Q242 / 270 · NDA 2022 (Phase 2) · Mathematics
Where does the function $f(x)=\sum_{j=1}^7(x-j)^2$ attain its minimum value $?$
- A$x=3.5$
- Correct answerB$x=4$
- C$x=4.5$
- D$x=4$
Solution
$\begin{aligned}
& f(x)=\sum_{j=1}^7(x-j)^2=\sum_{j=1}^7\left(x^2-2 x j+j^2\right) \\
& x^2 \sum_{j=1}^7 1-2 x \sum_{j=1}^7 j+\sum_{j=1}^7 j^2 \\
& =7 x^2-2 x \frac{7(7+1)}{2}+\frac{7(7+1)(14+1)}{6} \\
& =7 x^2-56 x+140 \\
& f^{\prime}(x)=14 x-56=0 \Rightarrow x=4 \\
& f^{\prime \prime}(x)=14 \gt 0
\end{aligned}$
$\therefore \quad f(x)$ is minimum at $x=4$.
Hence, the correct answer is option: 'B'.