QJEEM22Maf04c: JEE Main 2022 Mathematics - Class 11 JEEM22Maf04c Solution: The number of values of a N such that the variance of 3, 7, 12, …

Given,

5 numbers are 3, 7, 12, a, 43 $-$ a

$\therefore$ Mean $(\overline x ) = {{3 + 7 + 12 + a + 43 - a} \over 5}$

$= {{65} \over 5}$

$= 13$

We know,

Variance $({\sigma ^2}) = {{\sum {x_1^2} } \over n} - {(\overline x )^2}$

$= {{{3^2} + {7^2} + {{12}^2} + {a^2} + {{(43 - a)}^2}} \over 5} - {(13)^2}$

$= {{9 + 49 + 144 + {a^2} + 1849 + {a^2} - 86a} \over 5} - 169$

$= {{2{a^2} - 86a + 2051} \over 5} - 169$

$= {{2{a^2} - 86a + 2051 - 845} \over 5}$

$= {{2{a^2} - 86a + 1206} \over 5}$

$= {2 \over 5}({a^2} - 43a + 603)$

Variance will be natural number if ${a^2} - 43a + 603$ in multiple of 5.

$\therefore$ ${a^2} - 43a + 603 = 5n$

$\Rightarrow {a^2} - 43a + 603 - 5n = 0$ ..... (1)

$\therefore$ $a = {{43\, \pm \,\sqrt D } \over 2}$

Now "a" will be natural number if

(1) D is a perfect square and

(2) $43\, \pm \,\sqrt D$ is multiple of 2

From equation (1),

$D = {( - 43)^2} - 4\,.\,1\,.\,(603 - 5n)$

$= 1849 - 2412 + 20n$

$= 20n - 563$

For any value of n, unit digit of 20n is always 0. Then 20n $-$ 563 will give a number whose unit digit is 7.

For perfect square numbers ex : 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100

So, unit digit is either 1, 4, 6, 5, 6, 9, 0 it can't be 7. So D can't be perfect square.