Consider the function f ( x )=( x /2 x ) and g ( x )= max . f ( t ): x ≤ t ≤ x +1 If f ( x )= k has 2 distinct real roots then range of k is equal to

Question

Consider the function f ( x )=( x /2 x ) and g ( x )= max . f ( t ): x ≤ t ≤ x +1 If f ( x )= k has 2 distinct real roots then range of k is equal to (A) (0, (1/ e )) (B) (0, (1/ e ln 2)) (C) ((1/ e ln 2), ∞) (D) (-∞, 0)

Tardigrade Admin · Accepted Answer

Clearly, f ( x )= k has two distinct roots for k ∈(0, (1/ e ln 2)).

Q. Consider the function $f (x) = \frac{x}{2 ^{x}}$ and $g (x) = max . {f (t) : x \leq t \leq x + 1}$
If $f (x) = k$ has 2 distinct real roots then range of $k$ is equal to

$(0, \frac{1}{e})$

$(0, \frac{1}{e l n 2})$

$(\frac{1}{e l n 2}, \infty)$

$(- \infty, 0)$

Clearly, $f (x) = k$ has two distinct roots for $k \in (0, \frac{1}{e l n 2})$ .

Q. Consider the function f(x)=2xx​ and g(x)=max.{f(t):x≤t≤x+1} If f(x)=k has 2 distinct real roots then range of k is equal to

(0,e1​)

(0,eln21​)

(eln21​,∞)

(−∞,0)