Determinants Matrices Question 76
Question: If $ g(x)= \begin{vmatrix} {a^{-x}} & {e^{x{\log _{e}}a}} & x^{2} \\ {a^{-3x}} & {e^{3x{\log _{e}}a}} & x^{4} \\ {a^{-5x}} & {e^{5x{\log _{e}}a}} & 1 \\ \end{vmatrix}, $ then
Options:
A) $ g(x)+g(-x)=0 $
B) $ g(x)-g(-x)=0 $
C) $ g(x)\times g(-x)=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ g(x)= \begin{vmatrix} {a^{-x}} & {e^{{\log _{e}}a^{x}}} & x^{2} \\ {a^{-3x}} & {e^{{\log _{e}}a^{3x}}} & x^{4} \\ {a^{-5x}} & {e^{{\log _{e}}a^{5x}}} & 1 \\ \end{vmatrix} $
$ = \begin{vmatrix} {a^{-x}} & e^{x} & x^{2} \\ {a^{-3x}} & e^{3x} & x^{4} \\ {a^{-5x}} & e^{5x} & 1 \\ \end{vmatrix}( {e^{\log a^{x}}}=a^{x} ) $
$ \Rightarrow g(-x)= \begin{vmatrix} a^{x} & {a^{-x}} & x^{2} \\ a^{3x} & {a^{-3x}} & x^{4} \\ a^{5x} & {a^{-5x}} & 1 \\ \end{vmatrix} $
$ =- \begin{vmatrix} {a^{-}}^{x} & a^{x} & x^{2} \\ {a^{-3x}} & a^{3x} & x^{4} \\ {a^{-5x}} & a^{5x} & 1 \\ \end{vmatrix}( \begin{aligned} & Interchaning \\ & andIIcolumns \\ \end{aligned} ) $
$ =-g(x) $
$ \Rightarrow g(x)+g(-x)=0 $
BETA
