Consider f:(-∞, 0) ∪((1/3 e), ∞) arrow R, defined by f(x)=(3 x/2) ln (e-(1/3 x)) then

Question

Consider f:(-∞, 0) ∪((1/3 e), ∞) arrow R, defined by f(x)=(3 x/2) ln (e-(1/3 x)) then (A)  f(x) has no point of inflection. (B)  f(x) is surjective but not injective. (C) f(x) is bijective function. (D) f(x)=(-1/2 e) has two distinct solutions.

Tardigrade Admin · Accepted Answer

Correct answer is (a) f(x) has no point of inflection.Correct answer is (b) f(x) is surjective but not injective.Correct answer is (d) f(x)=(-1/2 e) has two distinct solutions.

Q. Consider $f : (- \infty, 0) \cup (\frac{1}{3 e}, \infty) \to R$ , defined by $f (x) = \frac{3 x}{2} ln (e - \frac{1}{3 x})$ then

$f (x)$ has no point of inflection.

$f (x)$ is surjective but not injective.

$f (x)$ is bijective function.

$f (x) = \frac{- 1}{2 e}$ has two distinct solutions.

Correct answer is (a) $f (x)$ has no point of inflection. Correct answer is (b) $f (x)$ is surjective but not injective. Correct answer is (d) $f (x) = \frac{- 1}{2 e}$ has two distinct solutions.

Q. Consider f:(−∞,0)∪(3e1​,∞)→R, defined by f(x)=23x​ln(e−3x1​) then

f(x) has no point of inflection.

f(x) is surjective but not injective.

f(x) is bijective function.

f(x)=2e−1​ has two distinct solutions.