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If n( ) = 50, n(A)=15, n(B)=13 and n(A∩B)=10, find n(A1), n(B1) and n(AUB).​

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993x 93x 293x 2 +ax−2=093x 2 +ax−2=0Since, ONE root is 1, then x=193x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−193x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−B=0 has equal roots.93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 293x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ SUBSTITUTING a=−1 ]93x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=093x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=3693x 2 +ax−2=0Since, one root is 1, then x=1⇒ 3(1) 2 +a(1)−2=0⇒ 3+a−2=0⇒ a+1=0⇒ a=−1⇒ Now, it is given that ax 2 +6ax−b=0 has equal roots.∴ b 2 −4ac=0⇒ (6a) 2 −4(a)(−b)=0⇒ 36a 2 +4ab=0⇒ 36(−1) 2 +4(−1)b=0 [ Substituting a=−1 ]⇒ 36−4b=0⇒ 4b=36∴ b=9



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