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If f(x) = 2ex - ae-x + (2a+1)x - 3 is monotonically increasing for all x € R and the range of values of ‘a’ are
a € [λ, ∞), then find the value of λ.
a € [λ, ∞), then find the value of λ.
Correct answer is '0'. Can you explain this answer?
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If f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x &...
f ’(x) must be positive for all x e r
∴ 2ex + ae-x + (2a + 1) > 0
⇒ e-x (2(ex)2 + (2a + 1)ex + a} > 0
⇒ {2t2 + (2a + 1)t + a} > 0 where t = ex, Possible graphs of lines are
∴ 2ex + ae-x + (2a + 1) > 0
⇒ e-x (2(ex)2 + (2a + 1)ex + a} > 0
⇒ {2t2 + (2a + 1)t + a} > 0 where t = ex, Possible graphs of lines are
Most Upvoted Answer
If f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x &...
To show that f(x) is monotonically increasing for all x, we need to show that its derivative f'(x) is always positive.
f(x) = 2ex - ae-x
f'(x) = 2ex + ae-x
We want to show that f'(x) > 0 for all x. We can start by factoring out 2e from f'(x):
f'(x) = 2e(x + a/e)
Since e is a positive constant, we know that the sign of f'(x) is determined by the sign of (x + a/e). If (x + a/e) > 0, then f'(x) > 0 and f(x) is increasing. If (x + a/e) < 0,="" then="" f'(x)="" />< 0="" and="" f(x)="" is="" decreasing.="" if="" (x="" +="" a/e)="0," then="" f'(x)="0" and="" f(x)="" has="" a="" critical="" />
Since we want to show that f(x) is increasing for all x, we need to show that (x + a/e) > 0 for all x. We know that a is a positive constant, so a/e is also positive. Therefore, we can choose any value of x that is greater than -a/e, and we will have (x + a/e) > 0. Since there are no restrictions on the domain of f(x), we can conclude that f(x) is monotonically increasing for all x.
f(x) = 2ex - ae-x
f'(x) = 2ex + ae-x
We want to show that f'(x) > 0 for all x. We can start by factoring out 2e from f'(x):
f'(x) = 2e(x + a/e)
Since e is a positive constant, we know that the sign of f'(x) is determined by the sign of (x + a/e). If (x + a/e) > 0, then f'(x) > 0 and f(x) is increasing. If (x + a/e) < 0,="" then="" f'(x)="" />< 0="" and="" f(x)="" is="" decreasing.="" if="" (x="" +="" a/e)="0," then="" f'(x)="0" and="" f(x)="" has="" a="" critical="" />
Since we want to show that f(x) is increasing for all x, we need to show that (x + a/e) > 0 for all x. We know that a is a positive constant, so a/e is also positive. Therefore, we can choose any value of x that is greater than -a/e, and we will have (x + a/e) > 0. Since there are no restrictions on the domain of f(x), we can conclude that f(x) is monotonically increasing for all x.
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If f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x €R and the range of values of ‘a’ area €[λ, ∞), then find the value of λ.Correct answer is '0'. Can you explain this answer?
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If f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x €R and the range of values of ‘a’ area €[λ, ∞), then find the value of λ.Correct answer is '0'. 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 f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x €R and the range of values of ‘a’ area €[λ, ∞), then find the value of λ.Correct answer is '0'. 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 f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x €R and the range of values of ‘a’ area €[λ, ∞), then find the value of λ.Correct answer is '0'. Can you explain this answer?.
If f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x €R and the range of values of ‘a’ area €[λ, ∞), then find the value of λ.Correct answer is '0'. 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 f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x €R and the range of values of ‘a’ area €[λ, ∞), then find the value of λ.Correct answer is '0'. 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 f(x) = 2ex-ae-x+ (2a+1)x -3 is monotonically increasing for all x €R and the range of values of ‘a’ area €[λ, ∞), then find the value of λ.Correct answer is '0'. Can you explain this answer?.
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