Question

50 points ,please solve only 2 questions,class 11th

Answer 1

Let  the  unknown observations x and  y

Observations:- 2, 4, 10 , 12, 14 , x , y

Total Number of  observation :  = 7

Sum of  observation = 2 + 4 + 10 + 12 + 14 + x + y = 42  + x + y

We know that ,

$(mean)\bar{x}=\frac{\text{Sum of obesevations}}{\text{Total Number of observations}}\\\\8=\frac{42+x+y}{7}\\\\x+y=56-42\\\\x+y=14.....................i)$

Calculate Variance:-

$\left|\begin{array}{ccc}x_i&d=x-\bar{x}&d^{2}\\2&-6&36\\4&-4&16\\10&2&4\\12&4&16\\14&6&36\\x&x-8&(x-8)^{2} \\y&y-8&(y-8)^{2}\end{array}\right|$

$\sum{d^{2}}=36+16+4+16+36+(x-8)^{2}+(y-8)^{2}\\\\=108+x^{2}+64-16x+y^{2}+64-16y\\ \\=236+x^{2}+y^{2}-16(x+y)\\\\=236+x^{2}+y^{2}-16\times14$

$Variance=\frac{\sum{d^{2}}}{n}\\\\16=\frac{x^{2}+y^{2}+12}{7}\\\\x^{2}+y^{2}+12=112\\\\x^{2}+y^{2}=100\\\\x^{2}+y^{2}+2xy-2xy=100\\\\(x+y)^{2}-2xy=100\\\\14^{2}-2xy=100\\ \\2xy=196-100\\\\2xy=96\\\\4xy=192\\\\(x-y)^{2}=(x+y)^{2} -4xy\\\\(x-y)^{2}=14^{2}-192\\ \\(x-y)^{2}=4\\\\x-y=4..................ii)$

Adding  equ i) and equ ii)

$2x=16\\\\x=8\\\\\text{Putting x=8 in equ i)}\\\\8+y=14\\\\y=6$

Hence,

That, observations are  8 and 6

Second  question:-2

Let  p(n): $10^{2n-1}+1$ ,

p(1): = 11 , which is  divisible by 11

p(m):  $10^{2m+1}+1$ , divisible by 11

$Let,\\\\\frac{10^{2m-1}+1}{11}=\lambda\\\\10^{2m-1}+1=11\lambda\\\\Now,\\p(m+1): 10^{2(m+1)-1}+1\\\\=10^{2m-1+2} +1\\\\=10^{2m-1} .10^{2}+1\\ \\=(11\lambda-1).100+1\\\\=11(100\lambda-9),\\\\\text{Which is divisible by 11}\\\\\text{hence, p(n) is true for all}\;n\in\;N$

Answer 2

Answer:

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