pls someone solve these problems

1. show that f:R-R defined by f(x) = 2x3 -7 for all x belonging to R is bijective

2. f(x) = x+7 g(x) = x-7 x belonging to R find (fog)(7)

3.f(x) = x3 g(x)= cos3x find fog(x)

4.f:R-R f(x) = x/x2 +1 find f (f (x))

5. f:R-R f(x) = x2 - 3x + 2 find f (f (x))

6.f(x) = mode x  g(x) = [x] f:R-R find  fog(5/ 2) gof ( - square root 2)  where [ greatest integer function]

2.)  fx =  x+7  ......(1)

gx  =  x-7

fog(x)  =  (x-7) +7  {  by substituting x by g(x) in (1) }

fog(x) = x

=>fog(7) = 7.

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3)  f(x) =x3  g(x) = cos3x

fog(x) =  cos33x  =  cos9x +3cos3x  / 4

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f(x)  = x /  (x2+ 1)

=> f(fx)  =  x/(x2+1)  /  (x/x2+1)2 +1

=  x/x2+1  /  x2 + x4+1+2x2 /( x2+1)2

=x(x2+1)  /  x4+3x2+1

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5.)  f(x)  = x2- 3x+ 2

=> f(fx) =  (x2-3x+2)2 -3(x2-3x+2)+2 .

simplify to get the ans.

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thanks ana and for answering me first  3 thumps up

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my pleasure!:)

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 1)f(x)=2x^3-7                                                                                                                                                                                                          let's take x1,x2 belongs to R such that,                                                                                                                                            f(x1)=f(x2)=>2x1^3-7=2x2^3-7

=>x1=x2

=>f is one- one

now for onto we show that every element of range "R"                                                                                                                                         is associated to atleast one element belongs to Domain

let y=2x^3-7=>x=[(y+7)/2]^1/3

now for every real y we get a real x                                                                                                                                                                 where  y belongs to Range,x belongs to Domain 

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THANK YOU AYUSH FOR ANSWERING. 1 THUMPS UP

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f(x) = mod. of x and g(x) = greatest integer of x,

then fog(5/2).gof(-squareroot of2)= f(g(5/2)).g(f(-squareroot of2)) = f(2). g(square root of 2) = 2.1 = 2

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thank you routhan.1 thumps up

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 2. f(x) = x+7 g(x) = x-7 x belonging to R find (fog)(7)

sol. (fog)(7)=f{g(7)}=f(0)=0+7=7  ans

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 3.f(x) = x3 g(x)= cos3x find fog(x)

sol.   fog(x)= f{g(x)}= f(cos3x)= cos33x  

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f(x)=2x+1 find domain

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1)one one:2x1^3-7=2x2^3-7

                  x1=x2

​         therfore, one one

onto:2x-7=y

        2x=y+7

         x=(y+7)/2

2((y+7)/2)-7=y

(2y+14)2-7=y

y+7-7=y

        therefore onto

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show that the function f:R-R defined by f(x) =|x|+x is neither one-one or onto
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