SATHEE: Functions 2 Question 18

Functions 2 Question 18

19.

Find the natural number $a$ for which

$\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1)$

where the function $f$ satisfies the relation $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and further $f (1) = 2$ .

(1992, 6M)

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Answer:

Correct Answer: 19. $(a = 3)$

Solution:

Let $f (n) = 2^{n}$ for all positive integers $n$ .

Now, for $n = 1, f (1) = 2 = 2$ !

$\Rightarrow$ It is true for $n = 1$ .

Again, let $f (k)$ is true.

$\begin{aligned} \Rightarrow f (k) & = 2^{k}, for some k \in N . \\ Again, f (k + 1) & = f (k) \cdot f (1) [by definition] \\ = 2^{k} \cdot 2 [from induction assumption] \\ = 2^{k + 1} \end{aligned}$

Therefore, the result is true for $n = k + 1$ . Hence, by principle of mathematical induction,

$f (n) = 2^{n}, \forall n \in N$

Now, $\sum_{k = 1}^{n} f (a + k) = \sum_{k = 1}^{n} f (a) f (k) = f (a) \sum_{k = 1}^{n} 2^{k}$

$\begin{aligned} = f (a) \cdot \frac{2 (2^{n} - 1)}{2 - 1} \\ = 2^{a} \cdot 2 (2^{n} - 1) = 2^{a + 1} (2^{n} - 1) \end{aligned}$

But $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1) = 2^{4} (2^{n} - 1)$

Therefore, $a + 1 = 4 \Rightarrow a = 3$

Functions 2 Question 18

19.

Answer:

Engineering Preparation

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Functions 2 Question 18

19.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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