Let f: N times N rightarrow N be a function s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the function $f: N 	imes N ightarrow N$ with the conditions:$f(1,1) = 2$$f(m+1, n) = f(m, n) + 2(m+n)$$f(m, n+1) = f(m, n) + 2(m+n-1)$To

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Given the function $f: N 	imes N ightarrow N$ with the conditions:$f(1,1) = 2$$f(m+1, n) = f(m, n) + 2(m+n)$$f(m, n+1) = f(m, n) + 2(m+n-1)$To find $f(2,2)$,we can use the given recurrence relations:First,calculate $f(2,1)$ using the first relation with $m=1, n=1$:$f(2,1) = f(1,1) + 2(1+1) = 2 + 2(2) = 2 + 4 = 6$Next,calculate $f(2,2)$ using the second relation with $m=2, n=1$:$f(2,2) = f(2,1) + 2(2+1-1) = 6 + 2(2) = 6 + 4 = 10$Alternatively,using the other path:$f(1,2) = f(1,1) + 2(1+1-1) = 2 + 2(1) = 4$$f(2,2) = f(1,2) + 2(1+2) = 4 + 2(3) = 4 + 6 = 10$Thus,$f(2,2) = 10$.

Let $f: N \times N \rightarrow N$ be a function such that $f(1,1)=2$,$f(m+1, n)=f(m, n)+2(m+n)$,and $f(m, n+1)=f(m, n)+2(m+n-1)$ for all $m, n \in N$. Find the value of $f(2,2)$.

Similar Questions

If $a$ and $b$ are two fixed positive integers such that $f(a + x) = b + [b^3 + 1 - 3b^2f(x) + 3b\{f(x)\}^2 - \{f(x)\}^3]^{1/3}$ for all real $x$,then $f(x)$ is a periodic function with period:

If $f(x) + 2f(1/x) = 3x$ for $x \neq 0$ and $S = \{x \in R : f(x) = f(-x)\}$,then $S$:

If a function $f(x)$ satisfies $f(x + y) = f(x) f(y)$ for all $x, y \in N$ such that $f(1) = 3$ and $\sum_{x=1}^n f(x) = 120$,then what is the value of $n$?

Suppose $f : R \rightarrow (0, \infty)$ be a differentiable function such that $5f(x + y) = f(x) \cdot f(y), \forall x, y \in R$. If $f(3) = 320$,then $\sum_{n=0}^5 f(n)$ is equal to:

If $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3, 4, 5, \ldots$,then $f(9)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module