If f: R arrow R is a function satisfying the property f (2 x +3)+ f (2 x +7)=2, x ∈ R, then the period of f ( x ) is

Question

If f: R arrow R is a function satisfying the property f (2 x +3)+ f (2 x +7)=2, x ∈ R, then the period of f ( x ) is (A) 2 (B) 4 (C) 5 (D) 10

Tardigrade Admin · Accepted Answer

We have, f(2 x+3)+f(2 x+7)=2 dots (1) Replace x by x+1, f(2 x+5)+f(2 x+9)=2 dots(2) Now replace x by x+2, f(2 x+7)+f(2 x+11)=2 dots(3) From (1)-(3), we get f(2 x+3)-f(2 x+11)=0 i.e., f(2 x+3)=f(2 x+11) ⇒ T=4

Q. If $f : R \to R$ is a function satisfying the property $f (2 x + 3) + f (2 x + 7) = 2, x \in R$ , then the period of $f (x)$ is

2

4

5

10

We have, $f (2 x + 3) + f (2 x + 7) = 2 \dots$ (1)
Replace $x$ by $x + 1, f (2 x + 5) + f (2 x + 9) = 2 \dots$ (2)
Now replace $x$ by $x + 2, f (2 x + 7) + f (2 x + 11) = 2 \dots$ (3)
From $(1) - (3)$ , we get $f (2 x + 3) - f (2 x + 11) = 0$
i.e., $f (2 x + 3) = f (2 x + 11) \Rightarrow T = 4$

Q. If f:R→R is a function satisfying the property f(2x+3)+f(2x+7)=2,x∈R, then the period of f(x) is

2

4

5

10

We have, f(2x+3)+f(2x+7)=2… (1) Replace x by x+1,f(2x+5)+f(2x+9)=2…(2) Now replace x by x+2,f(2x+7)+f(2x+11)=2…(3) From (1)−(3), we get f(2x+3)−f(2x+11)=0 i.e., f(2x+3)=f(2x+11)⇒T=4

Q. If $f : R \to R$ is a function satisfying the property $f (2 x + 3) + f (2 x + 7) = 2, x \in R$ , then the period of $f (x)$ is

We have, $f (2 x + 3) + f (2 x + 7) = 2 \dots$ (1)
Replace $x$ by $x + 1, f (2 x + 5) + f (2 x + 9) = 2 \dots$ (2)
Now replace $x$ by $x + 2, f (2 x + 7) + f (2 x + 11) = 2 \dots$ (3)
From $(1) - (3)$ , we get $f (2 x + 3) - f (2 x + 11) = 0$
i.e., $f (2 x + 3) = f (2 x + 11) \Rightarrow T = 4$