April 2026

Examples of Two Surjective Functions Whose Sum is Not Surjective 📺 Video Explanation 📝 Question Give examples of two surjective functions: \[ f_1,f_2:\mathbb{Z}\to\mathbb{Z} \] such that: \[ (f_1+f_2)(x)=f_1(x)+f_2(x) \] is not surjective. ✅ Solution Take: \[ f_1(x)=x \] and: \[ f_2(x)=x \] 🔹 Check \(f_1\) For any: \[ y\in\mathbb{Z} \] choose: \[ x=y \] Then: […]

Examples of Two One-One Functions Whose Sum is Not One-One 📺 Video Explanation 📝 Question Give examples of two one-one functions: \[ f_1,f_2:\mathbb{R}\to\mathbb{R} \] such that: \[ (f_1+f_2)(x)=f_1(x)+f_2(x) \] is not one-one. ✅ Solution Take: \[ f_1(x)=x \] and: \[ f_2(x)=-x \] 🔹 Check \(f_1\) Function: \[ f_1(x)=x \] is one-one because different inputs give

Number of Onto Functions from \(A=\{1,2,\dots,n\}\) to Itself 📺 Video Explanation 📝 Question Find the number of all onto functions: \[ f:A\to A \] where: \[ A=\{1,2,3,\dots,n\} \] ✅ Solution Set \(A\) has: \[ n \] elements. 🔹 Key Fact For finite sets with same number of elements: \[ \text{onto} \iff \text{one-one} \] So every

Show Onto Function on Finite Set is One-One 📺 Video Explanation 📝 Question Let: \[ A=\{1,2,3\} \] Show that if: \[ f:A\to A \] is onto, then \(f\) must be one-one. ✅ Solution Set: \[ A=\{1,2,3\} \] has 3 elements. 🔹 Onto Property Since function is onto: Every element of codomain has pre-image. So all:

Show One-One Function on Finite Set is Onto 📺 Video Explanation 📝 Question Let: \[ A=\{1,2,3\} \] Show that if: \[ f:A\to A \] is one-one, then \(f\) must be onto. ✅ Solution Set \(A\) has: \[ 3 \] elements. 🔹 One-One Property If function is one-one: Different inputs have different outputs. So images of:

Prove \(f(x)=\log_a x\) is a Bijection 📺 Video Explanation 📝 Question Show that: \[ f:\mathbb{R}_0^+\to\mathbb{R},\quad f(x)=\log_a x \] where: \[ a>0,\ a\neq1 \] is a bijection. ✅ Solution 🔹 Step 1: Prove One-One (Injective) Assume: \[ f(x_1)=f(x_2) \] Then: \[ \log_a x_1=\log_a x_2 \] Therefore: \[ x_1=x_2 \] ✔ Hence, one-one. 🔹 Step 2: Prove

Show \(f(x)=e^x\) is One-One but Not Onto 📺 Video Explanation 📝 Question Show that: \[ f:\mathbb{R}\to\mathbb{R},\quad f(x)=e^x \] is one-one but not onto. Also discuss what happens if codomain is: \[ \mathbb{R}_0^+=\{y\in\mathbb{R}:y>0\} \] ✅ Solution 🔹 Step 1: Prove One-One (Injective) Assume: \[ f(x_1)=f(x_2) \] Then: \[ e^{x_1}=e^{x_2} \] Taking logarithm: \[ x_1=x_2 \] ✔

Prove \(f(x)=4x^3+7\) is a Bijection 📺 Video Explanation 📝 Question Show that: \[ f:\mathbb{R}\to\mathbb{R},\quad f(x)=4x^3+7 \] is a bijection. ✅ Solution 🔹 Step 1: Prove One-One (Injective) Assume: \[ f(x_1)=f(x_2) \] Then: \[ 4x_1^3+7=4x_2^3+7 \] Simplify: \[ x_1^3=x_2^3 \] Since cube function is strictly increasing: \[ x_1=x_2 \] ✔ Hence, \(f\) is one-one. 🔹 Step

All One-One Functions from \(A=\{1,2,3\}\) to Itself 📺 Video Explanation 📝 Question Let: \[ A=\{1,2,3\} \] Write all one-one functions from \(A\) to itself. ✅ Solution Since domain and codomain both have 3 elements, every one-one function is a permutation of \(\{1,2,3\}\). Total number of one-one functions: \[ 3!=6 \] 🔹 All One-One Functions 1.

Is Person to Ancestor Relation a Function? 📺 Video Explanation 📝 Question Consider the relation: \[ \{(a,b): a \text{ is a person, } b \text{ is an ancestor of } a\} \] Check: Is it a function? If yes, is it injective? Is it surjective? ✅ Solution 🔹 Is it a Function? A relation is