Let f: R+ arrow -1,0,1 defined by f(x)=sgn(x-x4+x7-x8-1) where sgn denotes signum function. Then f(x) is

Question

Let f: R+ arrow -1,0,1 defined by f(x)=sgn(x-x4+x7-x8-1) where sgn denotes signum function. Then f(x) is (A) many-one and onto (B) many-one and into (C) one-one and onto (D) one-one and into

Tardigrade Admin · Accepted Answer

f(x)=sgn(x-x4+x7-x8-1) For x ∈(0,1) ; x-1 < 0, x7-x4 < 0 ∴ x-x4+x7-x8-1 < 0 For x ∈(1, ∞) ; x < x4, x7 < x8 ∴ x-x4+x7-x8-1 < 0 Also for x=1 ; x-x4+x7-x8-1=-1 Thus x-x4+x7-x8-1 < 0 for all x ∈ R+ Hence sgn(x-x4+x7-x8-1)=-1 ∀ x ∈ R+. Therefore f(x) is many-one and into.

Q. Let f:R+→{−1,0,1} defined by f(x)=sgn(x−x4+x7−x8−1) where sgn denotes signum function. Then f(x) is

many-one and onto

many-one and into

one-one and onto

one-one and into

f(x)=sgn(x−x4+x7−x8−1) For x∈(0,1);x−1<0,x7−x4<0 ∴x−x4+x7−x8−1<0 For x∈(1,∞);x<x4,x7<x8 ∴x−x4+x7−x8−1<0 Also for x=1;x−x4+x7−x8−1=−1 Thus x−x4+x7−x8−1<0 for all x∈R+ Hence sgn(x−x4+x7−x8−1)=−1∀x∈R+. Therefore f(x) is many-one and into.

Q. Let $f : R^{+} \to {- 1, 0, 1}$ defined by $f (x) = s g n (x - x^{4} + x^{7} - x^{8} - 1)$ where sgn denotes signum function. Then $f (x)$ is