Telangana Open School (TOSS) Class 12 Mathematics — Flashcards

A: Step 1: Base case — For $n=1$: LHS = 1, RHS = $\frac{1(1+1)}{2} = 1$. So, true for $n=1$.<br>Step 2: Inductive hypothesis — Assume true for $n=k$: $1+2+\dots+k = \frac{k(k+1)}{2}$.<br>Step 3: Prove fo

A: Step 1: Base case — $n=1$: LHS = $1^2 = 1$, RHS = $\frac{1(2)(3)}{6} = 1$. True.<br>Step 2: Assume true for $n=k$: $\sum k^2 = \frac{k(k+1)(2k+1)}{6}$.<br>Step 3: For $n=k+1$: Add $(k+1)^2$ to both si

A: Step 1: $n=1$: LHS = $1^3 = 1$, RHS = $\left(\frac{1\cdot2}{2}\right)^2 = 1$. True.<br>Step 2: Assume true for $n=k$.<br>Step 3: For $n=k+1$: Add $(k+1)^3$ to both sides:<br>LHS becomes $\left(\frac{k

A: Step 1: Form all ordered pairs where first element is from A and second from B. Step 2: (1,3), (1,4), (2,3), (2,4). Answer: A × B = {(1,3), (1,4), (2,3), (2,4)}

A: Step 1: Domain = set of first elements → {1,2,3}. Step 2: Range = set of second elements → {2,3,4}. Answer: Domain = {1,2,3}, Range = {2,3,4}

A: Step 1: Check if any input has more than one output. Step 2: Input '1' maps to both 2 and 4. Step 3: A function must have unique output for each input. Answer: No, because 1 has two images.

A: Step 1: $\sqrt{-48} = \sqrt{48 \cdot (-1)} = \sqrt{48} \cdot \sqrt{-1}$ Step 2: $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$ Step 3: $\sqrt{-1} = i$ Therefore, $\sqrt{-48} = 4\sqrt{3}i$

A: Step 1: $\sqrt{-5} = \sqrt{5}i$, $\sqrt{-20} = \sqrt{20}i = 2\sqrt{5}i$ Step 2: Multiply: $(\sqrt{5}i)(2\sqrt{5}i) = 2 \cdot 5 \cdot i^2 = 10 \cdot (-1) = -10$ Note: $\sqrt{a} \cdot \sqrt{b} = \sqrt

A: Step 1: Divide 27 by 4 → Quotient = 6, Remainder = 3 Step 2: $i^{27} = i^{4\cdot6 + 3} = (i^4)^6 \cdot i^3 = (1)^6 \cdot i^3 = i^3$ Step 3: $i^3 = i^2 \cdot i = (-1) \cdot i = -i$ Answer: $-i$