Let S={1,2,3,4,5,6,7} . Then number of possible functions f: S → S such that f(m

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1.Relation and Function

hard

Let $S=\{1,2,3,4,5,6,7\} .$ Then the number of possible functions $f: S \rightarrow S$ such that $f(m \cdot n)=f(m) \cdot f(n)$ for every $m, n \in S$ and $m . n \in S$ is equal to $......$

$500$

$600$

$570$

$490$

(JEE MAIN-2021)

Solution

$f(m n)=f(m) \cdot f(n)$

Put $m=1 f(n)=f(1) \cdot f(n) \Rightarrow f(1)=1$

Put $m=n=2$

$f(4)=f(2) \cdot f(2)$

$f(2)=1 \Rightarrow f(4)=1 \text { or } f(2)=2 \Rightarrow f(4)=4$

Put $m=2, n=3$

$f(6)=f(2) \cdot f(3)$

$\text { when } f(2)=1 \Rightarrow f(3)=1 \text { to } 7$

$f(2)=2 \Rightarrow f(3)=1 \text { or } 2 \text { or } 3$

$f(5), f(7)$ can take any value

Total $=(1 \times 1 \times 7 \times 1 \times 7 \times 1 \times 7)$

$+(1 \times 1 \times 3 \times 1 \times 7 \times 1 \times 7)$

$=490$

Standard 12

Mathematics

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as

$f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1 .$ Then, the value of $\sum_{\mathrm{k}=1}^{20} \frac{1}{\sin (\mathrm{k}) \sin (\mathrm{k}+\mathrm{f}(\mathrm{k}))}$ is equal to:

hard

(JEE MAIN-2021)

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normal

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hard

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normal

The range of $f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$ is

medium

(JEE MAIN-2023)

Let $S=\{1,2,3,4,5,6,7\} .$ Then the number of possible functions $f: S \rightarrow S$ such that $f(m \cdot n)=f(m) \cdot f(n)$ for every $m, n \in S$ and $m . n \in S$ is equal to $......$

Solution

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Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as

$f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1 .$ Then, the value of $\sum_{\mathrm{k}=1}^{20} \frac{1}{\sin (\mathrm{k}) \sin (\mathrm{k}+\mathrm{f}(\mathrm{k}))}$ is equal to: