Question

If \[f(x) = \left| {\begin{array}{*{20}{c}}
  {{x^n}}&{\sin x}&{\cos x} \\
  {n!}&{\sin (n\dfrac{\pi }{2})}&{\cos (n\dfrac{\pi }{2})} \\
  a&{{a^2}}&{{a^3}}
\end{array}} \right|\] then the value of :
\[\dfrac{{{d^n}}}{{d{x^n}}}(f(x))\] at x=0, for n=2m+1 is;

Answer 1

Hint: Here the given question is to solve for the matrix given, first the question needs to differentiate the term and then solve for the given values of the variables as given in the question. To solve the question we need to understand that when two rows or columns of a matrix are the same then the value of the determinant is zero for the matrix.

Formulae Used: The value of determinant is zero when any two or three rows or columns are the same in a matrix.

Complete step-by-step answer:
Here to solve the question first we need to differentiate the given matrix and thus solve further, with the given values in the question, on solving we get:
\[
   \Rightarrow f(x) = \left| {\begin{array}{*{20}{c}}
  {{x^n}}&{\sin x}&{\cos x} \\
  {n!}&{\sin (n\dfrac{\pi }{2})}&{\cos (n\dfrac{\pi }{2})} \\
  a&{{a^2}}&{{a^3}}
\end{array}} \right| \\
   \Rightarrow \dfrac{{{d^n}}}{{d{x^n}}}(f(x)) = \left| {\begin{array}{*{20}{c}}
  {n!}&{\sin \left( {x + \dfrac{{n\pi }}{2}} \right)}&{\cos \left( {x + \dfrac{{n\pi }}{2}} \right)} \\
  {n!}&{\sin (n\dfrac{\pi }{2})}&{\cos (n\dfrac{\pi }{2})} \\
  a&{{a^2}}&{{a^3}}
\end{array}} \right| \\
   \Rightarrow for\,x = 0,\,and\,n = 2m + 1 \\
 \]
Here when differentiating the other two rows then also the same result will be obtained, that is the determinant value of the matrix will be zero, hence here we ignore the other two determinants which will have determinant value of zero.

\[
   \Rightarrow \dfrac{{{d^n}}}{{d{x^n}}}(f(x)) = \left| {\begin{array}{*{20}{c}}
  {n!}&{\sin \left( {0 + \dfrac{{(2m + 1)\pi }}{2}} \right)}&{\cos \left( {0 + \dfrac{{(2m + 1)\pi }}{2}} \right)} \\
  {n!}&{\sin ((2m + 1)\dfrac{\pi }{2})}&{\cos ((2m + 1)\dfrac{\pi }{2})} \\
  a&{{a^2}}&{{a^3}}
\end{array}} \right| \\
   \Rightarrow \dfrac{{{d^n}}}{{d{x^n}}}(f(x)) = \left| {\begin{array}{*{20}{c}}
  {n!}&{\sin \left( {\dfrac{{(2m + 1)\pi }}{2}} \right)}&{\cos \left( {\dfrac{{(2m + 1)\pi }}{2}} \right)} \\
  {n!}&{\sin ((2m + 1)\dfrac{\pi }{2})}&{\cos ((2m + 1)\dfrac{\pi }{2})} \\
  a&{{a^2}}&{{a^3}}
\end{array}} \right| = 0 \\
 \]
Here the determinant value is zero because the two rows of the matrix is same, and we know the property of determinant that when two rows will be same then the value gives zero

Note: The given question is about the combination of matrix and differentiation, here first we need to differentiate. We know that to find the differentiation of the matrix, we need to differentiate each and every rows, or columns individually.