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If y = ax3 + 3x2 + (2a + 1)x + 1000 is strictly increasing function for all values of x, then -
- a)–3/2 < a < 1
- b)a > 1
- c)a < –3/2
- d)a > 1 or a < – 3/2
Correct answer is option 'B'. Can you explain this answer?
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If y = ax3 + 3x2 + (2a + 1)x + 1000 is strictly increasing function fo...
If the function is strictly increasing for all values of x, then the derivative of the function must be positive for all values of x.
Taking the derivative of the function, we get:
dy/dx = 3ax^2 + 6x(a-1) + 2a - 1
To determine the values of a that make the derivative positive for all values of x, we can set the derivative equal to zero and solve for x:
3ax^2 + 6x(a-1) + 2a - 1 = 0
This is a quadratic equation in terms of x. The discriminant of the quadratic equation is:
D = (6(a-1))^2 - 4(3a)(2a-1)
For the derivative to be positive for all values of x, the discriminant must be negative. Therefore:
(6(a-1))^2 - 4(3a)(2a-1) < />
Simplifying this inequality, we get:
36(a^2 - 2a + 1) - 24a(2a-1) < />
36a^2 - 72a + 36 - 48a^2 + 24a < />
-12a^2 - 48a + 36 < />
Dividing through by -12, we get:
a^2 + 4a - 3 > 0
Factoring the quadratic, we get:
(a+3)(a-1) > 0
This inequality is true when a < -3="" or="" a="" /> 1.
Therefore, the values of a that make the function strictly increasing for all values of x are a < -3="" or="" a="" /> 1.
Taking the derivative of the function, we get:
dy/dx = 3ax^2 + 6x(a-1) + 2a - 1
To determine the values of a that make the derivative positive for all values of x, we can set the derivative equal to zero and solve for x:
3ax^2 + 6x(a-1) + 2a - 1 = 0
This is a quadratic equation in terms of x. The discriminant of the quadratic equation is:
D = (6(a-1))^2 - 4(3a)(2a-1)
For the derivative to be positive for all values of x, the discriminant must be negative. Therefore:
(6(a-1))^2 - 4(3a)(2a-1) < />
Simplifying this inequality, we get:
36(a^2 - 2a + 1) - 24a(2a-1) < />
36a^2 - 72a + 36 - 48a^2 + 24a < />
-12a^2 - 48a + 36 < />
Dividing through by -12, we get:
a^2 + 4a - 3 > 0
Factoring the quadratic, we get:
(a+3)(a-1) > 0
This inequality is true when a < -3="" or="" a="" /> 1.
Therefore, the values of a that make the function strictly increasing for all values of x are a < -3="" or="" a="" /> 1.
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If y = ax3 + 3x2 + (2a + 1)x + 1000 is strictly increasing function for all values of x, then -a)–3/2 < a < 1b)a > 1c)a < –3/2d)a > 1 or a < – 3/2Correct answer is option 'B'. Can you explain this answer?
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If y = ax3 + 3x2 + (2a + 1)x + 1000 is strictly increasing function for all values of x, then -a)–3/2 < a < 1b)a > 1c)a < –3/2d)a > 1 or a < – 3/2Correct answer is option 'B'. 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 If y = ax3 + 3x2 + (2a + 1)x + 1000 is strictly increasing function for all values of x, then -a)–3/2 < a < 1b)a > 1c)a < –3/2d)a > 1 or a < – 3/2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If y = ax3 + 3x2 + (2a + 1)x + 1000 is strictly increasing function for all values of x, then -a)–3/2 < a < 1b)a > 1c)a < –3/2d)a > 1 or a < – 3/2Correct answer is option 'B'. Can you explain this answer?.
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