- Free Download 11 Class Rb Tripathi Maths Book Pdf Download
- Free Download 11 Class RB Tripathi Maths Book Pdf
- Free Download 11 Class Rb Tripathi Maths Book Pdf Free
- Free Download 11 Class Rb Tripathi Maths Book Pdf Class
DISCLAIMER: This website is created solely for Jee aspirants to download pdf, eBooks, study materials for free. Jeemain.guru is trying to help the students who cannot afford buying books is our aim. 0 Results for Pdf file of r b tripathi maths 12 class complete solution in Bhopal.
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
Free Download 11 Class Rb Tripathi Maths Book Pdf Download
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
Free Download 11 Class RB Tripathi Maths Book Pdf
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
Free Download 11 Class Rb Tripathi Maths Book Pdf Free
(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