Educational

Check Function Type Check Injective / Surjective / Bijective 🎥 Video Explanation 📝 Question Given \( f:\mathbb{R} \to \mathbb{R} \), \[ f(x) = x + \sqrt{x^2} \] Find whether the function is: A. Injective B. Surjective C. Bijective D. None of these ✅ Solution 🔹 Step 1: Simplify \[ \sqrt{x^2} = |x| \] So, \[ […]

Function from Relation x^2 + y^2 = 1 Check Which Relation Defines a Function 🎥 Video Explanation 📝 Question Let \(A = B = \{x \in \mathbb{R} : -1 \le x \le 1\}\) and \(C = \{x \in \mathbb{R} : x \ge 0\}\). \[ S = \{(x, y) \in A \times B : x^2 +

Find \(f^{-1}(x)\) 📝 Question Let: \[ f(x)=4-(x-7)^3 \] Find \(f^{-1}(x)\). ✅ Solution 🔹 Step 1: Let \[ y=4-(x-7)^3 \] — 🔹 Step 2: Solve for \(x\) \[ (x-7)^3=4-y \] Take cube root: :contentReference[oaicite:0]{index=0} \[ x=7+\sqrt[3]{4-y} \] — 🔹 Step 3: Write inverse Interchange \(x\) and \(y\): \[ f^{-1}(x)=7+\sqrt[3]{4-x} \] — 🎯 Final Answer \[ \boxed{f^{-1}(x)=7+\sqrt[3]{4-x}}

Find \(\alpha\) and \(\beta\) 📝 Question Given: \[ g=\{(1,1),(2,3),(3,5),(4,7)\} \] and \(g(x)=\alpha x+\beta\). Find \(\alpha\) and \(\beta\). ✅ Solution 🔹 Step 1: Form equations Using \((1,1)\): \[ \alpha(1)+\beta=1 \] \[ \alpha+\beta=1 \quad …(1) \] Using \((2,3)\): \[ 2\alpha+\beta=3 \quad …(2) \] — 🔹 Step 2: Solve equations Subtract (1) from (2): :contentReference[oaicite:0]{index=0} Substitute into (1):

Find \(f\circ g\) 📝 Question Let: \[ f=\{(1,2),(3,5),(4,1)\} \] \[ g=\{(2,3),(5,1),(1,3)\} \] Find \(f\circ g\). ✅ Solution 🔹 Step 1: Definition \[ (f\circ g)(x)=f(g(x)) \] — 🔹 Step 2: Check domain compatibility Range of \(g = \{1,3\}\) Domain of \(f = \{1,3,4\}\) So composition is defined for inputs where \(g(x)\in\{1,3\}\). — 🔹 Step 3: Compute

Find \(g\circ f(x)\) 📝 Question Let: \[ f(x)=2x+1,\quad g(x)=x^2-2 \] Find \(g\circ f(x)\). ✅ Solution 🔹 Step 1: Use definition \[ (g\circ f)(x)=g(f(x)) \] — 🔹 Step 2: Substitute \(f(x)\) \[ g(f(x))=g(2x+1) \] — 🔹 Step 3: Apply function \(g\) \[ g(2x+1)=(2x+1)^2-2 \] — 🔹 Step 4: Expand :contentReference[oaicite:0]{index=0} — 🎯 Final Answer \[ \boxed{g\circ

Find \(f^{-1}\) 📝 Question Let: \[ A=\{a,b,c,d\} \] \[ f=\{(a,b),(b,d),(c,a),(d,c)\} \] Find \(f^{-1}\). ✅ Solution 🔹 Step 1: Definition of inverse The inverse of a function is obtained by reversing each ordered pair: \[ (x,y) \Rightarrow (y,x) \] — 🔹 Step 2: Reverse each pair \[ (a,b) \Rightarrow (b,a) \] \[ (b,d) \Rightarrow (d,b) \]

Find Domain of \(f(x)=\sqrt{25-x^2}\) 📝 Question Find the domain of the function: \[ f(x)=\sqrt{25-x^2} \] ✅ Solution 🔹 Step 1: Condition for square root For the function to be defined: \[ 25-x^2 \ge 0 \] — 🔹 Step 2: Solve inequality \[ x^2 \le 25 \] \[ -5 \le x \le 5 \] — 🎯

Which Relation is a Function? 📝 Question Let: \[ A=\{1,2,3\} \] Given relations: \[ f=\{(1,3),(2,3),(3,2)\} \] \[ g=\{(1,2),(1,3),(3,1)\} \] Determine which of these is a function. ✅ Solution 🔹 Step 1: Definition of function A relation is a function if every element of domain has exactly one image. — 🔹 Step 2: Check relation \(f\)

Find \(f^{-1}(x)\) for \(f(x)=4x-3\) 📝 Question Let: \[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=4x-3 \] Find \(f^{-1}(x)\). ✅ Solution 🔹 Step 1: Check invertibility The function is linear with non-zero slope (4). Hence, it is one-one and onto, so inverse exists. — 🔹 Step 2: Let \[ y=4x-3 \] Interchange \(x\) and \(y\): \[ x=4y-3 \] — 🔹