Functions Question 396
Question: Which of the following functions is even,
Options:
A) $ f(x)=\sqrt{1+x+x^{2}}-\sqrt{1-x+x^{2}} $
B) $ f(x)=log( \frac{1-x}{1+x} ) $
C) $ f(x)=log( x+\sqrt{1+x^{2}} ) $
D) $ f(x)=\frac{e^{x}+{e^{-x}}}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ f(-x)=\sqrt{1+(-x)+{{(-x)}^{2}}}-\sqrt{1-(-x)+{{(-x)}^{2}}} $ $ =\sqrt{1-x+x^{2}}-\sqrt{1+x+x^{2}}=-f(x) $ Hence, $ f(x) $ is odd. [b] $ f(-x)=log{ \frac{1-(-x)}{1+(-x)} }=\log ( \frac{1+x}{1-x} )=-f(x) $ Hence, f(x) is odd. [c] $ f(-x)=log( -x+\sqrt{1+{{(-x)}^{2}}} ) $ $ =\log { \frac{(-x+\sqrt{1+x^{2}})(x+\sqrt{1+x^{2}})}{(x+\sqrt{1+x^{2}})} } $ $ =\log ( \frac{1}{x+\sqrt{1+x^{2}}} )=-f(x) $ Hence, f(x) is odd. [d] $ f(-x)=\frac{{e^{-x}}+{e^{-(-x)}}}{2}=\frac{{e^{-x}}+e^{x}}{2}=f(x) $ Hence, f(x) is even.
BETA
