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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Application of Derivatives

NDA 2022 · Mathematics · Application of Derivatives

Where does the function f(x)=∑_ j=1 ^7(x-j)^2 attain its minimum...

Where does the function f(x)=∑j=17(x−j)2f(x)=\sum_{j=1}^7(x-j)^2 attain its minimum value ??

Options

  1. Ax=3.5x=3.5
  2. Bx=4x=4
  3. Cx=4.5x=4.5
  4. Dx=4x=4

Correct answer

B. x=4x=4

Step-by-step solution

f(x)=∑j=17(x−j)2=∑j=17(x2−2xj+j2)x2∑j=171−2x∑j=17j+∑j=17j2=7x2−2x7(7+1)2+7(7+1)(14+1)6=7x2−56x+140f′(x)=14x−56=0⇒x=4f′′(x)=14>0\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)isminimumatis minimum atx=4$.
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