Question 1: If $f (x) = x^2 - 3x +4$ , then find the values of x satisfying the equation. $f (x) = f (2x + 1)$.

Given $f (x) = x^2 - 3x +4$

$\therefore f(2x+1) = (2x+1)^2 - 3(2x+1) + 4 = 4x^2 - 2x + 2$

Given $f(x) = f(2x+1)$

$\Rightarrow x^2 - 3x +4 = 4x^2 - 2x + 2$

$\Rightarrow 3x^2 + x - 2 = 0$

$\Rightarrow 3x^2 + 3x - 2x - 2 = 0$

$\Rightarrow 3x(x+1) - 2 (x+ 1) = 0$

$\Rightarrow (x+1)(3x-2) = 0$

$\Rightarrow x = -1 \ or\ x =$ $\frac{2}{3}$

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Question 2: If $f(x) = (x-a)^2 (x-b)^2$, find $f(a+b)$.

Given $f(x) = (x-a)^2 (x-b)^2$

$\therefore f(a+b) = (a+b -a)^2 (a+b-b)^2 = b^2 a^2$

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Question 3: If $y = f(x) =$ $\frac{ax-b}{bx-a}$, show that $x= f(y)$.

Given $y = f(x) =$ $\frac{ax-b}{bx-a}$

$\Rightarrow y =$ $\frac{ax-b}{bx-a}$

$\Rightarrow y(bx-a) = (ax-b)$

$\Rightarrow xyb -ya = ax - b$

$\Rightarrow xyb - ax = ay - b$

$\Rightarrow x(by-a) = ay-b$

$\Rightarrow x =$ $\frac{ay - b}{by -a}$

$\Rightarrow \therefore x = f(y)$. Hence proved.

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Question 4: If $f(x) =$ $\frac{1}{1-x}$, show that $f[f \{f(x) \} ]= x$

$f(x) =$ $\frac{1}{1-x}$

$\therefore f(f(x)) =$ $\frac{1}{1-\frac{1}{1-x}}$ $=$ $\frac{1-x}{1-x-1}$ $=$ $\frac{x-1}{x}$

$f(f(f(x))) =$ $\frac{1}{1-\frac{x-1}{x}}$ $=$ $\frac{x}{x-x+1}$ $= x$

$\therefore f(f(f(x))) = x$. Hence proved.

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Question 5: If $f(x) =$ $\frac{x-1}{x+1}$ show that $f[f(x)] = x$.

Given $f(x) =$ $\frac{x-1}{x+1}$

$\therefore f(f(x)) = f \Big($ $\frac{x-1}{x+1}$ $\Big)$

$=$ $\frac{ \frac{x+1}{x-1} + 1}{ \frac{x+1}{x-1}- 1}$

$=$ $\frac{x+1 + x -1 }{x+1 - (x-1)}$

$=$ $\frac{2x}{x+1 - x + 1}$ $=$ $\frac{2x}{2}$ $= x$

$\therefore f(f(x)) = x$ . Hence proved.

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Question 6: If $f(x) = \Bigg\{$ $\begin{array}{lll} x^2, \hspace*{0.5cm} \ when \ x <0 \\ x, \hspace*{0.5cm} \ when \ 0 \leq x < 1 \\ \frac{1}{x}, \hspace*{0.5cm} \ when \ x > 1 \end{array}$

Find i) $f(\frac{1}{2})$     ii) $f(-2)$     iii) $f(1)$     iv) $f(\sqrt{3})$     v) $f(-\sqrt{3})$

Given $f(x) = \Bigg\{$ $\begin{array}{lll} x^2, \hspace*{0.5cm} \ when \ x <0 \\ x, \hspace*{0.5cm} \ when \ 0 \leq x < 1 \\ \frac{1}{x}, \hspace*{0.5cm} \ when \ x > 1 \end{array}$

i) $f \Big($ $\frac{1}{2}$ $\Big)$ $=$ $\frac{1}{2}$   [ Using $f(x) = x, 0 \leq x \leq 1$ ]

ii) $f(-2) = (-2)^2 = 4$    [ Using $f(x) = x^2 , x < 0$ ]

iii) $f(1) =$ $\frac{1}{1}$ $=$ $1$    [ Using $f(x) =$ $\frac{1}{x}$ $, x \geq 1$  ]

iv) $f(\sqrt{-3})$ does not exist since $x$ is not defined in $R$.

