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Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer?.

Solutions for Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Let a be a real number such that the function f(x) = ax2 + 6x - 15, x ∈R is increasing in (-∞, 3/4) and decreasing in (3/4, ∞) . Then the function g(x) = ax2 - 6x + 15, x ∈R, has aa)local minimum at x = -3/4b)local maximum at x = 3/4c)local minimum at x = 3/4d)local maximum at x = -3/4Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.