Question

If $ f(x) $ and $ g(x) $ are two functions such that $ f(x) + g(x) = e^x $ and $ f(x) - g(x) = e^{-x} $, then:

• A. \( f(x) \) is odd function, \( g(x) \) is odd function
• B. \( f(x) \) is even function, \( g(x) \) is even function
• C. \( f(x) \) is even function, \( g(x) \) is odd function
• D. \( f(x) \) is odd function, \( g(x) \) is even function

Hint:

For exponential functions, remember that the sum and difference of \( e^x \) and \( e^{-x} \) correspond to the hyperbolic cosine and sine functions, which have known even and odd properties.

Answer 1

The Correct Option is C

We are given the following two equations:
1. \( f(x) + g(x) = e^x \) 2. \( f(x) - g(x) = e^{-x} \) 
Step 1: Add the two equations.
By adding the two 
given equations, we get:
\[ (f(x) + g(x)) + (f(x) - g(x)) = e^x + e^{-x} \] \[ 2f(x) = e^x + e^{-x} \] Recall that the sum of exponential terms is the definition of the hyperbolic cosine function: \[ e^x + e^{-x} = 2\cosh(x), \] so we have: \[ 2f(x) = 2\cosh(x) \quad \Rightarrow \quad f(x) = \cosh(x). \] Since \( \cosh(x) \) is an even function (i.e., \( \cosh(-x) = \cosh(x) \)), it follows that \( f(x) \) is an even function. 
Step 2: Subtract the two equations.
Next, subtract the second equation from the first: \[ (f(x) + g(x)) - (f(x) - g(x)) = e^x - e^{-x} \] \[ 2g(x) = e^x - e^{-x} \] The difference of the exponential terms is the definition of the hyperbolic sine function: \[ e^x - e^{-x} = 2\sinh(x), \] so we have: \[ 2g(x) = 2\sinh(x) \quad \Rightarrow \quad g(x) = \sinh(x). \] Since \( \sinh(x) \) is an odd function (i.e., \( \sinh(-x) = -\sinh(x) \)), it follows that \( g(x) \) is an odd function. 
Step 3: Conclusion
Thus, we have determined that \( f(x) \) is an even function and \( g(x) \) is an odd function.