Question

Let f(x) be defined on R, such that $f(x) + x^2$ is an odd function, and $f(x) + 2^x$ is an even function. Then 8|f(1)| equals

Answer 1

14

Answer 2

To solve this problem, we will use the definitions of odd and even functions.

A function $g(x)$ is odd if $g(-x) = -g(x)$. A function $h(x)$ is even if $h(-x) = h(x)$.

Given conditions:

1. $f(x) + x^2$ is an odd function. Let $g(x) = f(x) + x^2$. Since $g(x)$ is odd, we have: $g(-x) = -g(x)$ $f(-x) + (-x)^2 = -(f(x) + x^2)$ $f(-x) + x^2 = -f(x) - x^2$ $f(-x) = -f(x) - 2x^2$ (Equation 1)

2. $f(x) + 2^x$ is an even function. Let $h(x) = f(x) + 2^x$. Since $h(x)$ is even, we have: $h(-x) = h(x)$ $f(-x) + 2^{-x} = f(x) + 2^x$ (Equation 2)

Now we have a system of two equations. Substitute the expression for $f(-x)$ from Equation 1 into Equation 2: $(-f(x) - 2x^2) + 2^{-x} = f(x) + 2^x$

Rearrange the terms to solve for $f(x)$: $2^{-x} - 2x^2 - 2^x = 2f(x)$ $f(x) = \frac{2^{-x} - 2^x - 2x^2}{2}$ $f(x) = \frac{2^{-x} - 2^x}{2} - x^2$

Now we need to find the value of $f(1)$. Substitute $x=1$ into the expression for $f(x)$: $f(1) = \frac{2^{-1} - 2^1}{2} - 1^2$ $f(1) = \frac{1/2 - 2}{2} - 1$ $f(1) = \frac{-3/2}{2} - 1$ $f(1) = -\frac{3}{4} - 1$ $f(1) = -\frac{3}{4} - \frac{4}{4}$ $f(1) = -\frac{7}{4}$

Finally, we need to calculate $8|f(1)|$: $8|f(1)| = 8 \left|-\frac{7}{4}\right|$ $8|f(1)| = 8 \times \frac{7}{4}$ $8|f(1)| = 2 \times 7$ $8|f(1)| = 14$