Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept 🎬 Absolute Max–Min Trick JEE mein jab question bole absolute maximum / minimum , toh sirf derivative nikaalna kaafi nahi hota . 👉 Real game interval checking ka hota hai. ✍️ Short Explanation Derivative hume possible candidates deta hai, lekin final decision hamesha function values se hota hai. Isliye JEE mein absolute max–min ke liye ek fixed rule follow kiya jaata hai. 🔹 Step 1 — Meaning of Critical Points Critical points woh x x -values hote hain jahan: f ′ ( x ) = 0 or f ′ ( x )  is undefined f'(x) = 0 \quad \text{or} \quad f'(x) \text{ is undefined} 📌 Important: Ye sirf candidates hote hain Guaranteed max/min nahi hote 🔹 Step 2 — Closed Interval Rule (MOST IMPORTANT 🔥) Agar interval ho: [ a , b ] [a, b] [ a , b ] Toh absolute maximum aur minimum ke liye sirf aur sirf yeh points check hote hain : 1️⃣ Har critical point jo ( a , b ) (a,b)  ke andar ho 2️⃣ Endpoint x = a x = a 3️⃣ Endpoint x = b x = b Th...

  ❓ Question Let f ( x ) = x 3 + a x 2 + b log ⁡ e ∣ x ∣ + 1 , x ≠ 0 f(x) = x^3 + ax^2 + b\log_e|x| + 1,\quad x \ne 0 Given that x = − 1  and  x = 2 x = -1 \text{ and } x = 2 are the critical points of f ( x ) f(x) . Let m m  and M M  respectively be the absolute minimum and absolute maximum values of f ( x ) f(x)  in the interval [ − 2 ,   − 1 2 ] . Find: ∣ M + m ∣ 🖼️ Question Image ✍️ Short Solution Critical points are obtained from f ′ ( x ) = 0 We’ll first find constants a a  and b b , then evaluate f ( x ) f(x)  at endpoints and critical points inside the interval to get absolute max and min. 🔹 Step 1 — Find the derivative f ( x ) = x 3 + a x 2 + b ln ⁡ ∣ x ∣ + 1 f(x) = x^3 + ax^2 + b\ln|x| + 1 Differentiate: f ′ ( x ) = 3 x 2 + 2 a x + b x f'(x) = 3x^2 + 2ax + \frac{b}{x} 🔹 Step 2 — Use critical points Given critical points: x = − 1 , x = 2 x = -1,\quad x = 2 So, f ′ ( − 1 ) = 0 , f ′ ( 2 ) = 0 f'(-1) = 0,\quad f'(...