Question

If ${\sum }_{i=1}^n{a}_i^2=\lambda ,$ where $a_i\ge 0$ and if the greatest and least values of ${\left({\sum }_{i=1}^n{a}_i^2\right)}^2$ are ${\lambda }_1$ and ${\lambda }_2$ respectively, then ${\lambda }_1-{\lambda }_2$ is:

• A. $n\lambda$
• B. $(n-1)\lambda$
• C. $(n+2)\lambda$
• D. $(n+1)\lambda$

Answer 1

Answer: (B)

$(n-1)\lambda$

$\because \frac{a_1^2+a_2^2+a_3^2+...a_n^2}{n}>{\left(\frac{a_1+a_2+a_3+...+a_n}{n}\right)}^2$

$\Rightarrow \frac{\lambda }{n}>{\left(\frac{a_1+a_2+a_3+...+a_n}{n}\right)}^2$

$\Rightarrow {\left(\frac{a_1+a_2+a_3+...+a_n}{n}\right)}^2<n\lambda$....$\left(1\right)$

Also, ${(a_1+a_2+a_3+...+a_n)}^2=a_1^2+a_2^2+a_3^2+...+a_2^2+2\sum a_1{a}_2$

or ${(a_1+a_2+a_3+...+a_n)}^2=\lambda +2\sum a_1{a}_2$

or ${(a_1+a_2+a_3+...+a_n)}^2>\lambda$ .........$\left(2\right)$

From eqns$\left(1\right)$ and $\left(2\right)$ we get

$\lambda <{(a_1+a_2+a_3+...+a_n)}^2<n\lambda$

$\therefore$ Greatest value${\lambda }_1=n\lambda$ and Least value ${\lambda }_2=\lambda$

Hence, Greatest value $-$ Least value $={\lambda }_1-{\lambda }_2=n\lambda -\lambda =(n-1)\lambda$