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Chapter 3: Functions – Exercise 3.4 Functions – Exercise 3.4 – Q.1 We have, Functions – Exercise 3.4 – Q.2 We have, f(x) = 2x + 5 and g(x) = x2 + x We observe that f(x) = 2x + 5 is defined for all x ϵ R So, domain (g) = R Clearly g(x) = x2 + x is defined for all x ϵ R So, domain (g) = R ∴ Domain (f) ∩ Domain (g) = R (i) Clearly, (f + g): R → R is given by (f + g)(x) = f(x) + g(x) = 2x + 5 + x2 + x = x2 + 3x + 5 Domain (f + g) = R (ii) We find that f - g : R → R is defined as (f - g)(x) = f(x) - g(x) = 2x + 5 - (x2 + x) = 2x + 5 - x2 - x = -x2 + x + 5 Domain (f - g) = R (iii) We find that fg : R → R is given by (fg)(x) = f(x) × g(x) = (2x + 5) × (x2 + x) = 2x3 + 2x2 + 5x2 + 5x = 2x3 + 7x2 + 5x Domain (fg) = R (iv) we have, g(x) = x2 + x ∴ f(x) = 0 ⟹ x2 + x = 0 ⟹ x(x + 1) = 0 ⟹ x = 0 or, x = -1 Functions – Exercise 3.4 – Q.3 We have, Functions – Exercise 3.4 – Q.4 We have, We have, We have,
We have,
f(x) = 2x + 5 and g(x) = x2 + x
We observe that f(x) = 2x + 5 is defined for all x ϵ R
So, domain (g) = R
Clearly g(x) = x2 + x is defined for all x ϵ R
∴ Domain (f) ∩ Domain (g) = R
(i) Clearly, (f + g): R → R is given by
(f + g)(x) = f(x) + g(x)
= 2x + 5 + x2 + x
= x2 + 3x + 5
Domain (f + g) = R
(ii) We find that f - g : R → R is defined as
(f - g)(x) = f(x) - g(x)
= 2x + 5 - (x2 + x)
= 2x + 5 - x2 - x
= -x2 + x + 5
Domain (f - g) = R
(iii) We find that fg : R → R is given by
(fg)(x) = f(x) × g(x)
= (2x + 5) × (x2 + x)
= 2x3 + 2x2 + 5x2 + 5x
= 2x3 + 7x2 + 5x
Domain (fg) = R
(iv) we have,
g(x) = x2 + x
∴ f(x) = 0 ⟹ x2 + x = 0
⟹ x(x + 1) = 0
⟹ x = 0 or, x = -1
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Chapter 3: Functions – Exercise 3.2...
Chapter 3: Functions – Exercise 3.1...
Chapter 3: Functions – Exercise 3.3...
