Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept Parameter-Based Quadratic Tricks (α, β) in 59 Sec Whenever you see a quadratic with p, k, λ in coefficients, panic mat karo . 👉 Parameter-based quadratics follow a fixed golden framework — once you apply it step-by-step, answers come automatically. 1️⃣ Quadratic with Parameter General form: a x 2 + b x + c = 0 ax^2 + bx + c = 0 Here, a ,   b ,   c depend on a parameter  p a,\ b,\ c \quad \text{depend on a parameter } p Roots are: α ,   β \alpha,\ \beta 👉 Goal in JEE questions: Find range of p such that roots satisfy a given condition. 2️⃣ Real Roots Condition (ALWAYS STEP 1) For real roots: D = b 2 − 4 a c ≥ 0 D = b^2 - 4ac \ge 0 📌 This step is non-negotiable . Most students forget it and lose marks. 👉 Solve: b 2 − 4 a c ≥ 0 ⇒ range of  p b^2 - 4ac \ge 0 \Rightarrow \text{range of } p 3️⃣ Sign of Roots (α, β) Use standard relations: α + β = − b a , α β = c a \alpha + \beta = -\frac{b}{a}, \quad \...

  ❓ Concept Integration Trick – tan⁻¹ Form in 60 Sec! Whenever you see an integral of the type ∫ sin ⁡ x a + b cos ⁡ 2 x   d x (or with extra terms in the numerator), understand one thing clearly 👇 👉 This is a pure tan⁻¹ game . ✍️ Short Explanation This type of integral is very common in JEE Main + Advanced . The key idea is: sin x dx is the derivative of cos x The denominator becomes a quadratic in cos x Final answer always involves tan⁻¹ 1️⃣ Step 1 — Identify the Pattern Integral of the form: ∫ (something) ⋅ sin ⁡ x a + b cos ⁡ 2 x   d x 👉 Always try substitution : u = cos ⁡ x u = \cos x d u = − sin ⁡ x   d x du = -\sin x\,dx This instantly simplifies the integral. 2️⃣ Step 2 — Apply Substitution From substitution: sin ⁡ x   d x = − d u \sin x\,dx = -du For definite integrals , limits change: x = 0 ⇒ u = cos ⁡ 0 = 1 x = 0 \Rightarrow u = \cos 0 = 1 x = π ⇒ u = cos ⁡ π = − 1 x = \pi \Rightarrow u = \cos \pi = -1 So the integral converts into: ...

  ❓ Question Evaluate the limit: lim ⁡ x → 0 + tan ⁡  ⁣ ( 5 x 1 / 3 )   ln ⁡  ⁣ ( 1 + 3 x 2 ) ( tan ⁡ − 1 ( 3 x ) ) 2   [   e 5 x 4 / 3 − 1   ]​ 🖼️ Question Image ✍️ Short Solution This is a standard JEE-type limit based on small-angle and small-x approximations . As x → 0 + x \to 0^+ , each function behaves like its first-order term . We’ll convert every function into its leading approximation and simplify. 🔹 Step 1 — Recall standard limits As t → 0 t \to 0 : tan ⁡ t ∼ t \tan t \sim t ln ⁡ ( 1 + t ) ∼ t \ln(1+t) \sim t tan ⁡ − 1 t ∼ t \tan^{-1} t \sim t e t − 1 ∼ t e^t - 1 \sim t These shortcuts are mandatory for fast JEE solving. 🔹 Step 2 — Apply approximations one by one 1️⃣ Tangent term tan ⁡ ( 5 x 1 / 3 ) ∼ 5 x 1 / 3 \tan(5x^{1/3}) \sim 5x^{1/3} 2️⃣ Logarithmic term ln ⁡ ( 1 + 3 x 2 ) ∼ 3 x 2 \ln(1 + 3x^2) \sim 3x^2 3️⃣ Inverse tangent term tan ⁡ − 1 ( 3 x ) ∼ 3 x So, ( tan ⁡ − 1 ( 3 x ) ) 2 ∼ ( 3 x ) 2 = 9 x 4️⃣ Exponential term e 5 x 4 / 3 − 1 ∼ ...

  ❓ Concept Goal: Count unpaired electrons in coordination complexes using a universal 3-step shortcut: 1️⃣ Oxidation state → 2️⃣ d-electron count → 3️⃣ Ligand strength → high/low spin → final unpaired electrons. This removes the need for drawing full MO diagrams every time. ✍️ Short Solution To find unpaired electrons in any complex: Oxidation state → d-electrons → High/Low spin \text{Oxidation state} \rightarrow \text{d-electrons} \rightarrow \text{High/Low spin} Everything depends on ligand strength (spectrochemical series). 1️⃣ Step 1 — Find the Oxidation State Use: Charge on complex = metal + ∑ ( ligand charges ) \text{Charge on complex} = \text{metal} + \sum(\text{ligand charges}) Then: d-electrons = atomic d-electrons − oxidation state \text{d-electrons} = \text{atomic d-electrons} - \text{oxidation state} Examples: Co (Z = 27) → atomic d = 9 Co³⁺ → d = 9 – 3 = 6 Fe (Z = 26) → atomic d = 8 Fe³⁺ → d = 8 – 3 = 5 Mn (Z...

  ❓ 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...