JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

f(x)=\left(x-a_{1}\right)^{2}+\left(x-a_{2}\right)^{2}+\cdots+\left(x-a_{n}\right)^{2}

, find

x

where

f(x)

is minimum.

View answer and explanation

Correct option: B

\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}

Option B (

\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}

) is correct because the minimum value of f(x) occurs at the average of the

a_i

values.

More Questions from QUADRATIC EQUATIONS

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\alpha ,\beta

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{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

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{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

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9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)