1. Free Download 11 Class Rb Tripathi Maths Book Pdf Class
2. Free Download 11 Class RB Tripathi Maths Book Pdf
3. Free Download 11 Class Rb Tripathi Maths Book Pdf Download

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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) = x² + 5
(ii) f(x)= x² +8, g(x) = 3x³ +1
(iii) f(x) = x, g(x) = |x|
(iv) f(x) = x² + 2x + 3, g(x) = 3x – 4
Solution:
(i) Given,
f(x) = 2x + 3 and g(x) = x² + 5
(fog)(x) = f(g(x))
= f(x² + 5)
= 2(x² + 5) + 3
= 2x² + 10 + 3
= 2x² + 13

(gof)(x) = g(x))
= g(2x + 3)
= (2x + 3)2 + 5
= 4x² + 9 + 12x + 5
= 4x² + 12x + 14

(ii) Given, f(x)= x² + 8 and g(x) = 3x³ + 1
(fog)(x) = f(g(x))
= f(3x³ + 1)
= (3x³ + 1)² +8
= 9x⁶ + 6x³ + 1 +8
= 9x⁶ + 6x³ +9

(gof)(x) = g(f(x))
= g(x² + 8)
= 3(x² + 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)= x² + 2x + 3 and g(x) = 3x – 4
(fog)(x) = f(g(x))
= f(3x – 4)
= (3x – 4)² + 2(3x – 4) + 3
= 9x² – 24x + 16 + 6x – 8 + 3
= 9x² – 18x + 11

(gof)(x) = g(f(x)) = g(x² + 2x + 3)
= 3(x² + 2x + 3) – 4 = 3x² + 6x + 9 – 4
= 3x² + 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) = x² and g(x)= √x, then find gof and fog. Are they identity functions ?
Solution:
Given,
f : R⁺ → R⁺, 4(x) = x²
g : R⁺ → R⁺, g(x) = √x
(gof)(x) = g[f(x)] = g(x²) = √x² = 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) = [latex]frac { 1 }{ 3 }[/latex] (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) = x², g(x) = cos x and h(x) = 2x + 3, then find {ho(gof)} (√2π).
Solution:
Given function,
f(x) = x², g(x) = cos x, h(x) = 2x + 3
∴ {ho(gof)}(x) = hog{f(x)}
= h[g{f(x)}]
= h[g(x²)] = h(cos x²)
= 2 cos x² + 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) = x⁴ + 2x + 4
Solution:
Given, f: R → R
f(x) = 2x + x^-2
g: R → M R, g(x)= x⁴ + 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) = x² + 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) = x² + 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
= 4x² – 12x + 9 + 6x – 9 + 1
= 4x² – 6x + 1

(ii) (gof)(x) = g{f(x)}
= g(x² + 3x + 1)
= 2(x² + 3x + 1) – 3
= 2x² + 6x + 2 – 3
= 2x² + 6x – 1

(iii) (fof)(x) = f{f(x)}
= f(x² + 3x + 1)
= (x² + 3x + 1)² + 3(x² + 3x + 1)+1
= x⁴ + 9x² + 1 + 6x³ + 6x + 2x² + 3x² + 9x + 3 + 1
= x⁴ +6x³ + 14x² + 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

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