Let f(x) + 2f(1/x) = x² + 5 and 2g(x) - 3g(1/2) = x, x 0. If α = ∫₁² f(x)dx, and β = ∫₁² g(x)dx, t

Q: Why can't we just integrate the functional equation directly?

That is a brilliant question! The reason is that $f(x)$ is trapped inside the equation with $f(\frac{1}{x})$. You cannot integrate $f(\frac{1}{x})$ without knowing the explicit form of $f(x)$ first. You must isolate the function explicitly before applying the integral operator. Think of it as cleaning the room before you can start decorating; you need the function in its simplest form first.

Q: Wait, why is $g(\frac{1}{2})$ a constant?

In the expression $2g(x) - 3g(\frac{1}{2}) = x$, the term $g(\frac{1}{2})$ is the value of the function $g$ at the specific point $x = 1/2$. Since $1/2$ is a fixed number, $g(1/2)$ is just a scalar value, not a function of $x$. It is a common trap to treat it as a variable, but once you plug in $x = 1/2$, it collapses into a simple number, making the equation solvable.

Q: Is there a shortcut to find $9\alpha + \beta$ without solving for the functions?

While sometimes possible in competitive exams to use properties of integrals, here the functions are simple enough that solving for them is the most robust and error-proof path. Don't look for shortcuts when the direct path is a straight line! Solving for the functions builds your intuition for future, more complex problems where shortcuts might not exist.

Question

Let $f(x) + 2f(\frac{1}{x}) = x^2 + 5$ and $2g(x) - 3g(\frac{1}{2}) = x, x > 0$. If $\alpha = \int_1^2 f(x)dx$, and $\beta = \int_1^2 g(x)dx$, then the value of $9\alpha + \beta$ is:

Options:

(A) 1

(B) 0

(C) 10

(D) 11

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with solving a system involving functional equations and definite integration. The first equation is given by:

$$f(x) + 2f\left(\frac{1}{x}ight) = x^2 + 5$$

To solve for $f(x)$, we apply the substitution $x 	o \frac{1}{x}$. This yields the transformed equation:

$$f\left(\frac{1}{x}ight) + 2f(x) = \frac{1}{x^2} + 5$$

### The Master Equation for $f(x)$

We now have a system of two linear equations. To isolate $f(x)$, we multiply the second equation by $2$:

$$2f\left(\frac{1}{x}ight) + 4f(x) = \frac{2}{x^2} + 10$$

Subtracting the original equation from this result eliminates the $f(\frac{1}{x})$ term:

$$3f(x) = \frac{2}{x^2} - x^2 + 5$$

Dividing by $3$, we obtain the explicit form of the function:

$$f(x) = \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3}$$

### Solving for $g(x)$

The second functional equation is $2g(x) - 3g\left(\frac{1}{2}ight) = x$. Note that $g\left(\frac{1}{2}ight)$ is a constant.

Substituting $x = \frac{1}{2}$ into the equation gives:

$$2g\left(\frac{1}{2}ight) - 3g\left(\frac{1}{2}ight) = \frac{1}{2} \Rightarrow -g\left(\frac{1}{2}ight) = \frac{1}{2} \Rightarrow g\left(\frac{1}{2}ight) = -\frac{1}{2}$$

Substituting this constant back into the original expression for $g(x)$:

$$2g(x) - 3\left(-\frac{1}{2}ight) = x \Rightarrow 2g(x) + \frac{3}{2} = x \Rightarrow g(x) = \frac{x}{2} - \frac{3}{4}$$

### The Integration Finale

We define $\alpha = \int_1^2 f(x) \, dx$ and $\beta = \int_1^2 g(x) \, dx$. For $\alpha$, we integrate the expression derived earlier:

$$\alpha = \int_1^2 \left( \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3} ight) \, dx = \left[ -\frac{2}{3x} - \frac{x^3}{9} + \frac{5x}{3} ight]_1^2$$

Evaluating at the limits $x=2$ and $x=1$:

$$\alpha = \left( -\frac{1}{3} - \frac{8}{9} + \frac{10}{3} ight) - \left( -\frac{2}{3} - \frac{1}{9} + \frac{5}{3} ight) = \frac{19}{9} - \frac{8}{9} = \frac{11}{9}$$

For $\beta$, we integrate $g(x)$:

$$\beta = \int_1^2 \left( \frac{x}{2} - \frac{3}{4} ight) \, dx = \left[ \frac{x^2}{4} - \frac{3x}{4} ight]_1^2 = \left( 1 - \frac{6}{4} ight) - \left( \frac{1}{4} - \frac{3}{4} ight) = -\frac{1}{2} - \left( -\frac{1}{2} ight) = 0$$

### Final Calculation

Finally, we compute the required value:

$$9\alpha + \beta = 9\left(\frac{11}{9}ight) + 0 = 11$$

The final answer is **11**.

Animated Solution for Mathematics - Definite Integration: Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x, x > 0$ . If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is:

Select Answer:

Visualized Solution

Introduction to Functional Equations

Analyzing the Equation for $f (x)$

Substitution Trick: $x \to \frac{1}{x}$

Eliminating $f (\frac{1}{x})$

Solving for $f (x)$

Setting up the Integral for $α$

Integrating $f (x)$

Evaluating $α$

Analyzing the Equation for $g (x)$

Finding the Constant $g (\frac{1}{2})$

Determining $g (x)$

Integrating $g (x)$ for $β$

Evaluating $β$

Final Calculation: $9 α + β$

Conclusion and Key Takeaways

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Master Equation for $f (x)$

Solving for $g (x)$

The Integration Finale

Final Calculation

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Select Answer:

Visualized Solution

Introduction to Functional Equations

Analyzing the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Substitution Trick: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→x1​

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x1​)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Setting up the Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Integrating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Analyzing the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(21​)

Determining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Integrating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9α+β