2^m=56+2ⁿ\(\Rightarrow \)2^m-2ⁿ=56
Ans. m-6,n=3

{0,1,2}

{\(\frac { 1 }{ 2 } ,1,\frac { 3 }{ 2 } \)}

{4,8,12}

n(A\(\cup \)B)\(\ge \)n(B)=6

Let M be the set of students who had taken mathematics, Pbe the set of students who had taken physics and C be the set of students who had taken chemistry.
Here n(U) = 25, n(M) = 15, n(P) = 12, n(C) = 11, n(M \(\cap\) C) = 5, n(M \(\cap\) P) = 9, n(P \(\cap\) C) = 4, n(M \(\cap\) P \(\cap\) C)= 3

From the Venn diagram, we have
n(M) = a + b + d + e = 15
n(P) = b + c + e + f = 12
n(C) = d + e + f + g = 11
n(M \(\cap\) C) = d + e = 5
n(M \(\cap\) P) = b + e = 9
n(P \(\cap\) C) = e + f = 4
n(M \(\cap\) P \(\cap\) C) = e = 3
Now e = 3
d + e = 5 \(\Rightarrow\) d + 3 = 5 \(\Rightarrow\) d = 5 - 3 \(\Rightarrow\) d = 2
b + e = 9 \(\Rightarrow\) b + 3 = 9 \(\Rightarrow\) b = 9 - 3 \(\Rightarrow\) b = 6
e + f = 4 \(\Rightarrow\) 3 + f = 4 \(\Rightarrow\) f = 4 - 3 \(\Rightarrow\) f = 1
a + b + d + e = 15
\(\Rightarrow\) a + 6 + 2 + 3 = 15 \(\Rightarrow\) a = 15 - 11 = 4
b + c + e + f = 12
\(\Rightarrow\) 6 + c + 3 + 1 = 12 \(\Rightarrow\) c = 12 - 10 = 2
d + e + f + g = l1
\(\Rightarrow\) 2 + 3 + 1 + g = 11\(\Rightarrow\) g = 11 - 6 = 5
\(\therefore\) (i) g = 5
(ii) c = 2
(iii) a = 4
(iv) f = 1
(v) b = 6
(vi) a + b + c + d + e + f + g = 4 + 6 + 2 + 2 + 3 + 1 + 5 = 23
(vii) a + c + g = 4 + 2 + 5 = 11
(viii) 25 - (a + b + c + d + e + f + g) = 25 - 23 = 2

(i) {1,7,9}
(ii) {1,3,5,7,9}
(iii) ф
(iv) {2,4,6,8,10}

(i) Here X = {1, 3, 5}and Y = {1, 2, 3}
\(\therefore\) X \(\cap\)Y = {1, 3}
(ii) Here A = {a, e, i, o, u} and B = {a, b, c}
\(\therefore\) A \(\cap\) B= {a}
(iii) Here A = {x : x is a natural number and multiple of 3}
= {3, 6, 9, 12,...........} and
B = {x : x is a natural number less than 6}
= {1, 2, 3, 4, 5}
\(\therefore\) A \(\cap\) B = {3}
(iv) Here A = {x : x is a natural number and 1 < x \(\le\) 6}
= {2, 3, 4, 5, 6}
and B = {x : x is a natural number and 6 < x < 10}
= {7, 8, 9}
\(\therefore\) A \(\cap\) B=ф
(v) Here A = {1, 2, 3} and B =ф
\(\therefore\) A \(\cap\) B=ф

(i) {3, 4} is a member of set A.
\(\therefore\) {3, 4} ∈ A
Hence {3, 4}⊂ A is incorrect.
(ii) {3, 4} is a member of set A.
\(\therefore\){3, 4} ∈ A is correct.
(iii) Here {3, 4} is a member of set A.
\(\therefore\) {{3,4}} is a set
\(\therefore\) {{3,4}} ⊂ A is correct.
(iv) 1 is a member of set A.
\(\therefore\) 1 ∈ A is correct.
(v) 1 is not a set, it is a member of set A.
\(\therefore\) 1⊂ A is incorrect.
(vi) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ⊂ A is correct.
(vii) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ∈ A is incorrect.
(viii) 3 is not a member of set A.
\(\therefore\) {1, 2, 3} is not a subset of set A.
\(\therefore\) {1, 2, 3} ⊂ A is incorrect.
(ix) ф is not a member of set A.
\(\therefore\) ф ∈ A is incorrect.
(x) Since ф is subset of every set,
\(\therefore\)ф ⊂ A is correct.
(xi)ф​​​​​​​ is not a member of set A.
\(\therefore\) {ф} ⊂ A is incorrect.