GATE CS 2015 Set 3 Q45 — Maxima and Minima, Mean Value Theorem, Integration

To find the value of the integral $\int_{1}^{2} f(x) dx$, we first need to determine the function $f(x)$ from the given functional equation:

$$\text{(1)}\quad a f(x) + b f\left(\frac{1}{x}\right) = \frac{1}{x} - 25$$

Since this relation holds for all non-zero $x$, we can substitute $x$ with $\frac{1}{x}$ to obtain a second equation:

$$\text{(2)}\quad a f\left(\frac{1}{x}\right) + b f(x) = x - 25$$

We now have a system of two linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$. To eliminate $f\left(\frac{1}{x}\right)$, multiply equation (1) by $a$ and equation (2) by $b$:

$$a^2 f(x) + ab f\left(\frac{1}{x}\right) = \frac{a}{x} - 25a$$

$$b^2 f(x) + ab f\left(\frac{1}{x}\right) = bx - 25b$$

Subtracting the second equation from the first gives:

$$(a^2 - b^2) f(x) = \frac{a}{x} - bx - 25a + 25b$$

$$f(x) = \frac{1}{a^2 - b^2} \left[ \frac{a}{x} - bx - 25(a - b) \right]$$

Now, we integrate $f(x)$ from $1$ to $2$:

$$\int_{1}^{2} f(x) dx = \frac{1}{a^2 - b^2} \int_{1}^{2} \left[ \frac{a}{x} - bx - 25(a - b) \right] dx$$

Let us evaluate each term of the integral individually:

1. $$\int_{1}^{2} \frac{a}{x} dx = a \Big[ \ln x \Big]_{1}^{2} = a \ln 2$$
2. $$\int_{1}^{2} bx dx = b \left[ \frac{x^2}{2} \right]_{1}^{2} = b \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{3b}{2}$$
3. $$\int_{1}^{2} 25(a - b) dx = 25(a - b) \Big[ x \Big]_{1}^{2} = 25(a - b)$$

Substituting these values back into the integral expression:

$$\int_{1}^{2} f(x) dx = \frac{1}{a^2 - b^2} \left[ a \ln 2 - \frac{3b}{2} - 25a + 25b \right]$$

Grouping the terms with $a$ and $b$:

$$\int_{1}^{2} f(x) dx = \frac{1}{a^2 - b^2} \left[ a(\ln 2 - 25) + b\left(25 - \frac{3}{2}\right) \right]$$

$$\int_{1}^{2} f(x) dx = \frac{1}{a^2 - b^2} \left[ a(\ln 2 - 25) + \frac{47b}{2} \right]$$

This matches Option A.