# RD Sharma Solutions For Class 12 Maths Exercise 2.3 Chapter 2 Function

RD Sharma Solutions for Class 12 Maths Exercise 2.3 Chapter 2 Function consists of problems on finding the composition of real functions. Class 12 is a major turning point in studentsâ€™ lives. It primarily gears them up to make critical decisions regarding their further studies and future goals. These problems are solved in a simple way to speed up the exam preparation of students. It mainly helps during revision and improves confidence among students before appearing for the board exam. To gain better knowledge about the concepts explained in this exercise, they can refer RD Sharma Solutions Class 12 Maths Chapter 2 Functions Exercise 2.3 free PDF, which are given below.

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### Exercise 2.3 Page No: 2.54

1. FindÂ fogÂ andÂ gof,Â if Â

(i) Â fÂ (x)Â =Â ex,Â g (x)Â =Â loge x

(ii) fÂ (x)Â =Â x2,Â g (x)Â =Â cosÂ x

(iii) fÂ (x)Â =Â |x|,Â gÂ (x)Â =Â sinÂ x

(iv) f (x) = x+1, g(x) = ex

(v) f (x)Â =Â sinâˆ’1Â x,Â g(x)Â =Â x2

(vi) fÂ (x)Â =Â x+1,Â gÂ (x)Â =Â sinÂ x

(vii) f(x)=Â xÂ +Â 1,Â gÂ (x)Â =Â 2xÂ +Â 3

(viii) f(x) = c, c âˆˆ R, g(x) = sin x2

(ix) f(x) = x2 + 2 , g (x) = 1 âˆ’ 1/ (1-x)

Solution:

(i) Given fÂ (x) = ex,Â g(x)Â =Â logeÂ x

Let f:Â RÂ â†’Â (0,Â âˆž);Â and g:Â (0,Â âˆž)Â â†’Â R

Now we have to calculate fog,

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â (Â 0,Â âˆž)Â â†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (logeÂ x)

=Â logeÂ ex

=Â x

Now we have to calculate gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (ex)

=Â logeÂ ex

=Â x

(ii) fÂ (x)Â =Â x2,Â g(x)Â =Â cosÂ x

f:Â Râ†’Â [0,Â âˆž)Â ;Â g:Â Râ†’[âˆ’1,Â 1]

Now we have to calculate fog,

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ notÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â DomainÂ (fog)Â =Â {x:Â x âˆˆ domainÂ ofÂ gÂ andÂ gÂ (x)Â âˆˆ domainÂ ofÂ f}

â‡’Â DomainÂ (fog)Â =Â x:Â xÂ âˆˆÂ RÂ andÂ cosÂ xÂ âˆˆÂ R}

â‡’Â DomainÂ ofÂ (fog)Â =Â R

(fog):Â Râ†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (cosÂ x)

=Â cos2 x

Now we have to calculate gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x2)

=Â cosÂ x2

(iii) Given fÂ (x)Â =Â |x|,Â g(x)Â =Â sinÂ x

f:Â RÂ â†’Â (0,Â âˆž)Â ;Â gÂ :Â Râ†’[âˆ’1,Â 1]

Now we have to calculate fog,

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (sinÂ x)

=Â |sinÂ x|

Now we have to calculate gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fogÂ :Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (|x|)

=Â sinÂ |x|

(iv) Given fÂ (x)Â =Â x + 1,Â g(x)Â =Â ex

f:Â Râ†’RÂ ;Â g:Â RÂ â†’Â [Â 1,Â âˆž)

Now we have calculate fog:

Clearly,Â rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (ex)

=Â ex + 1

Now we have to compute gof,

Clearly,Â rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x+1)

=Â ex+1

(v) Given fÂ (x)Â =Â sinÂ âˆ’1Â x,Â g(x)Â =Â x2

f:Â [âˆ’1,1]â†’ [(-Ï€)/2 ,Ï€/2];Â gÂ :Â RÂ â†’Â [0,Â âˆž)

Now we have to computeÂ fog:

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ notÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

DomainÂ (fog)Â =Â {x:Â xÂ âˆˆÂ domainÂ ofÂ gÂ andÂ gÂ (x)Â âˆˆÂ domainÂ ofÂ f}

DomainÂ (fog) = {x:Â xÂ âˆˆÂ RÂ andÂ x2Â âˆˆÂ [âˆ’1, 1]}

DomainÂ (fog) = {x:Â xÂ âˆˆÂ RÂ andÂ xÂ âˆˆÂ [âˆ’1, 1]}

DomainÂ ofÂ (fog) =Â [âˆ’1, 1]

fog:Â [âˆ’1,1]Â â†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (x2)

