If f and g two function defined on N such that f x - 2n-1if n isven 2n-2if is odd and g n -f n -f n-1 - Then range of g is

The correct option is A { mϵ N:m= multiple of 4 } ( 1/5)^20÷( 1/5)^15=( 1/5)^5x ∵ on L.H.S . bases are same the operation is division, so p ...

474

You visited us 474 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Composite Function

If f and ...

Question

If $f$ and $g$ two function defined on $N$ such that $f (x) = {\begin{matrix} 2 n - 1 i f n i s v e n 2 n + 2 i f i s o d d \end{matrix} a n d g (n) = f (n) + f (n + 1) .)$
Then range of $g$ is

{m ϵ N : m = m u l t i p l e o f 4}

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

s e t o f e v e n n a t u r a l n u m b e r

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

{m ϵ N : m = 4 k + 3, k i s a n a t u r a l n u m b e r}

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

m ϵ N : m = m u l t i p l e o f 3 o r m u l t i p l e o f 4

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

Open in App

Solution

The correct option is A ${m ϵ N : m = m u l t i p l e o f 4}$
${(\frac{- 1}{5})}^{20} \div {(\frac{- 1}{5})}^{15} = {(\frac{- 1}{5})}^{5 x}$
$∵$ on $L . H . S$ . bases are same the operation is division, so powers are subtracted, i.e
${(\frac{- 1}{5})}^{20 - 15} = {(\frac{- 1}{5})}^{5 x}$
${(\frac{- 1}{5})}^{5} = {(\frac{- 1}{5})}^{5 x}$
$∵$ bases are same , so powers are equated,
$5 = 5 x$
$x = 1$ Answer.

Suggest Corrections

Similar questions

Q. Let

f, g : N \to N

such that

f (n + 1) = f (n) + f (1), \forall n \in N

and

g

be any arbitrary function. Which of the following statements is

NOT

true

?

Q. If

g : N \to N

is given by

g (n) = 2 n + 3

, and

f : N \to N

is given by

f (n) = n + 1

, then find

g \circ f

f \circ g

and

g \circ g

Q. Let

N

be the set of natural numbers and two functions

f

and

g

be defined as

f, g : N \to N

such that :

f (n) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{n + 1}{2} & if n is odd \frac{n}{2} & in n is even \end{matrix}

and

g (n) = n - (- 1)^{n}

. The fog is:

Q. Let

f, g : N \to N

such that

f (n + 1) = f (n) + f (1), \forall n \in N

and

g

be any arbitrary function. Which of the following statements is

NOT

true

?

Q. Let

N

be the set of natural numbers and two functions

f

and

g

be defined as

f, g : N \to N

such that

f (n) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{n + 1}{2} & if n is odd \frac{n}{2} & if n is even \end{matrix}

and

g (n) = n - (- 1)^{n}

. Then

f \circ g

is :

Join BYJU'S Learning Program

If f and g two function defined on N such that f(x)={2n−1if n isven2n+2if is odd and g(n)=f(n)+f(n+1).) Then range of g is

The correct option is A {mϵN:m= multiple of 4}(−15)20÷(−15)15=(−15)5x∵ on L.H.S. bases are same the operation is division, so powers are subtracted, i.e(−15)20−15=(−15)5x(−15)5=(−15)5x∵ bases are same , so powers are equated,5=5xx=1 Answer.

If $f$ and $g$ two function defined on $N$ such that $f (x) = {\begin{matrix} 2 n - 1 i f n i s v e n 2 n + 2 i f i s o d d \end{matrix} a n d g (n) = f (n) + f (n + 1) .)$
Then range of $g$ is