Exams › JEE Advanced › Maths
Given that a² + b² + c² = 9*(x² + y² + z²), find the maximum value of (a*x + b*y + c*z) / (x² + y² + z²).
- 3
- 4
- 5
- 6
Correct answer: 3
Solution
By Cauchy-Schwarz, (ax+by+cz)² <= (a²+b²+c²)(x²+y²+z²) = 9(x²+y²+z²)². Dividing both sides by (x²+y²+z²)² gives [(ax+by+cz)/(x²+y²+z²)]² <= 9, so the maximum is 3.
Related JEE Advanced Maths questions
- In a chemistry class, there are 20 students, while a physics class has 30 students. If 10 students are enrolled in both classes, and the two classes are held at separate times, determine the value of k/8, where k represents the total number of students attending either class.
- Given that A represents the divisors of 15, B contains prime numbers less than 10, and C includes even numbers less than 9, how many elements are there in the intersection of (A ∪ C) and B?
- Identify the periodic function among the following:
- The function f(x) = √|x² - 5| x + 6 + √8 + 2|x| - |x|² is defined as a real number for values of x within which range?
- Let f(x) = 4x / (4x² + 2). If the sum of the integrals ∫(1/1997) + ∫(2/1997) +... + ∫(1196/1997) equals 499q, what is the value of q?
- Given that α, β, γ, and θ are the smallest positive angles in increasing order for which their sine values equal a positive constant k, what is the result of 4 sin(α/2) + 3 sin(β/2) + 2 sin(γ/2) + sin(θ/2)?
⚔️ Practice JEE Advanced Maths free + battle 1v1 →