Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question Let A A be a 3 × 3 3 \times 3 3 × 3 matrix such that ∣ adj ⁡ ( adj ⁡ ( adj ⁡ A ) ) ∣ = 81. If S = { n ∈ Z : ∣ adj ⁡ ( adj ⁡ A ) ∣ ( n − 1 ) 2 2 = ∣ A ∣   3 n 2 − 5 n − 4 } , then ∑ n ∈ S ∣ A ∣   n 2 + n is equal to ? 🖼️ Question Image ✍️ Short Solution We use two standard facts for an invertible n × n n \times n  matrix A A : det ⁡ ( adj ⁡ A ) = ( det ⁡ A ) n − 1 \det(\operatorname{adj} A) = (\det A)^{n-1} For 3×3, n = 3 ⇒ det ⁡ ( adj ⁡ A ) = ∣ A ∣ 2 n = 3 \Rightarrow \det(\operatorname{adj}A) = |A|^2 From ∣ adj ⁡ ( adj ⁡ ( adj ⁡ A ) ) ∣ = 81 |\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 81 , we express everything in terms of ∣ A ∣ |A| , then solve the given equation for integers n n . Finally we compute ∑ n ∈ S ∣ A ∣ n 2 + n \sum_{n\in S} |A|^{n^2+n} 🔹 Step 1 — Express all determinants in terms of ∣ A ∣ |A| Let ∣ A ∣ = D . For a 3×3 matrix: First adjoint: B 1 = adj ⁡ A , ∣ B 1 ∣ = ∣ A ∣ 2 = D 2 . Second adjoint...

  ❓ Question A basic buffer contains: NH 3 = 0.10  mol , NH 4 + = 0.10  mol Strong acid added: HCl = 0.05  mol Given: p K b ( NH 3 ) = 4.745 Find the change in pH after adding HCl. ✍️ Short Solution This is a weak base buffer (NH₃–NH₄Cl). When HCl is added, it reacts completely with NH₃, converting it into NH₄⁺. We then apply the Henderson–Hasselbalch equation for basic buffers to find the new pH. 🔹 Step 1 — Buffer System NH 3 + NH 4 + \text{NH}_3 + \text{NH}_4^+ Both initially = 0.10 mol → Perfect weak base buffer. 🔹 Step 2 — Henderson–Hasselbalch Equation (Basic Buffer) pOH = p K b + log ⁡ ( salt base ) \text{pOH} = pK_b + \log\left(\frac{\text{salt}}{\text{base}}\right) pH = 14 − pOH \text{pH} = 14 - \text{pOH} Given: p K b = 4.745 🔹 Step 3 — Reaction With Strong Acid HCl reacts fully with NH₃: NH 3 + HCl → NH 4 + \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ HCl added = 0.05 mol New moles: NH 3 = 0.10 − 0.05 = 0.05  mol \text...

  ❓ Question FOR: If for θ ∈ [ − π 3 ,   0 ] , ( x , y ) = ( 3 tan ⁡  ⁣ ( θ + π 3 ) ,   2 tan ⁡  ⁣ ( θ + π 6 ) ) lie on the curve x y + α x + β y + γ = 0 , xy + \alpha x + \beta y + \gamma = 0, then the value of α 2 + β 2 + γ 2 \alpha^2 + \beta^2 + \gamma^2 is equal to ? 🖼️ Question Image ✍️ Short Solution We are told that for every θ \theta  in the interval, x = 3 tan ⁡ ( θ + π 3 ) , y = 2 tan ⁡ ( θ + π 6 ) x = 3\tan\left(\theta + \frac{\pi}{3}\right), \quad y = 2\tan\left(\theta + \frac{\pi}{6}\right) always satisfies x y + α x + β y + γ = 0. xy + \alpha x + \beta y + \gamma = 0. That means there is a fixed relation between x x x and y y y which does not involve θ \theta . We’ll eliminate θ \theta  using a standard tangent identity. 🔹 Step 1 — Define new angles and variables Let A = θ + π 3 , B = θ + π 6 . A = \theta + \frac{\pi}{3}, \quad B = \theta + \frac{\pi}{6}. Then, x = 3 tan ⁡ A , y = 2 tan ⁡ B . x = 3\tan A,\quad y = 2\tan B. Compu...

  ❓ Question Dumas Method – Nitrogen % in 60 Sec! A quick, exam-oriented breakdown of how Dumas Method helps you find percentage of nitrogen in an organic compound using only the nitrogen gas collected . 📝 1️⃣ Dumas Method Concept Organic compound is heated strongly → Nitrogen present in compound converts into N₂ gas → This N₂ is collected over water and its volume, temperature, pressure are measured. ✔ No chemical titration ✔ 100% physical calculation ✔ Based on ideal gas equation 📝 2️⃣ Correcting the Pressure (MOST Important Step) Gas is collected over water , so total pressure = Pressure of dry N₂ + water vapour pressure. Corrected pressure: P N 2 = P total − P aq. tension​ Given: P total = 715  mmHg , P aq. tension = 15  mmHg So, P N 2 = 700  mmHg 📝 3️⃣ Ideal Gas Equation (Find moles of N₂) P V = n R T Convert units: P = 700  mmHg = 700 / 760  atm P = 700\ \text{mmHg} = 700/760\ \text{atm} V = 50  mL = 0.0...

  ❓ Question Let the set of all values of p ∈ R p \in \mathbb{R}  for which both the roots of the equation x 2 − ( p + 2 ) x + ( 2 p + 9 ) = 0 are negative real numbers be the interval ( α , β ] (\alpha, \beta] . Then the value of β − 2 α is equal to ? 🖼️ Question Image ✍️ Short Solution We have a quadratic: x 2 − ( p + 2 ) x + ( 2 p + 9 ) = 0 Compare with x 2 − S x + P = 0 x^2 - Sx + P = 0 , where Sum of roots = S = p + 2 = S = p+2 Product of roots = P = 2 p + 9 = P = 2p+9 For both roots to be negative real numbers , we need: Real roots → Discriminant Δ ≥ 0 Both roots negative → Sum of roots < 0 <0 Product of roots > 0 >0 🔹 Step 1 — Conditions from sum and product Sum < 0: p + 2 < 0 ⇒ p < − 2 Product > 0: 2 p + 9 > 0 ⇒ p > − 9 2​ Together: − 9 2 < p < − 2 🔹 Step 2 — Discriminant condition Δ = ( p + 2 ) 2 − 4 ( 2 p + 9 ) ≥ 0 Compute: Δ = ( p 2 + 4 p + 4 ) − ( 8 p + 36 ) = p 2 − 4 p − 32 So...