Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question Two projectiles are fired from the ground with the same initial speed from the same point , at angles ( 45 ∘ + α ) and ( 45 ∘ − α ) with the horizontal. Find the ratio of their times of flight . 🖼️ Question Image ✍️ Short Solution This is a classic JEE symmetry question from projectile motion. The trick is to remember what depends on sine and what depends on sine double-angle . 🔹 Step 1 — Time of flight formula For a projectile fired from ground with speed u u u at angle θ \theta : T = 2 u sin ⁡ θ g 📌 Time of flight depends on sin ⁡ θ \sin\theta sin θ (not on sin ⁡ 2 θ \sin 2\theta sin 2 θ ). 🔹 Step 2 — Write times for both projectiles Let: T 1 T_1 ​ = time of flight at angle ( 45 ∘ + α ) (45^\circ+\alpha) T 2 T_2 ​ = time of flight at angle ( 45 ∘ − α ) Using the formula: T 1 = 2 u sin ⁡ ( 45 ∘ + α ) g T_1=\frac{2u\sin(45^\circ+\alpha)}{g} T 2 = 2 u sin ⁡ ( 45 ∘ − α ) g T_2=\frac{2u\sin(45^\circ-\alpha)}{g} 🔹 Step 3 — Take the r...

❓ Concept 🎬 Infinite Solutions + Geometry Trick in 59 Sec Teen equations, teen variables… par answer bole “infinitely many solutions” ? 👉 Samajh jao — determinant zero ho chuka hai aur geometry ka scene start ho gaya hai 🔥 ✍️ Short Explanation Aise questions mein students sirf algebra pe atak jaate hain, jabki JEE ka smart move hota hai: 👉 Algebra se parameters nikaalo 👉 Geometry se final answer (radius, distance, etc.) 🔹 Step 1 — Infinite Solutions Condition (CORE RULE 💯)** System: A X = B AX = B Infinite solutions tab milte hain jab: ∣ A ∣ = 0 and rank ⁡ ( A ) = rank ⁡ ( A ∣ B ) < number of variables |A| = 0 \quad \text{and} \quad \operatorname{rank}(A)=\operatorname{rank}(A|B)<\text{number of variables} 📌 Jaise hi yeh dikhe: System dependent hai Parameters (λ, μ, …) enter karte hain 🔹 Step 2 — Geometric Meaning (VERY IMPORTANT 🔥)** 3 variables ⇒ 3D space Har linear equation ⇒ ek plane 👉 Infinite solutions ka matlab: Pla...

  ❓ Concept 🎬 Remainder Tricks for Huge Powers Kabhi-kabhi JEE mein aise questions milte hain jahan power itna bada hota hai ki calculator bhi give-up kar de 😅 Tab kaam aata hai modulo + cyclic pattern — aur poora question 1-min trick se solve ho jaata hai. 🖼️ Concept Image ✍️ Short Explanation Basic funda: 👉 Remainder depends on modulo behaviour — not on size of the number. Isliye pehle number ko chhota banao, phir power handle karo. 🔹 Step 1 — Remainder Concept If a ≡ r ( m o d n ) a \equiv r \pmod n then a k ≡ r k ( m o d n ) a^k \equiv r^k \pmod n 📌 Matlab: Pehle base ko modulo se reduce karo Phir power lagao 🔹 Step 2 — Reduce the Base First (Always!) Large number ko directly power nahi karte. Pehle: a m o d     n = r a \mod n = r Phir sirf r ke saath kaam karo — calculation ultra-simple ho jaata hai. Example: 388 64 m o d     7 ⇒ 388 ≡ 3 m o d     7 388^{64} \mod 7 \Rightarrow 388 \equiv 3 \mod 7 So: 388 64 ≡ 3 64 m o d     7 38...

❓ Question Let a n a_n a n ​ be the n th n^{\text{th}}  term of an A.P. If S n = a 1 + a 2 + ⋯ + a n = 700 ,    a 6 = 7 S_n=a_1+a_2+\cdots+a_n=700,\; a_6=7  and S 7 = 7 S_7=7 , then find a n a_n ​ . 🖼️ Question Image ✍️ Short Solution Let the A.P. have first term a 1 a_1 ​ and common difference d d . Then: a 6 = a 1 + 5 d = 7. Sum of first 7 terms: S 7 = 7 2 ( 2 a 1 + 6 d ) = 7. S_7 = \frac{7}{2}\big(2a_1 + 6d\big) = 7. Divide both sides by 7: 1 2 ( 2 a 1 + 6 d ) = 1 ⇒ a 1 + 3 d = 1. \frac{1}{2}\big(2a_1 + 6d\big) = 1 \quad\Rightarrow\quad a_1 + 3d = 1. Now subtract the two linear equations: ( a 1 + 5 d ) − ( a 1 + 3 d ) = 7 − 1    ⇒    2 d = 6    ⇒    d = 3. (a_1+5d) - (a_1+3d) = 7 - 1 \;\Rightarrow\; 2d = 6 \;\Rightarrow\; d = 3. Then a 1 + 3 d = 1 ⇒ a 1 + 9 = 1 ⇒ a 1 = − 8. a_1 + 3d = 1 \Rightarrow a_1 + 9 = 1 \Rightarrow a_1 = -8. So the A.P. is − 8 ,    − 5 ,    − 2 ,    1 ,    4 ,    7 ,    10 , … Next, use the condition S n = 700 S_n = 700 . Sum ...

