JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard

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Analyzing +4 | -1

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {{x^3}}&{{x^2}}&{3{x^2}}\\ 1&{ - 6}&4\\ P&{{P^2}}&{{P^3}} \end{array}} \right|,

here

p

is a constant, then

\frac{{{d^3}f\left( x \right)}}{{d{x^3}}}

View answer and explanation

Correct option: D

A constant

Option D (A constant) is correct because after taking the third derivative of the determinant function, the result is a constant value.

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