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Chapter 8: Problem 668

If the function \(\mathrm{f}\) satisfies the relation \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})\) for all \(\mathrm{x}, \mathrm{y} \in \mathrm{N}\), Further if \(\mathrm{f}(1)=3\) and \(n_{r=1} f(a+r)=(81 / 2)\left(3^{n}-1\right)\) then \(a=\) (A) 4 (B) 2 (C) 1 (D) 3

Short Answer

Expert verified
Based on the given information, the value of 'a' is either (A) 4 or (D) 3.

Step by step solution

01

Analyzing the functional equation

First let's analyze the functional equation: \(f(x+y) = f(x)f(y)\) for all \(x, y \in N\). Now, let's find \(f(2)\): To find \(f(2)\), we will let \(x = y = 1\) and use the given, \(f(1) = 3\). So, \(f(2) = f(1 + 1) = f(1)f(1) = 3 \cdot 3 = 9\).
02

Find \(f(3)\)

Next, let's find \(f(3)\): To find \(f(3)\), we will let \(x=1, y=2\). Then \(f(3) = f(1 + 2) = f(1)f(2) = 3 \cdot 9 = 27\)
03

Using the summation condition

Now let's use the condition \(\sum_{r=1}^n f(a+r) = \frac{81}{2}(3^n - 1)\). We will start by plugging in \(n=1\) and see if we can find the value of 'a'. For \(n=1\), the condition becomes: \(f(a+1) = \frac{81}{2}(3^1-1)\) \(\Rightarrow f(a+1) = 81\)
04

Finding the value of 'a'

We're given that the function is defined for all natural numbers, so let's check the possible cases for the value of 'a'. 1) If \(a=1\), then \(f(2) = 81\), but we know that \(f(2) = 9\). So, 'a' cannot be 1. 2) If \(a=2\), then \(f(3) = 81\), but we know that \(f(3) = 27\). So, 'a' cannot be 2. 3) If \(a=3\), then \(f(4) = 81\). We don't have any information on \(f(4)\), so it's possible that \(f(4) = 81\) and 'a' = 3. 4) If \(a=4\), then \(f(5) = 81\), but we don't have any information on \(f(5)\). So, it's possible that 'a' = 4.
05

Conclusion

Based on the given options (A) 4, (B) 2, (C) 1, (D) 3, the correct answer corresponds to cases 3 and 4. So, the value of 'a' is either (A) 4 or (D) 3. Considering the given context of the exercise, we have enough information to eliminate the other options, so we're left with two possible solutions.

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Most popular questions from this chapter

If the \(1^{\text {st }}\) term and common ratio of a G. P. are 1 and 2 respectively then \(\mathrm{s}_{1}+\mathrm{s}_{3}+\mathrm{s}_{5}+\ldots+\mathrm{s}_{2 \mathrm{n}-1}=\) (A) \((1 / 3)\left(2^{2 n}-5 n+4\right)\) (B) \((1 / 3)\left(2^{2 n+1}-5 n\right)\) (C) \((1 / 3)\left(2^{2 n+1}-3 n-2\right)\) (D) \((1 / 3)\left(2^{2 n+1}-5 n^{2}\right)\)

If \(\left\\{a_{n}\right\\}\) is an A. P. then \(a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+\ldots+a_{99}^{2}-a_{100}^{2}\) (A) \((50 / 99)\left(\mathrm{a}_{1}^{2}-\mathrm{a}_{100}^{2}\right)\) (B) \([(1000) /(99)]\left(\mathrm{a}_{100}^{2}-\mathrm{a}_{1}^{2}\right)\) (C) \((50 / 51)\left(\mathrm{a}_{1}^{2}+\mathrm{a}_{100}^{2}\right)\) (D) None of this

\(0.4+0.44+0.444+\ldots\) to \(2 \mathrm{n}\) terms \(=\) (A) \((4 / 81)\left(18 \mathrm{n}+1+100^{-\mathrm{n}}\right)\) (B) \((4 / 81)\left(18 \mathrm{n}-1+100^{-\mathrm{n}}\right)\) (C) \((4 / 81)\left(18 \mathrm{n}-1+10^{-\mathrm{n}}\right)\) (D) \((4 / 81)\left(18 n-1+100^{n}\right)\)

In a G. P., the last term is 1024 and the common ratio is \(2 .\) Its 20 th term from the end is (A) \([1 /(512)]\) (B) \([1 /(1024)]\) (C) \([1 /(256)]\) (D) 512

If \([1 /(\mathrm{b}-\mathrm{c})],[1 /(2 \mathrm{~b}-\mathrm{x})]\) and \([1 /(\mathrm{b}-\mathrm{a})]\) are in A. P., then \(\mathrm{a}-(\mathrm{x} / 2), \mathrm{b}-(\mathrm{x} / 2), \mathrm{c}-(\mathrm{x} / 2)\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G P.
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