Relations and Functions Test - 20

$$n\left ( B \right )=4$$
$$n\left ( A \right )=3$$
Therefore there exists $$2^{3 \times 4}$$ relations from $$A$$ to $$B$$
i.e $$2^{12}$$ relations

$$f\left ( x^{2}-3x-2 \right )=\left ( x^{2}-3x+2 \right )^{2}-3\left ( x^{2}-3x+2 \right )+2$$
$$x^{4}+9x^{2}+4-6x^{3}-12x+4x^{2}-3x^{2}+9x-6+2$$
$$=x^{4}-6x^{3}+2x^{2}+21x+12$$

$$f(n+1)=2f(n)+1$$        .....(1)
Substitute $$n=1$$ in (1)
$$f(2)=2f(1)+1=3=2^2-1$$

Substitute $$n=2$$ in (1)
$$f(3)=2f(2)+1$$
$$\Rightarrow f(3)=7=2^3-1$$

Substitute $$n=3$$ in (1)
$$f(4)=2f(3)+1$$
$$\Rightarrow f(4)=15=2^{4}-1$$

So, for all $$n \ge 1$$, $$f(n)=2^{n}-1$$

$$f=\{(a,0),(b,-2),(c,3)\}$$ and $$g=\{(a,-2),(b,0),(c,1)\}$$

$$f(0)=g(0)=1$$
$$f(x)+g(x)=(f+g)(x)$$
$$\Rightarrow \dfrac{1+1}{1.1}=2$$

$$f(0)=1;$$
$$g(0)=2$$
$$(f+g)(x)=f(x)+g(x)$$, $$fg(x)=f(x)g(x)$$
$$\dfrac{1+2}{1.2}=\dfrac{3}{2}$$

Suppose $$f(x)$$ is a polynomial of degree 'n'.
$$ \displaystyle f(x)=a_{ 0 }+a_{ 1 }x+........{ a }_{ n-1 }{ x }^{ n-1 }+a_{ n }x^{ n }\\ \displaystyle \Rightarrow f\left( \frac { 1 }{ x }  \right) ={ a }_{ 0 }+\frac { { a }_{ 1 } }{ x } +\frac { { a }_{ 2 } }{ { x }^{ 2 } } +........+\frac { { a }_{ n-1 } }{ { x }^{ n-1 } } +\frac { { a }_{ n } }{ { x }^{ n } } $$

Given that $$ \displaystyle f\left ( x \right )+f\left ( \frac{1}{x} \right )=f\left ( x \right ).f \left(\frac{1}{x} \right)$$
$$ \displaystyle \Rightarrow \left( a_{ 0 }+a_{ 1 }x+........{ a }_{ n-1 }{ x }^{ n-1 }+a_{ n }x^{ n } \right) +\left( { a }_{ 0 }+\frac { { a }_{ 1 } }{ x } +\frac { { a }_{ 2 } }{ { x }^{ 2 } } +........+\frac { { a }_{ n-1 } }{ { x }^{ n-1 } } +\frac { { a }_{ n } }{ { x }^{ n } }  \right) \\  \displaystyle =\left( a_{ 0 }+a_{ 1 }x+........{ a }_{ n-1 }{ x }^{ n-1 }+a_{ n }x^{ n } \right) \left( { a }_{ 0 }+\frac { { a }_{ 1 } }{ x } +\frac { { a }_{ 2 } }{ { x }^{ 2 } } +........+\frac { { a }_{ n-1 } }{ { x }^{ n-1 } } +\frac { { a }_{ n } }{ { x }^{ n } }  \right) $$
.
Comparing the coefficients of $$x^n$$ on both sides, we get $$a_n=a_na_0 \Rightarrow a_0=1$$ ( $$\because a_n \neq 0$$)
Similarly comapring the coefficients of $$x^{n-1}$$ on both sides, we get $$a_{n-1}= a_{n}a_{1}+a_{n-1}a_{0} \Rightarrow a_na_1=0 \Rightarrow a_1 = 0$$
($$ \because a_0=1$$)
Similarly comparing the coefficients of $$x^{n-2},\; x^{n-3},.....x$$, we get 
$$ a_2=a_3=a_4=..........a_{n-1}=0$$
$$\Rightarrow f\left ( x \right )=a_nx^{n}+1$$
$$ \displaystyle \Rightarrow f\left( \frac { 1 }{ x }  \right) =\frac { { a }_{ n } }{ { x }^{ n } } +1\\ \displaystyle \Rightarrow \left( a_{ n }x^{ n }+1 \right) +\left( \frac { { a }_{ n } }{ { x }^{ n } } +1 \right) =\left( a_{ n }x^{ n }+1 \right) \left( \frac { { a }_{ n } }{ { x }^{ n } } +1 \right) $$
Comparing the constant terms on both sides we get $$ 2= a_n^{2}+1 \Rightarrow a_n = \pm 1$$
 $$ \Rightarrow \boxed { f(x)=\pm x^{ n }+1 } $$
Given $$f\left ( 2 \right )=9\Rightarrow f\left ( 2 \right )= \pm 2^{n}+1=9$$
$$\Rightarrow 2^{n}=8$$ ( $$ \because -ve$$ is not possible)
$$\Rightarrow n=3$$
$$f\left ( x \right )=x^{3}+1$$
$$\therefore f\left ( 3 \right )=27+1=28$$

$$g(2)=g(3)=0  ; g (0)=6$$
$$f(0)=0;f(1)=1;f(2)=4$$
$$\Rightarrow 6-1-4=1$$