April 2026

Define Bijections from \(A\) to \(B\) and Find Their Inverses 📝 Question Let: \[ A=\{1,2,3,4\}, \quad B=\{a,b,c,d\} \] Define any four bijections from \(A\) to \(B\). Also find their inverse functions. ✅ Solution A function is bijective if it is both one-one and onto. :contentReference[oaicite:0]{index=0} 🔹 Bijection 1 \[ f_1=\{(1,a),(2,b),(3,c),(4,d)\} \] Inverse: \[ f_1^{-1}=\{(a,1),(b,2),(c,3),(d,4)\} \] […]

Is \(f(x)=\cos(x+2)\) Invertible? 📝 Question Let: \[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=\cos(x+2) \] Is \(f\) invertible? Justify your answer. ✅ Solution 🔹 Step 1: Check whether \(f\) is one-one Assume: \[ f(x_1)=f(x_2) \] \[ \cos(x_1+2)=\cos(x_2+2) \] Using identity: \[ \cos A=\cos B \Rightarrow A=B \ \text{or} \ A=2\pi-B \] So, \[ x_1+2=x_2+2 \quad \text{or} \quad x_1+2=2\pi-(x_2+2) \]

Show \(g^{-1}\) Exists but \(f^{-1}\) Does Not Exist and Find \(g^{-1}\) 📝 Question Let: \[ A=\{x\in \mathbb{R} \mid -1 \le x \le 1\} \] Define functions: \[ f:A\to A,\quad f(x)=x^2 \] \[ g:A\to A,\quad g(x)=\sin\left(\frac{\pi x}{2}\right) \] Show that \(g^{-1}\) exists but \(f^{-1}\) does not exist. Also find \(g^{-1}\). ✅ Solution 🔹 Step 1: Check

Show \(f(x)=(x+1)^2-1\) is Invertible and Find Set \(S\) 📝 Question Let: \[ f:[-1,\infty)\to[-1,\infty), \quad f(x)=(x+1)^2-1 \] Show that \(f\) is invertible. Also find the set: \[ S=\{x: f(x)=f^{-1}(x)\} \] ✅ Solution 🔹 Step 1: Prove that \(f\) is one-one The function \(f(x)=(x+1)^2-1\) is increasing on \([-1,\infty)\). Hence, \(f\) is one-one (injective). 🔹 Step 2: Range

Show \(f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}\) is Invertible and Find \(f^{-1}\) 📝 Question Let: \[ f:\mathbb{R}\to(-1,1), \quad f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \] Show that \(f\) is invertible and find \(f^{-1}\). ✅ Solution 🔹 Step 1: Prove that \(f\) is one-one The function is strictly increasing since exponential functions are increasing. Hence, \(f\) is one-one (injective). 🔹 Step 2: Range of the function

Show \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is Invertible and Find \(f^{-1}\) 📝 Question Let: \[ f:\mathbb{R}\to(-1,1), \quad f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}} \] Show that \(f\) is invertible and find \(f^{-1}\). ✅ Solution 🔹 Step 1: Prove that \(f\) is one-one The function is strictly increasing because exponential functions are increasing. Hence, \(f\) is one-one (injective). 🔹 Step 2: Range of the function

Show \(f(x)=\frac{4x}{3x+4}\) is Invertible and Find \(f^{-1}\) 📝 Question Let: \[ f:\mathbb{R}\setminus\left\{-\frac{4}{3}\right\}\to \mathbb{R}, \quad f(x)=\frac{4x}{3x+4} \] Show that \(f:\mathbb{R}\setminus\{-\frac{4}{3}\}\to \text{Range}(f)\) is one-one and onto. Hence, find \(f^{-1}\). ✅ Solution 🔹 Step 1: Prove that \(f\) is one-one Assume: \[ f(x_1)=f(x_2) \] \[ \frac{4x_1}{3x_1+4}=\frac{4x_2}{3x_2+4} \] Cross-multiplying: \[ 4x_1(3x_2+4)=4x_2(3x_1+4) \] \[ 12x_1x_2+16x_1=12x_1x_2+16x_2 \] \[ 16x_1=16x_2 \Rightarrow x_1=x_2

Show \(f(x)=9x^2+6x-5\) is Invertible on \(\mathbb{N}\to S\) and Find \(f^{-1}\) 📺 Video Explanation 📝 Question Let: \[ f:\mathbb{N}\to\mathbb{N},\qquad f(x)=9x^2+6x-5 \] Let \(S\) be the range of \(f\). Show that: \[ f:\mathbb{N}\to S \] is invertible. Find: \[ f^{-1} \] Also find: \[ f^{-1}(43),\qquad f^{-1}(163) \] ✅ Solution 🔹 Step 1: Show that \(f:\mathbb{N}\to S\) is

Show \(f(x)=5x^2+6x-9\) is Invertible on \(\mathbb{R}_+\) and Find \(f^{-1}\) 📺 Video Explanation 📝 Question Let: \[ f:\mathbb{R}_+\to[-9,\infty),\qquad f(x)=5x^2+6x-9 \] where \(\mathbb{R}_+\) denotes the set of all non-negative real numbers. Prove that \(f\) is invertible and: \[ f^{-1}(y)=\frac{\sqrt{54+5y}-3}{5} \] ✅ Solution 🔹 Step 1: Show that \(f\) is one-one Take: \[ x_1,x_2\in\mathbb{R}_+ \] Assume: \[ f(x_1)=f(x_2)

Show \(f(x)=\frac{x-2}{x-3}\) is Bijective and Find \(f^{-1}\) 📺 Video Explanation 📝 Question Let: \[ A=\mathbb{R}\setminus\{3\},\qquad B=\mathbb{R}\setminus\{1\} \] and: \[ f:A\to B,\qquad f(x)=\frac{x-2}{x-3} \] Show that \(f\) is one-one and onto, and hence find: \[ f^{-1}(x) \] ✅ Solution 🔹 Step 1: Show that \(f\) is one-one Assume: \[ f(x_1)=f(x_2) \] Then: \[ \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \] Cross