Question

Range of the function f (x)= -lx+1l is (a) [0, infinity ) (b) ( - infinity , 0]


FIRST ANSWER WILL BE MARKED AS BRAINLEIST AMSWER​

Answer 1

Answer:

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x l + lx-1l=

Algebraic approach:

Let's check in which ranges of expression holds true:

Two check points for : 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:

A. --> --> , not good as we are checking for ;

B. --> --> , which is true. This means that in this range given equation holds true for all x-es;

C. --> --> , not good as we are checking for .

So we got that the equation holds true only in the range .

(1) . Not sufficient.

(2) . Not sufficient.

(1)+(2) gives us the range , which is exactly the range for which given equation holds true. Sufficient.

Answer 2

Here, f:[0,∞)→[0,∞) i.e, domain is [0,∞) and codomain is [0,∞)

For one-one

f(x)=1+x

xf,(x)=

(x)= (1+x)21

>0,∀x∈[0,∞)

∴f(x) is increasing in this domain. Thus f(x) is one-one in its domain

For onto (we find range)

f(x)= 1+x

x

i.e. y= 1+x

⇒y+yx=x

⇒x=

1−y

y

⇒x=

1−y

y

≥0 as x≥0

∴0≤y



=1 and y<1

∴ Range



= Codomain

∴f(x) is one-one but not onto

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