RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 3. Functions

Subject: Maths - Class 10th

Q. No. 5 of Exercise 3E

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5

. Let f:(2, ∞) → R: f(x) = and g: (2, ∞) → R: g(x) = Find:

(i) (f + g) (x)


(ii) (f - g) (x)


(iii) (fg) (x)


Given:


f(x)=: x > 2 and g(x)= :x > 2


(i) To find: (f + g) (x)


Domain(f) = (2, ∞)


Range(f) = (0, ∞)


Domain(g) = (2, ∞)


Range(g) = (2, ∞)


(f + g) (x) = f(x) + g(x)


=


Therefore,


(f + g) (x) =


(ii) To find:(f - g)(x)


Range(g) Domain(f)


Therefore,


(f - g)(x) exists.


(f - g)(x) = f(x) – g(x)


=


Therefore,


(f - g) (x) =


(iii) To find:(fg)(x)


(fg)(x) = f(x).g(x)


=


=


=


=


Therefore,


(fg)(x) =


Chapter Exercises

Exercise 3B
Exercise 3C
Exercise 3D
Exercise 3E
Exercise 3F
Exercise 3A

More Exercise Questions

1

Let f : R → R : f(x) = x + 1 and g : R → R : g(x) = 2x – 3.Find

(i) (f + g) (x)


(ii) (f – g) (x)


(iii) (fg) (x)


(iv)(f/g) (x)
2

Let f : R → R : f(x) = 2x + 5 and g : R → R : g(x) = x2 + x.

Find


(i) (f + g) (x)


(ii) (f – g) (x)


(iii) (fg) (x)


(iv) (f/g)(x)
3

. Let f: R → R: f(x) = x3 + 1 and g: R → R: g(x) = (x + 1). Find:

(i) (f + g) (x)


(ii) (f – g) (x)


(iii) (1/f) (x)


(iv) (f/g) (x)
4

. Let f: R → R; f(x)= ,where c is a constant. Find

(i) (cf) (x)


(ii) (c2 f) (x)


(iii)


5

. Let f:(2, ∞) → R: f(x) = and g: (2, ∞) → R: g(x) = Find:

(i) (f + g) (x)


(ii) (f - g) (x)


(iii) (fg) (x)

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