=Â sinâˆ’1Â (x2)

Now we have to computeÂ gof:

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

fog:Â [âˆ’1, 1]Â â†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (sinâˆ’1Â x)

=Â (sinâˆ’1Â x)2

(vi) Given f(x)Â =Â x+1,Â g(x)Â =Â sinÂ x

f:Â Râ†’RÂ ;Â g:Â Râ†’[âˆ’1,Â 1]

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

SetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (sinÂ x)

=Â sinÂ xÂ +Â 1

Now we have to compute gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â RÂ â†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x+1)

=Â sinÂ (x+1)

(vii) Given fÂ (x)Â =Â x+1,Â gÂ (x)Â =Â 2xÂ +Â 3

f:Â Râ†’RÂ ;Â g:Â RÂ â†’Â R

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â Râ†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (2x+3)

=Â 2xÂ +Â 3Â +Â 1

=Â 2xÂ +Â 4

Now we have to computeÂ gof

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â RÂ â†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x+1)

=Â 2Â (xÂ +Â 1)Â +Â 3

=Â 2xÂ +Â 5

(viii) Given fÂ (x)Â =Â c,Â gÂ (x)Â =Â sinÂ x2

f:Â RÂ â†’Â {c}Â ;Â g:Â Râ†’Â [Â 0,Â 1Â ]

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â Râ†’R

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (sinÂ x2)

=Â c

Now we have to compute gof,

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â fog:Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (c)

=Â sinÂ c2

(ix) Given fÂ (x)Â =Â x2+Â 2 and gÂ (x)Â =Â 1 – 1 / (1 – x)

f:Â RÂ â†’Â [Â 2,Â âˆžÂ )

ForÂ domainÂ ofÂ g:Â 1âˆ’Â xÂ â‰ Â 0

â‡’Â xÂ â‰ Â 1

â‡’Â DomainÂ ofÂ g =Â R âˆ’ {1}

gÂ (xÂ )= 1 â€“ [1/(1 – x)] = (1 â€“ x – 1)/ (1 – x) = -x/(1 – x)

ForÂ rangeÂ ofÂ g

y = (- x)/ (1 – x)

â‡’Â yÂ â€“Â x yÂ =Â âˆ’Â x

â‡’Â yÂ =Â x yÂ âˆ’Â x

â‡’Â yÂ =Â xÂ (yâˆ’1)

â‡’ x = y/(y – 1)

RangeÂ ofÂ gÂ = R âˆ’ {1}

So,Â g:Â R âˆ’ {1} â†’ R âˆ’ {1}

Now we have to computeÂ fog

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

â‡’Â fog:Â RÂ âˆ’Â {1} â†’Â R

(fog)Â (x)Â =Â fÂ (gÂ (x))

= f (-x/ (1 – x))

= ((-x)/ (1 – x))2 + 2

= (x2 + 2x2 + 2 – 4x) / (1 – x)2

= (3x2 – 4x + 2)/ (1 – x)2

Now we have to computeÂ gof

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

â‡’Â gof:Â Râ†’R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x2Â +Â 2)

= 1 – 1 / (1 – (x2 + 2))

= – 1/ (1 – (x2 + 2))

= (x2 + 2)/ (x2 + 1)

2. Â LetÂ f(x) =Â x2Â +Â xÂ + 1 andÂ g(x) = sinÂ x. Show thatÂ fogÂ â‰ Â gof.

Solution:

Given f(x) =Â x2Â +Â xÂ + 1 andÂ g(x) = sinÂ x

Now we have to prove fogÂ â‰ Â gof

(fog)Â (x)Â =Â fÂ (gÂ (x))

=Â f (sinÂ x)

=Â sin2Â xÂ +Â sinÂ xÂ +Â 1

AndÂ (gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (x2+Â xÂ +Â 1)

=Â sinÂ (x2+Â xÂ +Â 1)

So,Â fogÂ â‰ Â gof.

3. IfÂ f(x) = |x|, prove thatÂ fofÂ =Â f.

Solution:

Given f(x) = |x|,

Now we have to prove thatÂ fofÂ =Â f.

