Published by:
CGP EDU Academic Team
Published on: August 13, 2026
If f(x) = 2e x – ae –x + (2a + 1) x – 3 monotonically increases for ∀ x ∈ R, then the minimum value of 'a' is
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
f(x) = 2e x – ae –x + (2a + 1) x – 3
f ′ (x) = 2e x + ae –x + (2a + 1)
Let e x = t ; t > 0
∴ 2t + 
+ (2a + 1) ⇒ 
⇒ Now f ′ (x) ≥ 0 ∴ 2t 2 + (2a + 1) t + a ≥ 0
Case-I D ≤ 0
(2a + 1) 2 – 8a ≤ 0
⇒ (2a – 1) 2 ≤ 0 ⇒ a = 
Case-II 
⇒ (i) D ≥ 0 ⇒ a ∈ R
(ii) f(0) ≥ 0 ⇒ a ≥ 0
(iii) – 
< 0 ⇒ 
< 0 ⇒ a > – 
∴ a ∈ [0, ∞ )
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
If f(x) = cos π x + 10x + 3x 2 + x 3 , – 2 ≤ x ≤ 3 then absolute minimum value of f(x) is-
The function f(x) = x 5 – 5x 4 + 5x 3 – 1 has-
The function f(x) = has no extrema. if (n ∈ z)
If x exceeds by its square by the greatest possible quantity then x is-
An extreme value of the function
f(x) = (sin –1 x) 3 + (cos –1 x) 3 (–1 < x < 1) is
Let f(x) = (t + 4) 4 log (t+ 1) dt, then -