JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 17)
Let X = {x
$$ \in $$ N : 1
$$ \le $$ x
$$ \le $$ 17} and
Y = {ax + b: x $$ \in $$ X and a, b $$ \in $$ R, a > 0}. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :
Y = {ax + b: x $$ \in $$ X and a, b $$ \in $$ R, a > 0}. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :
7
9
-7
-27
Explanation
Mean of X = $${{\sum\limits_{x = 1}^{17} x } \over {17}}$$ = $${{17 \times 18} \over {17 \times 2}}$$ = 9
Mean of Y = $${{\sum\limits_{x = 1}^{17} {\left( {ax + b} \right)} } \over {17}}$$ = 17
$$ \Rightarrow $$ $$a{{\sum\limits_{x = 1}^{17} x } \over {17}} + b$$ = 17
$$ \Rightarrow $$ 9a + b = 17 ....(1)
Given Var(Y) = 216
$$ \Rightarrow $$ $${{\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} } \over {17}} - {\left( {17} \right)^2}$$ = 216
$$ \Rightarrow $$ $${\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} }$$ = 8585
$$ \Rightarrow $$ (a + b)2 + (2a + b)2 +....+ (17a + b)2 = 8585
$$ \Rightarrow $$ 105a2 + b2 + 18ab = 505 ....(2)
From equation (1) & (2)
a = 3 & b = -10
$$ \therefore $$ a + b = –7
Mean of Y = $${{\sum\limits_{x = 1}^{17} {\left( {ax + b} \right)} } \over {17}}$$ = 17
$$ \Rightarrow $$ $$a{{\sum\limits_{x = 1}^{17} x } \over {17}} + b$$ = 17
$$ \Rightarrow $$ 9a + b = 17 ....(1)
Given Var(Y) = 216
$$ \Rightarrow $$ $${{\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} } \over {17}} - {\left( {17} \right)^2}$$ = 216
$$ \Rightarrow $$ $${\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} }$$ = 8585
$$ \Rightarrow $$ (a + b)2 + (2a + b)2 +....+ (17a + b)2 = 8585
$$ \Rightarrow $$ 105a2 + b2 + 18ab = 505 ....(2)
From equation (1) & (2)
a = 3 & b = -10
$$ \therefore $$ a + b = –7
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