Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 12 Maths NCERT Solutions for the Chapter 1 Miscellaneous Exercise — all 7 questions solved, bringing together injective/onto proofs, equivalence relations on a power set, and two classic counting problems.
Questions 1 and 2 are harder one-one/onto proofs than anything in Exercise 1.2 — Q1 needs a full case split by sign, and Q2 needs factoring a cubic. Question 3 is a conceptual trap: the subset relation "looks" like it should be an equivalence relation but fails symmetry, which is exactly the kind of question CBSE likes to test understanding (not just mechanics) with. Questions 4 and 5 are a counting result and a direct equality check, and the exercise closes with two of the most commonly asked MCQs in this entire chapter — counting how many relations or equivalence relations satisfy a given constraint, which trips up students who don't reason through why certain extensions are impossible.
Write the function piecewise: f(x)=\frac{x}{1+x} for x\ge0, and f(x)=\frac{x}{1-x} for x \lt 0. Within either piece, the same algebra used in Exercise 1.2 Q1 gives one-one directly: \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}\Rightarrow x_1=x_2 (and similarly for the negative piece).
For mixed signs, if x_1\ge0 and x_2 \lt 0, then f(x_1)\ge0 while f(x_2) \lt 0 — they can never coincide except trivially at 0, already covered. So f is one-one throughout.
For onto, given 0\le y \lt 1, solving y=\frac{x}{1+x} for x\ge0 gives x=\frac{y}{1-y}\ge0. Given -1 \lt y \lt 0, solving y=\frac{x}{1-x} for x \lt 0 gives x=\frac{y}{1+y} \lt 0. Every y\in(-1,1) has a valid pre-image.
If f(x_1)=f(x_2), then x_1^3-x_2^3=0, which factors as (x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0. The second factor rewrites as \left(x_1+\frac{x_2}{2}\right)^2+\frac{3x_2^2}{4} — a sum of squares, so it's zero only when x_1=x_2=0.
Either the second factor is strictly positive (forcing x_1=x_2 from the first factor), or both variables are already zero (also giving x_1=x_2).
Every set is a subset of itself — reflexive.
Taking X=\{1,2\}, A=\{1\}, B=\{1,2\}: A\subset B holds, but B\subset A fails since 2\in B but 2\notin A — not symmetric.
If A\subset B and B\subset C, then A\subset C — transitive.
For a function f:X\to X on a finite set, onto is equivalent to one-one (the chapter's own characteristic property of finite sets). So counting onto functions on \{1,\ldots,n\} is the same as counting one-one functions — which are exactly the permutations of the n elements.
Two functions on the same domain are equal exactly when they agree at every point, so check all four elements of A.
At x=-1: f(-1)=1+1=2, g(-1)=2(\tfrac32)-1=2 — match.
At x=0: f(0)=0, g(0)=2(\tfrac12)-1=0 — match.
At x=1: f(1)=0, g(1)=2(\tfrac12)-1=0 — match.
At x=2: f(2)=2, g(2)=2(\tfrac32)-1=2 — match.
Reflexivity forces (1,1),(2,2),(3,3) in; including (1,2) and (1,3) forces their reverses (2,1),(3,1) too (symmetry). The smallest such relation, R_1, already fails transitivity on its own — (2,1),(1,3)\in R_1 would need (2,3), which is missing.
The only pairs left to add are (2,3) and (3,2). Adding just one breaks symmetry; adding both gives the full universal relation, which — containing every pair — is automatically transitive, so it no longer qualifies as "not transitive." There's no way to extend R_1 while keeping it non-transitive.
Reflexivity and symmetry force (1,1),(2,2),(3,3),(2,1) in alongside (1,2); this smallest relation, R_1, is already transitive — it corresponds to the partition \{1,2\},\{3\}.
Every equivalence relation is entirely determined by how it partitions A, and any relation containing (1,2) must keep 1 and 2 in the same class. There are only two such partitions: \{1,2\},\{3\} (giving R_1) or all of \{1,2,3\} together (the universal relation).
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