Ravi Kant Kumar

Find \(f^{-1}\) for \(f(x)=3x\) on Given Finite Sets 📺 Video Explanation 📝 Question Let: \[ A=\{0,-1,-3,2\},\qquad B=\{-9,-3,0,6\} \] and: \[ f:A\to B,\qquad f(x)=3x \] Find: \[ f^{-1} \] ✅ Solution 🔹 Step 1: Find image of each element of \(A\) Using: \[ f(x)=3x \] We get: \(f(0)=0\) \(f(-1)=-3\) \(f(-3)=-9\) \(f(2)=6\) So: \[ f=\{(0,0),(-1,-3),(-3,-9),(2,6)\} \] 🔹 […]

Check Whether the Given Function Has an Inverse 📺 Video Explanation 📝 Question State with reasons whether the following function has inverse: \[ h:\{2,3,4,5\}\to\{7,9,11,13\} \] defined by: \[ h=\{(2,7),(3,9),(4,11),(5,13)\} \] ✅ Solution 🔹 Condition for inverse function A function has an inverse if and only if it is bijective. 🔹 Check one-one property Given: \[

Check Whether the Given Function Has an Inverse 📺 Video Explanation 📝 Question State with reasons whether the following function has inverse: \[ g:\{5,6,7,8\}\to\{1,2,3,4\} \] defined by: \[ g=\{(5,4),(6,3),(7,4),(8,2)\} \] ✅ Solution 🔹 Condition for inverse function A function has an inverse if and only if it is bijective. That means: one-one onto 🔹 Check

Check Whether the Given Function Has an Inverse 📺 Video Explanation 📝 Question State with reasons whether the following function has inverse: \[ f:\{1,2,3,4\}\to\{10\} \] defined by: \[ f=\{(1,10),(2,10),(3,10),(4,10)\} \] ✅ Solution 🔹 Condition for inverse function A function has an inverse if and only if it is: one-one (injective), and onto (surjective) That is,

Find \(f \circ g\) and \(g \circ f\) for Given Absolute Value Functions 📺 Video Explanation 📝 Question Let: \[ f(x)=|x|+x,\qquad g(x)=|x|-x,\qquad x\in\mathbb{R} \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) Also find: \[ (f\circ g)(-3),\quad (f\circ g)(5),\quad (g\circ f)(-2) \] ✅ Solution 🔹 Step 1: Simplify \(f(x)\) and \(g(x)\) Using modulus: For \(x\ge0\): \[ |x|=x \]

Find \(f \circ f\) for Given Piecewise Function 📺 Video Explanation 📝 Question Let: \[ f(x)= \begin{cases} 1+x, & 0\le x\le 2 \\[4pt] 3-x, & 2

Find \((f \circ f)\), \((f \circ f \circ f)\), \((f \circ f \circ f)(38)\), and \(f^2\) for \(f(x)=\sqrt{x-2}\) 📺 Video Explanation 📝 Question Let: \[ f(x)=\sqrt{x-2} \] Find: \((f\circ f)(x)\) \((f\circ f\circ f)(x)\) \((f\circ f\circ f)(38)\) \(f^2(x)\) Also, show that: \[ f\circ f\ne f^2 \] ✅ Solution 🔹 (i) Find \((f\circ f)(x)\) By definition: \[

Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\sqrt{x+3}\) and \(g(x)=x^2+1\) 📺 Video Explanation 📝 Question Let: \[ f(x)=\sqrt{x+3},\qquad g(x)=x^2+1 \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) ✅ Solution 🔹 Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute \(g(x)=x^2+1\): \[ (f\circ g)(x)=f(x^2+1) \] Since: \[ f(x)=\sqrt{x+3} \] So: \[ (f\circ g)(x)=\sqrt{(x^2+1)+3} \] \[

Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\tan x\) and \(g(x)=\sqrt{1-x^2}\) 📺 Video Explanation 📝 Question Let: \[ f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to\mathbb{R},\qquad f(x)=\tan x \] and \[ g:[-1,1]\to\mathbb{R},\qquad g(x)=\sqrt{1-x^2} \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) ✅ Solution 🔹 Step 1: Find range of \(g(x)\) Given: \[ g(x)=\sqrt{1-x^2},\qquad x\in[-1,1] \] Since: \[ 0\le 1-x^2\le 1 \] So:

Find \(f \circ g\) and \(g \circ f\) for \(f(x)=\sqrt{1-x}\) and \(g(x)=\ln x\) 📺 Video Explanation 📝 Question Let functions \(f\) and \(g\) be defined as: \[ f(x)=\sqrt{1-x},\qquad g(x)=\ln x \] Find: \((f\circ g)(x)\) \((g\circ f)(x)\) ✅ Solution 🔹 Find \((f\circ g)(x)\) By definition: \[ (f\circ g)(x)=f(g(x)) \] Substitute \(g(x)=\ln x\): \[ (f\circ g)(x)=f(\ln x)=\sqrt{1-\ln