =$

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A
$2 : x$
B
$x^2 : x$
C
$3x : 2$
D
$6 : x$

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Rank of Matrices , Some special determinants, differentiation and integration of determinants

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If $f(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}$,then the ratio $f^{\prime \prime}(x) : f^{\prime}(x) =$

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