Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for the Chapter 1 Miscellaneous Exercise — all 10 questions solved, moving from concrete numeric subsets to full proof-based reasoning about subset relations and set identities.
Unlike the earlier exercises in this chapter, this one is almost entirely proof-based — you're asked to justify subset and equality claims using the definitions themselves, not just compute an answer. Question 2 is a good self-check: six short statements where a single counterexample sinks a claim, but a full argument is needed to establish a true one. Questions 3–9 build a connected set of identities (A = (A∩B)∪(A−B), the four equivalent subset conditions, and the difference-of-differences result), all provable the same way — pick an arbitrary element and track where it must land. Question 10 closes with a construction problem: build three sets with every pair overlapping but no common element to all three.
First solve for A: x^2-8x+12=0 \Rightarrow (x-2)(x-6)=0 \Rightarrow x=2,6, so A=\{2,6\}.
Now compare elements: D=\{6\} is contained in every other set here, since 6 belongs to A, B and C.
A=\{2,6\}: both 2 and 6 appear in B and in C, so A\subset B and A\subset C.
B=\{2,4,6\}: every element of B is an even natural number, so B\subset C.
None of the reverse containments hold — for instance B\not\subset A since 4\in B but 4\notin A.
Let A=\{1,2\} and B=\{A,3\}=\{\{1,2\},3\}. Here 1\in A and A\in B. But the elements of B are \{1,2\} and 3 — neither equals 1 — so 1\notin B.
Let A=\{1\}, B=\{1,2\}, C=\{\{1,2\},3\}. Then A\subset B and B\in C (since B=\{1,2\} is an element of C). But the elements of C are \{1,2\} and 3, and A=\{1\} equals neither, so A\notin C.
Let x\in A. Since A\subset B, this gives x\in B. Since B\subset C, this gives x\in C. As x was an arbitrary element of A, every element of A is in C.
Let A=\{1,2\}, B=\{2,3\}, C=\{1,2,4\}. Here A\not\subset B (since 1\notin B) and B\not\subset C (since 3\notin C). But A\subset C, since both 1 and 2 lie in C.
Let A=\{1,2\}, B=\{2,3\}. Take x=1\in A. Since 1\notin B, indeed A\not\subset B. But x=1\notin B, so the conclusion fails.
This is the contrapositive of the definition of subset. Suppose, for contradiction, that x\in A. Since A\subset B, this would force x\in B, contradicting x\notin B. So x\notin A.
Since B\subseteq A\cup B, we have B=B\cap(A\cup B). Substituting the given conditions:
B=B\cap(A\cup B)=B\cap(A\cup C) [using A\cup B=A\cup C]
=(B\cap A)\cup(B\cap C) [distributive law]
=(A\cap B)\cup(B\cap C)=(A\cap C)\cup(B\cap C) [using A\cap B=A\cap C]
=C\cap(A\cup B) [distributive law, reversed]
=C\cap(A\cup C)=C [using A\cup B=A\cup C, then since C\subseteq A\cup C]
To show all four are equivalent, it suffices to prove the cycle (i)\Rightarrow(ii)\Rightarrow(iii)\Rightarrow(iv)\Rightarrow(i).
If A\subset B, every element of A lies in B, so there is no element of A outside B. Hence A-B=\phi.
If A-B=\phi, no element of A lies outside B, so A\subseteq B. Then A\cup B adds nothing new to B, so A\cup B=B.
If A\cup B=B, then since A\subseteq A\cup B=B, we get A\subseteq B. For any subset A of B, A\cap B=A.
If A\cap B=A, then every element of A is an element of A\cap B, hence an element of B. So A\subset B.
Let x\in C-B. Then x\in C and x\notin B.
Suppose, for contradiction, that x\in A. Since A\subset B, this would force x\in B — contradicting x\notin B. So x\notin A.
Thus x\in C and x\notin A, i.e., x\in C-A. As x was an arbitrary element of C-B, every such element also lies in C-A.
Let x\in A. Either x\in B or x\notin B. If x\in B, then x\in A\cap B; if x\notin B, then x\in A-B. Either way, x\in(A\cap B)\cup(A-B), so A\subseteq(A\cap B)\cup(A-B).
Conversely, both A\cap B and A-B consist entirely of elements of A, so (A\cap B)\cup(A-B)\subseteq A.
Let x\in A\cup(B-A). Then x\in A, or x\in B-A (so x\in B). Either way x\in A\cup B, giving A\cup(B-A)\subseteq A\cup B.
Conversely, let x\in A\cup B. If x\in A, then trivially x\in A\cup(B-A). If x\notin A, then since x\in A\cup B, we must have x\in B; combined with x\notin A, this gives x\in B-A, so again x\in A\cup(B-A). Hence A\cup B\subseteq A\cup(B-A).
Since A\cap B\subseteq A, taking the union of A with one of its own subsets adds no new elements. Formally, A\subseteq A\cup(A\cap B) always holds; and for the reverse, any x\in A\cup(A\cap B) is either in A directly, or in A\cap B\subseteq A — either way x\in A.
Since A\subseteq A\cup B, we get A\cap(A\cup B)\supseteq A\cap A=A. Also, an intersection can never contain more than either of its sets, so A\cap(A\cup B)\subseteq A.
Take A=\{1,2\}, B=\{2,3\}, C=\{2,4\}.
Then A\cap B=\{2\} and A\cap C=\{2\}, so A\cap B=A\cap C.
But B=\{2,3\}\neq\{2,4\}=C.
Since A\subseteq A\cup X, we have A=A\cap(A\cup X). Substituting the given A\cup X=B\cup X:
A=A\cap(B\cup X)=(A\cap B)\cup(A\cap X) [distributive law]
=(A\cap B)\cup\phi=A\cap B [using the given A\cap X=\phi]
So A=A\cap B, which means A\subseteq B.
By an identical argument starting from B=B\cap(B\cup X)=B\cap(A\cup X)=(A\cap B)\cup(B\cap X)=(A\cap B)\cup\phi=A\cap B, we get B=A\cap B, which means B\subseteq A.
We need each pair of sets to overlap, but no single element common to all three. A convenient construction: let each element be shared by exactly two of the three sets.
Take A=\{1,2\}, B=\{2,3\}, C=\{1,3\}.
Then A\cap B=\{2\} (non-empty), B\cap C=\{3\} (non-empty), and A\cap C=\{1\} (non-empty).
Checking the triple intersection: element 1 is in A and C but not B; element 2 is in A and B but not C; element 3 is in B and C but not A. No element belongs to all three sets simultaneously.
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