KVPY Sample Paper KVPY Stream-SX Model Paper-18

If \[U=\left\{ x:{{x}^{5}}-6{{x}^{4}}+11{{x}^{3}}-6{{x}^{2}}=0 \right\},\]\[A=\{x:{{x}^{2}}-5x+6=0)\] and \[B=\{x:{{x}^{2}}-3x+2=0\},then\,n(A\cap B)'\]is equal to:

A) 2

B) 3

C) 4

D) 5

Correct Answer: B

Solution :

we have, \[\operatorname{U}=\left\{ x:{{x}^{5}}-6{{x}^{4}}+11{{x}^{3}}-6{{x}^{2}}=0 \right\}\]\[=\{0,1,2,3\}\]

\[\operatorname{A}=\left\{ x:{{x}^{2}}-5x+6=0 \right\}=\left\{ 2,3 \right\}\]

and \[\operatorname{B}=\left\{ x:{{x}^{2}}-3x+2=0 \right\}=\left\{ 1,2 \right\}\]\[\therefore \operatorname{A}\cap B=\{2\}\]

Hence, \[\left( \operatorname{A}\cap \operatorname{B} \right)=\operatorname{U}-\left( \operatorname{A}\cap \operatorname{B} \right)\]

\[=\left\{ 0,1,2,3 \right\}-\left\{ 2 \right\}\]

\[=\left\{ 0,1,3 \right\}\]

\[\therefore n\left( A\cap B \right)'=3\]