Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept 🎬 Purely Real / Imaginary Trick Complex numbers ke questions JEE mein scary lagte hain — lekin purely real / purely imaginary ka poora game 👉 sirf ek condition + ek rationalizing trick pe based hota hai. 🖼️ Concept Image ✍️ Short Explanation Jab question bole: expression purely real hai expression purely imaginary hai Toh yaad rakho 👇 👉 Real part aur Imaginary part ka zero hona hi sab kuch decide karta hai. 🔹 Step 1 — Basic Form of a Complex Number Har complex number likha ja sakta hai: z = x + i y z = x + iy Where: Real part = x = ℜ ( z ) x = \Re(z) Imaginary part = y = ℑ ( z ) y = \Im(z) Ye breakdown har question ka base hai. 🔹 Step 2 — Condition for Purely Real A complex number purely real tab hota hai jab: ℑ ( w ) = 0 \Im(w) = 0 👉 Matlab imaginary part bilkul zero. Examples: 5 5  ✔️ purely real 3 − 0 i 3 - 0i  ✔️ purely real 2 + i 2 + i ❌ not purely real 🔹 Step 3 — Condition for Purely Imagin...

  ❓ 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 P P  be the parabola whose focus is ( − 2 ,   1 ) (-2,\,1) and whose directrix is 2 x + y + 2 = 0. 2x + y + 2 = 0. Find the sum of the ordinates of the points on P P P whose abscissa is x = − 2. 🖼️ Question Image ✍️ Short Solution A parabola is defined as the locus of a point whose distance from the focus equals its perpendicular distance from the directrix . We will: 1️⃣ Write the focus–directrix condition 2️⃣ Substitute x = − 2 x = -2 3️⃣ Solve for y y 4️⃣ Add the ordinates 🔹 Step 1 — Use focus–directrix definition Let ( x , y ) (x,y)  be any point on the parabola. Distance from focus ( − 2 , 1 ) (-2,1) : ( x + 2 ) 2 + ( y − 1 ) 2 \sqrt{(x+2)^2 + (y-1)^2} ​ Distance from directrix 2 x + y + 2 = 0 2x + y + 2 = 0 : ∣ 2 x + y + 2 ∣ 2 2 + 1 2 = ∣ 2 x + y + 2 ∣ 5 \frac{|2x + y + 2|}{\sqrt{2^2 + 1^2}} = \frac{|2x + y + 2|}{\sqrt{5}} ​ Equating squares: ( x + 2 ) 2 + ( y − 1 ) 2 = ( 2 x + y + 2 ) 2 5 (x+2)^2 + (y-1)^2 = \frac{...

  ❓ Question FOR: Let C 1 C_1 ​ be the circle in the third quadrant of radius 3 , that touches both coordinate axes . Let C 2 C_2  be the circle with centre ( 1 ,   3 ) (1,\,3)  that touches C 1 C_1   externally at the point ( α ,   β ) (\alpha,\,\beta) . If ( β − α ) 2 = m n , gcd ⁡ ( m , n ) = 1 , (\beta - \alpha)^2 = \frac{m}{n}, \quad \gcd(m,n)=1, then the value of m + n m+n is equal to ? 🖼️ Question Image ✍️ Short Solution This problem uses pure coordinate geometry logic : 👉 A circle touching both axes has its centre fixed by symmetry. 👉 When two circles touch externally , the point of contact lies on the line joining their centres . 👉 We find that point using section formula , then compute ( β − α ) 2 (\beta-\alpha)^2 . 🔹 Step 1 — Equation and centre of C 1 C_1 ​ Circle C 1 C_1 C 1 ​ : Lies in third quadrant Touches x-axis and y-axis Radius = 3 Hence, its centre must be: ( − 3 ,   − 3 ) (-3,\,-3) (Only this point keeps t...

  ❓ Question FOR: The mean and standard deviation of 100 observations are 40 and 5.1 , respectively. By mistake, one observation is taken as 50 instead of 40 . If the correct mean and correct standard deviation are μ \mu μ and σ \sigma σ , respectively, then the value of 10 ( μ + σ ) 10(\mu + \sigma) is equal to ? 🖼️ Question Image ✍️ Short Solution This is a standard JEE correction-type statistics question . The idea is simple: 👉 Use the given (wrong) mean and SD to compute the wrong total and wrong sum of squares . 👉 Correct the single wrong observation. 👉 Recalculate the correct mean and correct standard deviation . 🔹 Step 1 — Find the wrong total (Σx) Given: Number of observations, n = 100 n = 100 Wrong mean = 40 = 40 So, Σ x wrong = 100 × 40 = 4000 \Sigma x_{\text{wrong}} = 100 \times 40 = 4000 One value was taken as 50 instead of 40 . Correct total: Σ x correct = 4000 − 50 + 40 = 3990 \Sigma x_{\text{correct}} = 4000 - 50 + 40 = 3990 ...

