Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. Define $f:S\rightarrow S$ as $f(n)=\begin{cases}2n, & \text{if }n=1,2,3,4,5\\2n-11 & \text{if }n=6,7,8,9,10\end{cases}$. Let $g:S\rightarrow S$ be a function such that $fog(n)=\begin{cases}n+1 & ,\text{if }n\text{ is odd}\\n-1 & ,\text{if }n\text{ is even}\end{cases}$, then $g(10)((g(1)+g(2)+g(3)+g(4)+g(5))$ is equal to:

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Sets Relations and Functions

Let

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

. Define

f : S \to S

f (n) = {2 n, 2 n - 11 if n = 1, 2, 3, 4, 5 if n = 6, 7, 8, 9, 10

.

Let

g : S \to S

be a function such that

f o g (n) = {n + 1 n - 1, if n is odd, if n is even

, then

g (10) ((g (1) + g (2) + g (3) + g (4) + g (5))

is equal to:

Answer: 190

First find

f^{- 1}

. From

f (n) = 2 n

for

n \in {1, 2, 3, 4, 5}

, we get

f^{- 1} (n) = \frac{n}{2}

for

n \in {2, 4, 6, 8, 10}

.

From

f (n) = 2 n - 11

for

n \in {6, 7, 8, 9, 10}

, we get

f^{- 1} (n) = \frac{n + 11}{2}

for

n \in {1, 3, 5, 7, 9}

.

Since

f (g (n)) = {n + 1 n - 1 n odd n even

, we have:

g (n) = {f^{- 1} (n + 1) f^{- 1} (n - 1) n odd n even

For odd

n

g (n) = f^{- 1} (n + 1) = \frac{n + 1}{2}

(since

n + 1

is even)

For even

n

g (n) = f^{- 1} (n - 1) = \frac{( n - 1 ) + 11}{2} = \frac{n + 10}{2}

(since

n - 1

is odd)

Therefore:

$g (1) = 1$ , $g (2) = 6$ , $g (3) = 2$ , $g (4) = 7$ , $g (5) = 3$ , $g (10) = 10$

g (10) [g (1) + g (2) + g (3) + g (4) + g (5)] = 10 [1 + 6 + 2 + 7 + 3] = 10 \times 19 = 190

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer