Let f ( x )=( e ln x ⋅ operatornamesgn( x 2+ x +1)/ ln x ). If f ( x )= k has one distinct solution then range of k is,

Question

Let f ( x )=( e ln x ⋅ operatornamesgn( x 2+ x +1)/ ln x ). If f ( x )= k has one distinct solution then range of k is,  (A) (-∞, 0) (B) (0, e ] (C) (-∞, 0) ∪ e  (D) [e, ∞)

Tardigrade Admin · Accepted Answer

Correct answer is (c) (-∞, 0) ∪ e

Q. Let $f (x) = \frac{e ^{l n x} \cdot sgn ( x ^{2} + x + 1 )}{l n x}$ . If $f (x) = k$ has one distinct solution then range of $k$ is, $

(-\infty, 0)$

$(0, e]$

$(- \infty, 0) \cup {e}$

$[e, \infty)$

Correct answer is (c) $(- \infty, 0) \cup {e}$

Q. Let f(x)=lnxelnx⋅sgn(x2+x+1)​. If f(x)=k has one distinct solution then range of k is, $

(-\infty, 0)$

(0,e]

(−∞,0)∪{e}

[e,∞)

Correct answer is (c) (−∞,0)∪{e}

Q. Let $f (x) = \frac{e ^{l n x} \cdot sgn ( x ^{2} + x + 1 )}{l n x}$ . If $f (x) = k$ has one distinct solution then range of $k$ is, $

$(0, e]$

$(- \infty, 0) \cup {e}$

$[e, \infty)$

Correct answer is (c) $(- \infty, 0) \cup {e}$