  ❓ Concept 🎬 Limits Approximation Trick Limits me jab x → 0 x \to 0 aur expression weird, lengthy, ya scary lage — toh ek hi cheez kaam aati hai 👇 👉 STANDARD APPROXIMATIONS . ✍️ Short Explanation JEE limits ka core idea yeh hai: Jab input bahut chhota ho jaata hai, functions apna linear behaviour dikhate hain. Isliye hume: exact values nahi sirf first term (leading term) chahiye hota hai 🔹 Step 1 — Angle Approximations (as x → 0) Sabse important JEE tools: sin ⁡ θ ≈ θ \sin \theta \approx \theta tan ⁡ θ ≈ θ \tan \theta \approx \theta tan ⁡ − 1 ( x ) ≈ x \tan^{-1}(x) \approx x 📌 Reason: Small angle ⇒ curve almost straight line ban jaati hai. 👉 Agar argument small hai, bina soche replace kar do. 🔹 Step 2 — Exponential Approximation For small x x : e k x − 1 ≈ k x e^{kx} - 1 \approx kx General rule: e ( something small ) − 1 ⇒ that small thing e^{(\text{something small})} - 1 \Rightarrow \text{that small thing} 📌 Higher powers...

  ❓ Question Find the remainder when ( ( 64 ) 64 ) 64 is divided by 7 . ✍️ Short Solution This is a base + remainder theorem question. JEE ka golden rule yaad rakho 👇 👉 Base-number ko pehle decimal me convert karo, phir modulo apply karo. 🔹 Step 1 — Convert base-64 number to decimal ( 64 ) 64 = 6 × 64 + 4 (64)_{64} = 6\times 64 + 4 = 384 + 4 = 388 = 384 + 4 = 388 So the expression becomes: 388 64 388^{64} 🔹 Step 2 — Reduce base modulo 7 Instead of handling big powers, take modulo early: 388   m o d   7 = 388 − 7 × 55 = 388 − 385 = 3 388 \bmod 7 = 388 - 7\times 55 = 388 - 385 = 3 So, 388 64 ≡ 3 64 ( m o d 7 ) 388^{64} \equiv 3^{64} \pmod{7} 🔹 Step 3 — Use Fermat’s Little Theorem Since 7 is prime and gcd ⁡ ( 3 , 7 ) = 1 \gcd(3,7)=1 : 3 6 ≡ 1 ( m o d 7 ) 3^6 \equiv 1 \pmod{7} Now reduce the exponent: 64   m o d   6 = 4 64 \bmod 6 = 4 So, 3 64 ≡ 3 4 ( m o d 7 ) 3^{64} \equiv 3^4 \pmod{7} 🔹 Step 4 — Final calculation 3 4 = 81 3^4 = 81 81   m o d   7 = 4 81 \...

  ❓ Concept 🎬 Adj(A) & Determinant Tricks Adjugate (adj) questions look scary in JEE — but in reality, sab kuch power rules pe based hota hai . Agar 4 identities yaad hain, toh 10 seconds mein answer . 1️⃣ What is adj(A)? Adjugate of a matrix A A A : adj ( A ) = transpose of cofactor matrix \text{adj}(A) = \text{transpose of cofactor matrix} ⭐ Most Important Identity: A ⋅ adj ( A ) = adj ( A ) ⋅ A = ∣ A ∣   I A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A|\,I 📌 Ye identity almost har JEE matrix question ka base hoti hai. 2️⃣ adj(A) for Invertible A If: ∣ A ∣ ≠ 0 |A| \ne 0 Then: adj ( A ) = ∣ A ∣   A − 1 \text{adj}(A) = |A|\,A^{-1} 👉 Matlab adj(A) behaves like det(A) × inverse . 3️⃣ Determinant of adj(A) — BIG TRICK 🔥 For any n × n n \times n  matrix: ∣ adj ( A ) ∣ = ∣ A ∣   n − 1 |\text{adj}(A)| = |A|^{\,n-1} 📌 Very powerful rule : adj lene se determinant power mein convert ho jaata hai For JEE (most common): n = 3 n =...