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Functions question

2020 · 6 Sep · Shift 1 · Q33
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Functions question

2020 · 6 Sep · Shift 1 · Q33

JEE MainMathematicsFunctionsMCQ+4 / −1
If f(x + y) = f(x)f(y) and ∑x=1∞f(x)=2\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2x=1∑∞​f(x)=2, x, y ∈\in∈ N, where N is the set of all natural number, then the value of f(4)f(2){{f\left( 4 \right)} \over {f\left( 2 \right)}}f(2)f(4)​ is :
  1. A
    23{2 \over 3}32​
  2. B
    19{1 \over 9}91​
  3. C
    13{1 \over 3}31​
  4. D
    49{4 \over 9}94​
View written solutionFree

Correct answer: D

  1. Let us use the given functional equation

f(x+y)=f(x)f(y),  x,y∈N.f(x+y)=f(x)f(y), \,\, x,y\in \mathbb N.f(x+y)=f(x)f(y),x,y∈N.

Since x,yx,yx,y are natural numbers, take y=1y=1y=1. Then

f(x+1)=f(x)f(1).f(x+1)=f(x)f(1).f(x+1)=f(x)f(1).

This shows the sequence is geometric. In fact, by repeated use,

f(n)=(f(1))nfor all n∈N.f(n)=\big(f(1)\big)^n \quad \text{for all } n\in \mathbb N.f(n)=(f(1))nfor all n∈N.

Let

f(1)=r.f(1)=r.f(1)=r.

Then

f(n)=rn.f(n)=r^n.f(n)=rn.

  1. Now use the given infinite series:

∑x=1∞f(x)=2.\sum_{x=1}^{\infty} f(x)=2.∑x=1∞​f(x)=2.

So,

∑x=1∞rx=2.\sum_{x=1}^{\infty} r^x=2.∑x=1∞​rx=2.

This is a geometric series with first term rrr and common ratio rrr. For convergence, ∣r∣<1|r|<1∣r∣<1, and

∑x=1∞rx=r1−r=2.\sum_{x=1}^{\infty} r^x=\frac{r}{1-r}=2.∑x=1∞​rx=1−rr​=2.

Solve:

r1−r=2\frac{r}{1-r}=21−rr​=2

r=2(1−r)r=2(1-r)r=2(1−r)

r=2−2rr=2-2rr=2−2r

3r=23r=23r=2

r=23.r=\frac{2}{3}.r=32​.

Hence,

f(n)=(23)n.f(n)=\left(\frac{2}{3}\right)^n.f(n)=(32​)n.

  1. Compute the required value:
=\left(\frac{2}{3}\right)^2 =\frac{4}{9}.$$ --- 4. Check options: - A: $\frac{2}{3}$ - B: $\frac{1}{9}$ - C: $\frac{1}{3}$ - D: $\frac{4}{9}$ So the correct option is **D**.
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