## Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Ex 1.1

Question 1.

If f : R → R and g : R → R be two functions defined as below, then find (fog) (x) and (gof)(x):

(i) f(x) = 2x + 3, g(x) = x^{2} + 5

(ii) f(x)= x^{2} +8, g(x) = 3x^{3} +1

(iii) f(x) = x, g(x) = |x|

(iv) f(x) = x^{2} + 2x + 3, g(x) = 3x – 4

Solution:

(i) Given,

f(x) = 2x + 3 and g(x) = x^{2} + 5

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

= f(x^{2} + 5)

= 2(x^{2} + 5) + 3

= 2x^{2} + 10 + 3

= 2x^{2} + 13

(gof)(x) = g(x))

= g(2x + 3)

= (2x + 3)2 + 5

= 4x^{2} + 9 + 12x + 5

= 4x^{2} + 12x + 14

(ii) Given, f(x)= x^{2} + 8 and g(x) = 3x^{3} + 1

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

= f(3x^{3} + 1)

= (3x^{3} + 1)^{2} +8

= 9x^{6} + 6x^{3} + 1 +8

= 9x^{6} + 6x^{3} +9

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

= g(x^{2} + 8)

= 3(x^{2} + 8)^{3} + 1

(iii) Given, f(x)=x and g(x) = |x|

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

= f(|x|) = |x|

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

= g(x) = |x|

(iv) Given, f(x)= x^{2} + 2x + 3 and g(x) = 3x – 4

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

= f(3x – 4)

= (3x – 4)^{2} + 2(3x – 4) + 3

= 9x^{2} – 24x + 16 + 6x – 8 + 3

= 9x^{2} – 18x + 11

(gof)(x) = g(f(x)) = g(x^{2} + 2x + 3)

= 3(x^{2} + 2x + 3) – 4 = 3x^{2} + 6x + 9 – 4

= 3x^{2} + 6x + 5.

Question 2.

If A = {a,b,c}, B = {u, v, w}.

If f: A → B and g : B → A, defined as

f = {(a, v), (b, u), (c, w)}

g = {(u, b), (v, a), (w, c)}

then find (fog) and (gof).

Solution:

Given, f= {(a, v), (b, u), (c, w)}

g= {(u, b), (v, a), (w, c)}

f(a)= v and g(u) = b

f(b)= u and g(v) = a

f(c)= w and g(w) = C

So, from fog(x) = f(g(x)]

fog(u) = f(g(u)] = f(b) = u

fog(v) = f(g(v)] = f(a) = v

fog(w)= f[g(w)] = f(c) = w

So, fog = {(u, u), (v, v), (w, w)}

gof(a) = g[f(a)] = g(v) = a

gof(b) = g[fb)] = g(u) = b

gof(c) = g[(c)] = g(w) = C

gof = {(a, a), (b, b), (c, c)}

Question 3.

If f: R^{+} → R^{+} and g: R^{+} → R^{+}, defined as f(x) = x^{2} and g(x)= √x, then find gof and fog. Are they identity functions ?

Solution:

Given,

f : R^{+} → R^{+}, 4(x) = x^{2}

g : R^{+} → R^{+}, g(x) = √x

(gof)(x) = g[f(x)] = g(x^{2}) = √x^{2} = x

(fog)(x) = f(g(x)] = f(√x) = (√x)^{2} = x

So, (fog)(x) = (gof)(x) = x, ∀ x ∈ R^{+}

Hence, (fog) and (gof) are identity function.

Question 4.

If f : R → Rand g : R → R be such two functions that defined as f(x) = 3x +4 and g(x) = \(\frac { 1 }{ 3 }\) (x – 4), then find (fog)(x) and (gof)(x), also find (gog)(1).

Solution:

Given, f : R → R, f(x) = 3x + 4

Question 5.

If f, g, h be three functions from R to R, defined as f(x) = x^{2}, g(x) = cos x and h(x) = 2x + 3, then find {ho(gof)} (√2π).

Solution:

Given function,

f(x) = x^{2}, g(x) = cos x, h(x) = 2x + 3

∴ {ho(gof)}(x) = hog{f(x)}

= h[g{f(x)}]

= h[g(x^{2})] = h(cos x^{2})

= 2 cos x^{2} + 3

{ho(gof)}√2π = 2 cos (√2π)^{2} + 3

= 2 cos 2π + 3

= 2 x 1 + 3 = 5

Question 6.

If functions f and g be defined as below, then find (gof)(x) :

f : R → R, f(x) = 2x + x^{-2}

g : R → R, g(x) = x^{4} + 2x + 4

Solution:

Given, f: R → R

f(x) = 2x + x^{-2}

g: R → M R, g(x)= x^{4} + 2x + 4.

∴ (gof)(x) = g(f(x)} = g{2x + x^{-2}} = (2x + x^{-2})^{4} + 2(2x + x^{-2}) + 4

Question 7.

If A = {1, 2, 3, 4}, f : R → R, f(x) = x^{2} + 3x + 1 g : R → R, 8(x) = 2x – 3, then find

(i) (fog)(x)

(ii) (gof)(x)

(iii) (fof)(x)

(iv) (gog)(x)

Solution:

Given,

f : R→ R, f(x) = x^{2} + 3x + 1

g : R → R, g(x) = 2x – 3

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

= f{2x – 3}

= (2x – 3)^{2} + 3(2x – 3) + 1

= 4x^{2} – 12x + 9 + 6x – 9 + 1

= 4x^{2} – 6x + 1

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

= g(x^{2} + 3x + 1)

= 2(x^{2} + 3x + 1) – 3

= 2x^{2} + 6x + 2 – 3

= 2x^{2} + 6x – 1

(iii) (fof)(x) = f{f(x)}

= f(x^{2} + 3x + 1)

= (x^{2} + 3x + 1)^{2} + 3(x^{2} + 3x + 1)+1

= x^{4} + 9x^{2} + 1 + 6x^{3} + 6x + 2x^{2} + 3x^{2} + 9x + 3 + 1

= x^{4} +6x^{3} + 14x^{2} + 15x + 5

(iv) (gog)(x) = g{g(x)}

= g(2x – 3)

= 2(2x – 3) – 3

= 4x – 6 – 3

= 4x – 9

##### RBSE Solutions for Class 12 Maths