Application of DerivativesmediumFree

Values of m for Which 3 + mx + e^(-x) Is Always Decreasing | JEE

JEE Maths question with a full step-by-step solution.

Question
The complete set of values of mm, for which the function
f(x)=3+mx+e−xf(x) = 3 + mx + e^{-x}
 is always decreasing, is
A[0,∞)[0, \infty)
B(−∞,0](-\infty, 0]correct
C[2,5][2, 5]
D[7,17][7, 17]
Solution
Step 1: Differentiate.
f′(x)=m−e−x.f'(x) = m - e^{-x} .
 Step 2: Write the condition for ff to decrease everywhere.
f′(x)≤0 ∀x  ⟹  m≤e−xfor every x∈R.f'(x) \le 0 \ \forall x \implies m \le e^{-x} \quad \text{for every } x \in \mathbb{R} .
 Step 3: So mm must be at most the *smallest* value that e−xe^{-x} takes. Examine that range: as xx runs over R\mathbb{R}, e−xe^{-x} takes every positive value, so
e−x∈(0,∞).e^{-x} \in (0, \infty) .
 Step 4: The infimum is 00 and it is never attained, so requiring m≤e−xm \le e^{-x} for every xx implies that
m≤0.m \le 0 .
 Step 5: Check that m≤0m \le 0 really works. Then m−e−x≤0−e−x<0m - e^{-x} \le 0 - e^{-x} < 0 for every xx, so ff is (strictly) decreasing throughout.
m∈(−∞, 0].m \in (-\infty,\ 0] .
 Answer: (2).
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