Question

B (i) IS f:[0,2] → [-1,3] WHERE 68)= x=4x +3 11? ONTO? EX PLAIN. (ii) IS g: [0,2] →R WHERE g®)=x?-4x+3 1-1? ONTO? EXPLAIN, hi [0,3] → [-1,3] WHERE hes=84x+3 1-1? OU TO? EXPLAIN (iii) IS

Answer 1

ANSWER:-

$f(x)=x^2-4x+3,f(0)=3$

$=(x-3)(x-1)$

3 21 -)

$f(x)=x^2-4x+3$

df (x) dr 2.C – 4

$\large 2x=4$

$\large \Rightarrow x=2$

Minimum value $\large =x^2-4x+3$

$\large =4-8+3$

$\large =-1$

i)

$\large f:[0,2]\rightarrow [-1,3]$

In the domain of [0, 2]the function f(x) is decreasing and it's range is [-1, 3]by graph so function is OnTo

ii)

In the domain of [0, 2]the function only take [-1,3]

So, $\large range \neq R$

So, function g(x) is not OnTo.

iii)

In the domain of [0, 3] function will take the range [-1, 3].