Consider (fof)Â (x)Â =Â fÂ (fÂ (x))

=Â fÂ (|x|)

=Â ||x||

=Â |x|

=Â fÂ (x)

So,

(fof)Â (x)Â =Â fÂ (x),Â âˆ€xÂ âˆˆÂ R

Hence,Â fofÂ =Â f

4. IfÂ f(x) = 2xÂ + 5 andÂ g(x) =Â x2Â + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii)Â gof
(iii)Â fof
(iv)Â f2
Also, show thatÂ fofÂ â‰ Â f2

Solution:

f(x)Â andÂ g(x)Â are polynomials.

â‡’ f:Â RÂ â†’Â RÂ andÂ g:Â RÂ â†’Â R.

So,Â fog:Â RÂ â†’Â RÂ andÂ gof:Â RÂ â†’Â R.

(i) (fog) (x) = f (g (x))

= f (x2Â + 1)

= 2 (x2 + 1) + 5

=2x2Â + 2 + 5

= 2x2Â +7

(ii) (gof) (x) = g (f (x))

= g (2x +5)

= (2x + 5)2Â + 1

= 4x2Â + 20x + 26

(iii) (fof) (x) = f (f (x))

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

(iv) f2Â (x) = f (x) x f (x)

= (2x + 5) (2x + 5)

= (2x + 5)2

= 4x2Â + 20x +25

Hence, from (iii) and (iv) clearly fofÂ â‰ Â f2

5. IfÂ f(x) = sinÂ xÂ andÂ g(x) = 2x be two real functions, then describeÂ gofÂ andÂ fog.Â Are these equal functions?

Solution:

Given f(x) = sinÂ xÂ andÂ g(x) = 2x

WeÂ knowÂ that

f:Â Râ†’Â [âˆ’1,Â 1]Â andÂ g:Â Râ†’Â R

Clearly,Â theÂ rangeÂ ofÂ fÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ g.

gof:Â Râ†’Â R

(gof)Â (x)Â =Â gÂ (fÂ (x))

=Â gÂ (sinÂ x)

=Â 2Â sinÂ x

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â RÂ â†’Â R

So,Â (fog)Â (x)Â =Â fÂ (gÂ (x))

=Â fÂ (2x)

=Â sinÂ (2x)

Clearly,Â fog â‰  gof

Hence they are not equal functions.

6. LetÂ f,Â g,Â hÂ be real functions given byÂ f(x) = sinÂ x,Â gÂ (x) = 2xÂ andÂ hÂ (x) = cosÂ x. Prove thatÂ fogÂ =Â goÂ (f h).

Solution:

Given that f(x) = sinÂ x,Â gÂ (x) = 2xÂ andÂ hÂ (x) = cosÂ x

WeÂ knowÂ thatÂ f:Â Râ†’ [âˆ’1,Â 1]Â andÂ g:Â Râ†’ R

Clearly,Â theÂ rangeÂ ofÂ gÂ isÂ aÂ subsetÂ ofÂ theÂ domainÂ ofÂ f.

fog:Â RÂ â†’Â R

Now,Â (f h)Â (x) = fÂ (x) hÂ (x)Â =Â (sinÂ x)Â (cosÂ x)Â = Â½ sinÂ (2x)

DomainÂ ofÂ f hÂ isÂ R.

SinceÂ rangeÂ ofÂ sinÂ xÂ isÂ [-1, 1], âˆ’1Â â‰¤Â sinÂ 2xÂ â‰¤Â 1

â‡’ -1/2 â‰¤ sin x/2 â‰¤ 1/2

RangeÂ ofÂ f hÂ = [-1/2, 1/2]

So,Â (f h):Â RÂ â†’ [(-1)/2, 1/2]

Clearly,Â rangeÂ ofÂ f hÂ isÂ aÂ subsetÂ ofÂ g.

â‡’Â goÂ (f h):Â RÂ â†’Â R

â‡’ DomainsÂ ofÂ fogÂ andÂ goÂ (f h)Â areÂ theÂ same.

So,Â (fog)Â (x) = fÂ (gÂ (x))

=Â fÂ (2x)

=Â sinÂ (2x)

AndÂ (goÂ (f h))Â (x)Â =Â gÂ ((f(x). h(x))

=Â gÂ (sin xÂ cosÂ x)

=Â 2sinÂ xÂ cosÂ x

=Â sinÂ (2x)

â‡’Â (fog)Â (x)Â =Â (go (f h))Â (x),Â âˆ€xÂ âˆˆÂ R

Hence,Â fogÂ =Â goÂ (f h)