Determinants Matrices Question 81
Question: If $ f(x)= \begin{vmatrix} {2^{-x}} & {e^{x{\log _{e}}2}} & x^{2} \\ {2^{-3x}} & {e^{3x{\log _{e}}2}} & x^{4} \\ {2^{-5x}} & {e^{5x{\log _{e}}2}} & 1 \\ \end{vmatrix}, $ then
Options:
A) $ f(x)+f(-x)=0 $
B) $ f(x)-f(-x)=0 $
C) $ f(x)+f(-x)=2 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ f(x)= \begin{vmatrix} {2^{-x}} & 2^{x} & x^{2} \\ {2^{-3x}} & 2^{3x} & x^{4} \\ {2^{-5x}} & 2^{5x} & 1 \\ \end{vmatrix} $
$ \therefore f(-x) \begin{vmatrix} 2^{x} & {2^{-x}} & x^{2} \\ 2^{3x} & {2^{-3x}} & x^{4} \\ 2^{5x} & {2^{-5x}} & 1 \\ \end{vmatrix}=-f(x) $
BETA
