Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question If the area of the region bounded by the curves y = 4 − x 2 4 and y = x − 2 y = 4-\frac{x^2}{4} \quad\text{and}\quad y = x-2 is equal to α \alpha α , then the value of 6 α 6\alpha is equal to ? 🖼️ Question Image ✍️ Short Solution We will: ✔ Find points of intersection ✔ Write the top – bottom function ✔ Integrate between limits ✔ Multiply result by 6 🔹 Step 1 — Find intersection points Solve 4 − x 2 4 = x − 2 4-\frac{x^2}{4}=x-2 Bring to one side: 4 − x 2 4 − x + 2 = 0 ⇒ 6 − x − x 2 4 = 0 4-\frac{x^2}{4}-x+2=0 \Rightarrow 6-x-\frac{x^2}{4}=0 Multiply by 4: 24 − 4 x − x 2 = 0 24-4x-x^2=0 Rewrite: x 2 + 4 x − 24 = 0 x^2+4x-24=0 Solve: x = − 4 ± 16 + 96 2 = − 4 ± 4 7 2 = − 2 ± 2 7 x=\frac{-4\pm\sqrt{16+96}}{2} =\frac{-4\pm 4\sqrt{7}}{2} =-2\pm 2\sqrt{7} ​ (If you used y = x − 2 y=x-2 y = x − 2 , the algebra simplifies the same way — both limits are correct for the standard statement of this problem.) 🔹 Step 2 — Identify which curve is on top...

  ❓ 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 \...