If the function f ( x )= cx ⋅ e - x -( x 2/2)+ x is decreasing for every -∞

Question

If the function f ( x )= cx ⋅ e - x -( x 2/2)+ x is decreasing for every -∞< x ≤ 0, then the least value of c 2 is equal to (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

f ( x )= c x e - x -( x 2/2)+ x, we must have, f prime( x ) ≤ 0 ∀ x ≤ 0 ⇒ (1- x )( c e - x +1) ≤ 0 ∀ x ≤ 0 ⇒ c ≤- e x ∀ x ≤ 0 so, c ∈(-∞,-1] ⇒ Least value of c 2=1

Q. If the function $f (x) = c x \cdot e^{- x} - \frac{x ^{2}}{2} + x$ is decreasing for every $- \infty < x \leq 0$ , then the least value of $c^{2}$ is equal to

1

2

3

4

$f (x) = c x e^{- x} - \frac{x ^{2}}{2} + x$ , we must have, $f^{'} (x) \leq 0\forall x \leq 0$
$\Rightarrow (1 - x) (c e^{- x} + 1) \leq 0\forall x \leq 0 \Rightarrow c \leq - e^{x} \forall x \leq 0$
so, $c \in (- \infty, - 1] \Rightarrow$ Least value of $c^{2} = 1$

Q. If the function f(x)=cx⋅e−x−2x2​+x is decreasing for every −∞<x≤0, then the least value of c2 is equal to

1

2

3

4

f(x)=cxe−x−2x2​+x, we must have, f′(x)≤0∀x≤0 ⇒(1−x)(ce−x+1)≤0∀x≤0⇒c≤−ex∀x≤0 so, c∈(−∞,−1]⇒ Least value of c2=1

Q. If the function $f (x) = c x \cdot e^{- x} - \frac{x ^{2}}{2} + x$ is decreasing for every $- \infty < x \leq 0$ , then the least value of $c^{2}$ is equal to

$f (x) = c x e^{- x} - \frac{x ^{2}}{2} + x$ , we must have, $f^{'} (x) \leq 0\forall x \leq 0$
$\Rightarrow (1 - x) (c e^{- x} + 1) \leq 0\forall x \leq 0 \Rightarrow c \leq - e^{x} \forall x \leq 0$
so, $c \in (- \infty, - 1] \Rightarrow$ Least value of $c^{2} = 1$