GATE CS 2015 Set 3 QGA-5 — General Aptitude

Since $f(x)$ is a linear function, its rate of change (slope) is constant. We can find the value at $x = 5$ using the property of constant slope:

$$\text{Slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$

Using the given points $x_1 = -2$ with $f(-2) = 29$, and $x_2 = 3, x_3 = 5$ with $f(3) = 39$:

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

Substitute the known values into the equation:

$$\frac{f(5) - 39}{2} = \frac{39 - 29}{3 - (-2)}$$

$$\frac{f(5) - 39}{2} = \frac{10}{5} = 2$$

$$f(5) - 39 = 2 \times 2 = 4$$

$$f(5) = 39 + 4 = 43$$

Thus, the value of the function at $x = 5$ is 43.