F - N- N is defined as fn- 2- n-3k- k- Z- 10-n-n-3k-1- k- Z- 0-n-3k-2- k- Z -then n- N-fn-2-

The correct option is A {1,4, 7 } f(n) gt;2 ⇒ 10 n gt;2 ⇒ n lt;8 now, for n=3k+1 , n=1,4,7 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Applications of Dot Product

f : N→ N is...

Question

$f$ : $N \to N$ is defined as $f (n) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2, n = 3 k, & k \in Z 10 - n, n = 3 k + 1, & k \in Z 0, n = 3 k + 2, & k \in Z \end{matrix}$ then ${n \in N : f (n) > 2} =$

{3, 6, 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{1, 4, 7}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{4, 7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

f : Z \to Z

f (n) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2, & if n = 3 k, & k \in Z 10 - n, & if n = 3 k + 1, & k \in Z 0, & if n = 3 k + 2, & k \in Z \end{matrix}

S = {n \in Z : f (n) > 2},

S

f : C \to C

f (z) = \frac{z + i}{z - i}

z \neq i

z_{n} = f (z_{n - 1})

n \in N

z_{0} = K + i

z_{2020} = 1 + 2020 i

(\frac{1}{K})

Conclusions:
I. K $ N
II. K @ Z

n \sum k = 1 k (k + 1) (3 k - 1)

n \sum k = 1 \frac{1}{\sqrt{3 k + 4} + \sqrt{3 k + 1}}

Join BYJU'S Learning Program