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A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is 1/2 Also assume that the probability of the answer for a question being guessed, given that the student’s answer is correct, is 1/6 Then the probability that the student knows the answer of a randomly chosen question is

on June 26, 2025

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How to Believe in Yourself and Your Practice to Overcome Anxiety Before an Exam

on June 25, 2025

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How to Ask Questions Without Irritating Your Teacher: A Guide to Solving Unlimited Doubts Smartly

on June 25, 2025

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How Bad Company Can Hurt Your JEE Preparation

on June 25, 2025

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The Role of a Friend and What Healthy Competition Looks Like

on June 25, 2025

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How to Crack JEE or Any Competitive Exam in 6 Months: A Strategic Guide

on June 25, 2025

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Math JEE Adv. 2025 paper 1

To solve the problem, we'll follow these steps: Step 1: Assign Coordinates Let’s place the triangle Δ P Q R Δ PQR in the plane for simplicity: Let P = ( 0 , 0 ) P = ( 0 , 0 ) , Q = ( 1 , 0 ) Q = ( 1 , 0 ) , and R = ( 0 , 1 ) R = ( 0 , 1 ) . Step 2: Find the Coordinates of S S Given the vector equation: S P ‾ + 5 S Q ‾ + 6 S R ‾ = 0 ‾ , SP + 5 SQ + 6 SR = 0 , we express the vectors in terms of coordinates: S P ‾ = P − S = ( 0 − x , 0 − y ) , SP = P − S = ( 0 − x , 0 − y ) , S Q ‾ = Q − S = ( 1 − x , 0 − y ) , SQ = Q − S = ( 1 − x , 0 − y ) , S R ‾ = R − S = ( 0 − x , 1 − y ) . SR = R − S = ( 0 − x , 1 − y ) . Substituting into the equation: ( 0 − x , 0 − y ) + 5 ( 1 − x , 0 − y ) + 6 ( 0 − x , 1 − y ) = ( 0 , 0 ) . ( 0 − x , 0 − y ) + 5 ( 1 − x , 0 − y ) + 6 ( 0 − x , 1 − y ) = ( 0 , 0 ) . This gives the system: − x + 5 ( 1 − x ) + 6 ( − x ) = 0 and − y + 5 ( − y ) + 6 ( 1 − y ) = 0. − x + 5 ( 1 − x ) + 6 ( − x ) = 0 and − y + 5 ( − y ) + 6 ...

Problem Statement: We are given: A parabola with equation y 2 = 4 a x y 2 = 4 a x where a > 0 a > 0 . A point P P on this parabola. The normal to the parabola at P P meets the x-axis at point Q Q . The area of triangle P F Q PFQ is 120, where F F is the focus of the parabola. The slope m m of the normal and a a are both positive integers. We need to find the correct pair ( a , m ) ( a , m ) from the given options. Step 1: Understand the Parabola and Focus The standard parabola y 2 = 4 a x y 2 = 4 a x has: Vertex at ( 0 , 0 ) ( 0 , 0 ) . Focus F F at ( a , 0 ) ( a , 0 ) . Step 2: Parametrize Point P P on the Parabola Let P P be a point on the parabola. We can parametrize P P as: P = ( a t 2 , 2 a t ) , P = ( a t 2 , 2 a t ) , where t t is a parameter. Step 3: Find the Equation of the No...

Math JEE Adv. 2025 paper 1

To solve the problem, we'll count the number of seven-digit numbers formed using digits {0, 1, 2} where at least one of the digits 0 or 1 appears exactly twice. Here's the step-by-step solution: Step 1: Total Possible Numbers Each of the 7 digits can be 0, 1, or 2. The first digit cannot be 0 (to ensure it's a seven-digit number). Total numbers: 2 × 3 6 = 2 × 729 = 1458 2 × 3 6 = 2 × 729 = 1458 . Step 2: Numbers Where 0 Appears Exactly Twice Choose positions for 0: ( 6 2 ) ( 2 6 ) (since the first digit cannot be 0). Fill the remaining 5 digits: Each can be 1 or 2 (since we're fixing exactly two 0s). Count: ( 6 2 ) × 2 5 = 15 × 32 = 480 ( 2 6 ) × 2 5 = 15 × 32 = 480 . Step 3: Numbers Where 1 Appears Exactly Twice Choose positions for 1: ( 7 2 ) ( 2 7 ) . Fill the remaining 5 digits: Each can be 0 or 2 (but the first digit cannot be 0). If the first digit is 1: ( 6 1 ) × 2 5 = 6 × 32 = 192 ( 1 6 ) × 2 5 = 6 × 32 = 192 . I...

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