If f (x )= (2k+1 )x-3-ke^-x+2e^x is monotonically increasing for all x ∈R, then the least value of k is

Question

Accepted Answer

Correct answer: B. (b) Given, function f (x )= (2k+1 )x-3-ke^-x+2e^x is monotonically increasing for all x ∈R. ⇒f (x ) 0 (2k+1 )+ke^-x+2e^x 0( ⇒e^-x (2k+1 )e^x+k+2e^2x) 0 or (2k+1 )e^x+k+2e^2x 0 ⇒2^2x(e^x+k)+1(e^x+k) 0 ⇒(2e^x+1)(e^x+k) 0( ⇒e^x+k) 0 or k 0 Hence, least value of k is zero.

$If f (x) = (2 k + 1) x - 3 - k e^{- x} + 2 e^{x} is monotonically increasing for all x \in R, then the least value of k is$

Solution

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