Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The number of values of  such that the variance of&nbsp

The number of values of \mathrm{a \in \mathbb{N}} such that the variance of \mathrm{3,7,12, a, 43-a} is a natural number is :

Option: 1

0


Option: 2

2


Option: 3

5


Option: 4

infinite


Answers (1)

best_answer

\mathrm{\text { Mean }=\frac{3+7+12+a+(43-a)}{5}}\\

               \mathrm{=13}\\

\mathrm{Variance =\frac{(3-13)^{2}+(7-13)^{2}+(12-13)^{2}+(a-13)^{2}+(43-a-13)^{2}}{5}}

                               \mathrm{=\frac{100+36+1+a^{2}+169-26 a+900+a^{2}-60 a}{5}}\\

                 \mathrm{=\frac{1206+2 a^{2}-86 a}{5}}\\

                 \mathrm{=\frac{2\left(a^{2}-43 a+603\right)}{5}}

Now we can check that using any \mathrm{a\in N}\\, unit digit of \mathrm{a^{2}-43 a+603}\\ can never be \mathrm{0\: or\: 5}\\

So numerator is not a multiple of \mathrm{ 5}\\ and variance, thus can never be a natural number.

Hence answer is option 1.

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question