GATE CS 2015 Set 3 Q.55 (CS Q.45): Calculus

If for non-zero $x$ ,

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

where $a \ne b$ then $\int_1^2 f(x) dx$ is

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

$\frac{1}{a^2 - b^2}\left[a(2\ln 2 - 25) - \frac{47b}{2}\right]$

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

$\frac{1}{a^2 - b^2}\left[a(\ln 2 - 25) - \frac{47b}{2}\right]$

Official answer and solution

Solution

Eliminate the composite term f(1/x) using functional equation substitution and integrate the isolated function f(x).

We are given the functional equation:

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

Replace $x$ with $\frac{1}{x}$ in (1):

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

Multiply (1) by $a$ and (2) by $b$ :

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

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

Subtract the second equation from the first:

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

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

Now evaluate the definite integral $\int_1^2 f(x) dx$ :

\int_1^2 f(x) dx = \frac{1}{a^2 - b^2} \left[a \int_1^2 \left(\frac{1}{x} - 25\right) dx - b \int_1^2 (x - 25) dx\right]

$\int_1^2 \left(\frac{1}{x} - 25\right) dx = [\ln x - 25x]_1^2 = (\ln 2 - 50) - (0 - 25) = \ln 2 - 25$
$\int_1^2 (x - 25) dx = \left[\frac{x^2}{2} - 25x\right]_1^2 = (2 - 50) - \left(\frac{1}{2} - 25\right) = -48 - (-24.5) = -23.5 = -\frac{47}{2}$

Substituting back:

\int_1^2 f(x) dx = \frac{1}{a^2 - b^2} \left[a(\ln 2 - 25) - b\left(-\frac{47}{2}\right)\right] = \frac{1}{a^2 - b^2} \left[a(\ln 2 - 25) + \frac{47b}{2}\right]

This matches Option A.

Final answer

Common trap

Sign error when distributing the minus sign on the b integral (-b * (-47/2) = +47b/2).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

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