❓ Question Let a random variable X X  take values 0 , 1 , 2 , 3 0,1,2,3  with P ( X = 0 ) = p , P ( X = 1 ) = p , P ( X = 2 ) = P ( X = 3 ) and it is given that    E ( X 2 ) = 2 E ( X ) . \;E(X^2)=2E(X). Find the value of 8 p − 1. 🖼️ Question Image ✍️ Short Solution Let q = P ( X = 2 ) = P ( X = 3 ) q = P(X=2)=P(X=3) . Since total probability is 1: 2 p + 2 q = 1 ⇒ p + q = 1 2 ⇒ q = 1 2 − p . Compute the expectations: E ( X ) = 0 ⋅ p + 1 ⋅ p + 2 ⋅ q + 3 ⋅ q = p + 5 q . E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q. E ( X 2 ) = 0 2 ⋅ p + 1 2 ⋅ p + 2 2 ⋅ q + 3 2 ⋅ q = p + 13 q . E(X^2) = 0^2\cdot p + 1^2\cdot p + 2^2\cdot q + 3^2\cdot q = p + 13q. Use the given relation E ( X 2 ) = 2 E ( X ) E(X^2)=2E(X) : p + 13 q = 2 ( p + 5 q ) . Simplify: p + 13 q = 2 p + 10 q ⇒ − p + 3 q = 0 ⇒ 3 q = p . Substitute q = 1 2 − p q = \tfrac12 - p  into 3 q = p 3q = p : 3 ( 1 2 − p ) = p ⇒ 3 2 − 3 p = p 3\left(\tfrac12 - p\right) = p \quad\Rightarr...

  Question A bag contains 19 unbiased coins and one coin with heads on both sides . One coin is drawn at random and tossed; a head turns up. If the probability that the drawn coin was unbiased is m n \dfrac{m}{n} n m ​ where gcd ⁡ ( m , n ) = 1 \gcd(m,n)=1 , then find: n 2 − m 2 Question Image Short Solution Define events: U U : chosen coin is unbiased D D : chosen coin is double-headed H H : head turns up Use Bayes’ theorem: P ( U ∣ H ) = P ( U ) P ( H ∣ U ) P ( U ) P ( H ∣ U ) + P ( D ) P ( H ∣ D )​ Compute each term: P ( U ) = 19 20 P(U)=\dfrac{19}{20} ​ P ( D ) = 1 20 P(D)=\dfrac{1}{20} ​ P ( H ∣ U ) = 1 2 P(H|U)=\dfrac{1}{2} fair coin) P ( H ∣ D ) = 1 P(H|D)=1 (double-headed always gives head) Plug values, simplify to find m m  and n n . Compute n 2 − m 2 n^{2}-m^{2} Image Solution Conclusion Step 1: Apply Bayes P ( U ∣ H ) = 19 20 ⋅ 1 2 19 20 ⋅ 1 2 + 1 20 ⋅ 1 ​ ​ Step 2: Simplify Numerator: 19 20 ⋅ 1 2 = 19...

  Question Let y = y ( x ) y = y(x) be the solution of the differential equation ( x 2 + 1 ) y ′ − 2 x y = ( x 4 + 2 x 2 + 1 ) cos ⁡ x , with y ( 0 ) = 1 y(0) = 1 . Then evaluate: ∫ − 3 3 y ( x )   d x Question Image Short Solution Rewrite the ODE in standard linear form: y ′ − 2 x x 2 + 1 y = ( x 4 + 2 x 2 + 1 ) cos ⁡ x x 2 + 1​ Find the integrating factor : μ ( x ) = e ∫ − 2 x x 2 + 1 d x = e − ln ⁡ ( x 2 + 1 ) = 1 x 2 + 1​ Multiply the equation by μ ( x ) \mu(x) : d d x ( y x 2 + 1 ) = cos ⁡ x Integrate: y x 2 + 1 = sin ⁡ x + C Apply y ( 0 ) = 1 y(0)=1 : 1 1 = 0 + C    ⟹    C = 1 Hence: y ( x ) = ( x 2 + 1 ) ( sin ⁡ x + 1 ) Compute the definite integral: I = ∫ − 3 3 y ( x )   d x = ∫ − 3 3 ( x 2 + 1 ) ( sin ⁡ x + 1 ) d x Image Solution Conclusion Break the integral: I = ∫ − 3 3 ( x 2 + 1 ) sin ⁡ x   d x + ∫ − 3 3 ( x 2 + 1 ) d x First term: ( x 2 + 1 ) sin ⁡ x (x^{2}+1)\sin x  is an odd function (because x 2 + 1 x^{2}+1  is eve...