AMU 2026 — Mathematics PYQ

1. Mathematical Definition

For a set of $n$ observations $\{x_1,x_2,\dots ,x_n\}$, the sum of squares of deviations from a value $a$ is defined as the function:

$$S(a)=\sum _{i=1}^n(x_i-a{)}^2$$

2. Minimization Principle

To find the value of $a$ that minimizes this sum, we take the derivative of $S(a)$ with respect to $a$ and set it to zero:

$$\frac{dS}{da}=\frac{d}{da}\left[\sum _{i=1}^n(x_i-a{)}^2\right]=0$$

$$-2\sum _{i=1}^n(x_i-a)=0$$

Dividing by $-2$:

$$\sum x_i-\sum a=0$$

$$\sum x_i-na=0$$

$$na=\sum x_i$$

$$a=\frac{\sum x_i}{n}=\bar{x}$$

3. Conclusion

The derivative test shows that the sum of squares of deviations is minimized specifically when $a$ is equal to the arithmetic mean ($\bar{x}$). Therefore, compared to any other constant value, the sum of squares of deviations from the mean is always the minimum.

This directly matches option (b).