Let ${a_1},{a_2},{a_3},{a_4}$ be real numbers such that $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 1$
Then the smallest possible value of the expression ${\left( {{a_1} - {a_2}} \right)^2} + {\left( {{a_2} - {a_3}} \right)^2} + {\left( {{a_3} - {a_4}} \right)^2} + {\left( {{a_4} - {a_1}} \right)^2}$ lies in interval
A. (0, 1.5)
B. (-1.5, 2.5)
C. (2.5, 3)
D. (3, 3.5)
Answer
614.4k+ views
Hint: Start by evaluating the lowest or minimum value of the given expression and accordingly assume the value of individual terms as variables. Substitute this variable in the equation in order to find its value and henceforth the value of all the 4 terms. Based on this value , look for the best suitable option which satisfies the minimum value range.
Complete step-by-step answer:
Given, $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 1$
When we observe this question carefully then we come to know that the expression ${\left( {{a_1} - {a_2}} \right)^2} + {\left( {{a_2} - {a_3}} \right)^2} + {\left( {{a_3} - {a_4}} \right)^2} + {\left( {{a_4} - {a_1}} \right)^2}$ is always positive . Since all the terms are squared and the square of the negative term will also be positive therefore the minimum value for this equation is 0. And this can only be true, when all have the same values i.e. ${a_1} = {a_2} = {a_3} = {a_4} = a$
On putting the value of ${a_1},{a_2},{a_3},{a_4}$ as a and in the equation $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 1$ we get the value of
$
{a^2} + {a^2} + {a^2} + {a^2} = 1 \\
\Rightarrow 4{a^2} = 1 \\
\Rightarrow {a^2} = \dfrac{1}{4} \\
\Rightarrow a = \dfrac{1}{2} \\
$
Therefore, we get the values of ${a_1},{a_2},{a_3},{a_4}$ as ${a_1} = {a_2} = {a_3} = {a_4} = a = \dfrac{1}{2}$
So, we found the value of ${a_1},{a_2},{a_3},{a_4}$which is 0.5 and we also came to know that the minimum value of the given equation will be 0, which would lie in the range of (-1.5, 2.5)
So, the correct answer is “Option B”.
Note: Similar questions can be solved using the above procedure. Students must read the question at least twice as many times so we can derive very useful information for a solution , whenever we feel we are stuck. Attention must be given while simplifying or computing as any mistake may lead to wrong answers.
Complete step-by-step answer:
Given, $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 1$
When we observe this question carefully then we come to know that the expression ${\left( {{a_1} - {a_2}} \right)^2} + {\left( {{a_2} - {a_3}} \right)^2} + {\left( {{a_3} - {a_4}} \right)^2} + {\left( {{a_4} - {a_1}} \right)^2}$ is always positive . Since all the terms are squared and the square of the negative term will also be positive therefore the minimum value for this equation is 0. And this can only be true, when all have the same values i.e. ${a_1} = {a_2} = {a_3} = {a_4} = a$
On putting the value of ${a_1},{a_2},{a_3},{a_4}$ as a and in the equation $a_1^2 + a_2^2 + a_3^2 + a_4^2 = 1$ we get the value of
$
{a^2} + {a^2} + {a^2} + {a^2} = 1 \\
\Rightarrow 4{a^2} = 1 \\
\Rightarrow {a^2} = \dfrac{1}{4} \\
\Rightarrow a = \dfrac{1}{2} \\
$
Therefore, we get the values of ${a_1},{a_2},{a_3},{a_4}$ as ${a_1} = {a_2} = {a_3} = {a_4} = a = \dfrac{1}{2}$
So, we found the value of ${a_1},{a_2},{a_3},{a_4}$which is 0.5 and we also came to know that the minimum value of the given equation will be 0, which would lie in the range of (-1.5, 2.5)
So, the correct answer is “Option B”.
Note: Similar questions can be solved using the above procedure. Students must read the question at least twice as many times so we can derive very useful information for a solution , whenever we feel we are stuck. Attention must be given while simplifying or computing as any mistake may lead to wrong answers.
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