Question

answer the 16 question

Answer 1

Answer:

Step-by-step explanation:

PHOTO UNCLEAR CANT SEE PROPERLY SSRY

Answer 2

$\huge\star\underline{\mathcal\color{brown}{HELLO\:MATE}}\star$

$\color{brown}{HERE\:IS\:YR\:ANS}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\underline\color{Green}{YUP}$

f:R->R f(x)=x3+x

Let x,y belongs to R

f(x)=f(y)

x3+x=y3+y

(x3-y3)+x-y=0

(x-y)(x2+xy+y2+1)=0

x=y

f(x)=f(y)

=) x=y for all x,y belongs to R

So, f is one-one function

Let y be any arbitrary element of R

f(x)=y

=) x3+x=y

x3+x-y=0

For every value of y, the equation x3+x-y=0 has a real root a (alpha)

a3+a-y=0

a3+a=y

f(a)=y

For every y belongs to R there exists a (alpha) belongs R such that f(a)=y

So, f is a onto function

Hence, f:R->R is a bijective function

$\fbox\color{brown}{HOPE\:IT\:HELPS}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\huge{\mathcal{THANKS}}$

$<marquee >☝️ABHI♥️☝️$