Question

If $2 f(x) - 3 f\left(\frac{1}{x}\right) = x$, then $\int_1^e f(x) dx =$

• A. $-\frac{2+e^2}{5}$
• B. $2+e^5$
• C. $\frac{2+e^2}{5}$
• D. $2-\frac{e^2}{5}$

Hint:

Whenever you see an equation involving $f(x)$ and $f(1/x)$, or $f(x)$ and $f(-x)$, the standard trick is always to substitute the argument with its counterpart to form a solvable system of linear equations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a functional equation. We must first solve this equation to find the explicit expression for $f(x)$ and then integrate it over the interval $[1, e]$.

Step 2: Key Formula or Approach:
To find $f(x)$, substitute $x$ with $\frac{1}{x}$ in the given equation to create a system of two linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$. Solve this system to eliminate $f\left(\frac{1}{x}\right)$.

Step 3: Detailed Explanation:
The given functional equation is:
$$2f(x) - 3f\left(\frac{1}{x}\right) = x \quad \text{--- (Equation 1)}$$ Substitute $x \rightarrow \frac{1}{x}$:
$$2f\left(\frac{1}{x}\right) - 3f(x) = \frac{1}{x} \quad \text{--- (Equation 2)}$$ To eliminate $f\left(\frac{1}{x}\right)$, multiply Equation 1 by $2$ and Equation 2 by $3$:
$$4f(x) - 6f\left(\frac{1}{x}\right) = 2x$$ $$-9f(x) + 6f\left(\frac{1}{x}\right) = \frac{3}{x}$$ Add the two equations together:
$$-5f(x) = 2x + \frac{3}{x}$$ $$f(x) = -\frac{2}{5}x - \frac{3}{5x}$$ Now, evaluate the definite integral of $f(x)$ from $1$ to $e$:
$$\int_1^e f(x) dx = \int_1^e \left( -\frac{2}{5}x - \frac{3}{5x} \right) dx$$ $$= \left[ -\frac{1}{5}x^2 - \frac{3}{5}\ln|x| \right]_1^e$$ Substitute the upper limit $e$ and lower limit $1$:
$$= \left( -\frac{1}{5}e^2 - \frac{3}{5}\ln e \right) - \left( -\frac{1}{5}(1)^2 - \frac{3}{5}\ln 1 \right)$$ Since $\ln e = 1$ and $\ln 1 = 0$:
$$= \left( -\frac{e^2}{5} - \frac{3}{5} \right) - \left( -\frac{1}{5} - 0 \right)$$ $$= -\frac{e^2}{5} - \frac{3}{5} + \frac{1}{5} = -\frac{e^2}{5} - \frac{2}{5}$$ $$= -\frac{e^2 + 2}{5}$$

Step 4: Final Answer:
The integral evaluates to $-\frac{2+e^2}{5}$, matching option (A).