Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 12 Maths NCERT Solutions for Chapter 1 Ex 1.2 — all 12 questions solved, checking functions for one-one (injective) and onto (surjective) behaviour, and identifying bijective functions.
Questions 1–5 build the core intuition using standard functions — reciprocal, power, greatest integer, modulus and signum — showing how the same domain can behave completely differently depending on which set you're mapping to or from (Q1 in particular shows the same function turning from bijective to merely injective just by shrinking the domain from ℝ* to ℕ). Questions 6–10 apply this to less familiar functions, including a rational function (Q10) where proving "onto" takes real algebraic work, before two MCQs close the exercise.
If f(x_1)=f(x_2), then \frac{1}{x_1}=\frac{1}{x_2}\Rightarrow x_1=x_2, so f is one-one.
Given any y\in\mathbb{R}_*, taking x=\frac{1}{y} (a valid non-zero real since y\neq0) gives f(x)=y, so f is onto.
The one-one argument still holds for f:\mathbb{N}\to\mathbb{R}_*. But onto fails: taking y=2\in\mathbb{R}_* would need x=\frac12\notin\mathbb{N}.
Injective on the positive naturals (x_1^2=x_2^2\Rightarrow x_1=x_2), but not surjective — 2 has no natural square root.
Not injective (f(-1)=1=f(1)), not surjective (2 has no integer square root).
Not injective (f(-1)=1=f(1)), not surjective (−1 has no real square root, since x^2\ge0 always).
Injective (cubing is strictly increasing on the naturals), not surjective — 2 has no natural cube root.
Injective (cubing is strictly increasing over all of ℤ), not surjective — 2 has no integer cube root.
f(1.2)=1=f(1.9) even though 1.2\neq1.9 — not one-one. Since [x] is always an integer, a value like y=0.5\in\mathbb{R} is never produced — not onto.
f(-1)=1=f(1) — not one-one. Since |x|\ge0 always, a negative value like -1\in\mathbb{R} is never produced — not onto.
f(1)=1=f(2), both positive give the same output — not one-one. The only possible outputs are \{-1,0,1\}, so a value like 2\in\mathbb{R} is never produced — not onto.
The three distinct inputs 1,2,3 map to the three distinct outputs 4,5,6 — no two different inputs share an output.
f(x_1)=f(x_2)\Rightarrow3-4x_1=3-4x_2\Rightarrow x_1=x_2 — one-one. Given any y\in\mathbb{R}, x=\frac{3-y}{4}\in\mathbb{R} solves f(x)=y — onto.
f(-1)=2=f(1) — not one-one. Range is [1,\infty) since x^2\ge0, so a value like -1 is never attained — not onto.
If f(a_1,b_1)=f(a_2,b_2), then (b_1,a_1)=(b_2,a_2)\Rightarrow a_1=a_2,\ b_1=b_2 — one-one.
Given any (b,a)\in B\times A, taking (a,b)\in A\times B gives f(a,b)=(b,a) — onto.
f(1)=\frac{1+1}{2}=1 (odd rule), and f(2)=\frac22=1 (even rule). So f(1)=f(2)=1 while 1\neq2 — f is not one-one.
If f(x_1)=f(x_2), cross-multiplying (x_1-2)(x_2-3)=(x_2-2)(x_1-3) and expanding both sides cancels x_1x_2 and the constant terms, leaving -3x_1-2x_2=-3x_2-2x_1\Rightarrow x_1=x_2 — one-one.
For onto, given y\in B (so y\neq1), solving y=\frac{x-2}{x-3} gives x=\frac{3y-2}{y-1}. Checking whether this x could equal 3: 3=\frac{3y-2}{y-1}\Rightarrow3y-3=3y-2\Rightarrow-3=-2, a contradiction — so x is never 3, meaning it's always a valid element of A.
f(-1)=1=f(1) — many-one, not one-one. Since x^4\ge0 always, negative values like -1 are never attained — not onto.
f(x_1)=f(x_2)\Rightarrow3x_1=3x_2\Rightarrow x_1=x_2 — one-one. Given any y\in\mathbb{R}, x=\frac{y}{3} gives f(x)=y — onto.
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