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Question 7: If $f(x) = x^3 -$ $\frac{1}{x^3}$  show that $f(x) + f($ $\frac{1}{x}$ $) = 0$

Given $f(x) = x^3 -$ $\frac{1}{x^3}$

$\therefore f \Big($ $\frac{1}{x}$ $\Big) = \Big($ $\frac{1}{x}$ $\Big)^3 -$ $\frac{1}{\frac{1}{x^3}}$  $=$ $\frac{1}{x^3}$ $- x^3$

$\therefore f(x) + f \Big($ $\frac{1}{x}$ $\Big) =$ $\frac{1}{x^3}$ $- x^3 +$ $\frac{1}{x^3}$ $- x^3$

$\Rightarrow f(x) + f \Big($ $\frac{1}{x}$ $\Big) = 0$

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Question 8: If $f(x) =$ $\frac{2x}{1+x^2}$, show that $f( \tan \theta) = \sin 2\theta$

Given $f(x) =$ $\frac{2x}{1+x^2}$

$\therefore f( \tan \theta) =$ $\frac{2 \tan \theta}{1 + \tan^2 \theta }$

$= 2$ $\frac{\sin \theta}{ \cos \theta}$ $\Big($ $\frac{\cos^2 \theta}{\cos^2 \theta + \sin^2 \theta}$ $\Big)$

$= 2 \sin \theta \cos \theta = \sin 2 \theta$

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Question 9: If $f(x) =$ $\frac{x-1}{x+1}$, then show that:

i) $f($ $\frac{1}{x}$ $) = -f(x)$     ii) $f(-$ $\frac{1}{x}$ $) = -$ $\frac{1}{f(x)}$

Given $f(x) =$ $\frac{x-1}{x+1}$

i) $f($ $\frac{1}{x}$ $) =$ $\frac{\frac{1}{x}-1}{\frac{1}{x}+1}$ $=$ $\frac{1-x}{1+x}$ $= -f(x)$

ii) $f(-$ $\frac{1}{x}$ $) =$ $\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}$ $=$ $\frac{-1-x}{-1+x}$ $= -\frac{x+1}{x-1}$ $= -$ $\frac{1}{f(x)}$

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Question 10: If $f(x) = (a -x^n)^{1/n} , a> 0$ and $n \in N$, then prove that $f(f(x)) = x$ for all $x$

Given $f(x) = (a -x^n)^{1/n} , a> 0$ and $n \in N$

$\therefore f(f(x)) = \Big( a - \Big\{ [ a-x^n]^{\frac{1}{n}} \Big\} \Big)^{\frac{1}{n}}$

$= \Big( a - ( a - x^n) \Big)^{\frac{1}{n}}$

$= (x^n)^{\frac{1}{n}}$

$= x$

Hence $f(f(x)) =x$ for all $x$

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Question 11: If for non-zero $x, \ \ a f(x) +b f($ $\frac{1}{x}$ $) =$ $\frac{1}{x}$ $-5$, where $a \neq b$, then find $f(x)$.

Given $a f(x) +b f($ $\frac{1}{x}$ $) =$ $\frac{1}{x}$ $-5$   … … … … … i)

$\Rightarrow a f($ $\frac{1}{x}$ $) +b f($ $x$ $) =$ $x$ $-5$   … … … … … ii)

$a f(x) +b f($ $\frac{1}{x}$ $) +$ $a f($ $\frac{1}{x}$ $) +b f($ $x$ $) =$ $\frac{1}{x}$ $-5+x-5$

$(a+b) f(x) +(a+b) f($ $\frac{1}{x}$ $) =$ $\frac{1}{x}$ $+x-10$

$f(x) + f($ $\frac{1}{x}$ $) =$ $\frac{1}{(a+b)}$ $\Big($ $\frac{1}{x}$ $+x-10 \Big)$   … … … … … iii)

Subtracting ii) from 1

$a f(x) +b f($ $\frac{1}{x}$ $) -$ $a f($ $\frac{1}{x}$ $) - b f($ $x$ $) =$ $\frac{1}{x}$ $-5-x+5$

$(a-b) f(x) - (a-b) f($ $\frac{1}{x}$ $) =$ $\frac{1}{x}$ $-x$

$f(x) - f($ $\frac{1}{x}$ $) =$ $\frac{1}{(a-b)}$ $\Big($ $\frac{1}{x}$ $-x \Big)$   … … … … … iv)

Adding iii) and iv) we get

$2f(x) =$ $\frac{1}{a+b}$ $\Big($ $\frac{1}{x}$ $+ x - 10 \Big) +$ $\frac{1}{a-b}$ $\Big($ $\frac{1}{x}$ $- x \Big)$

$=$ $\frac{ (a-b) [ \frac{1}{x}+x-10 ] + (a+b) [ \frac{1}{x}-x ] }{a^2 - b^2}$

$=$ $\frac{\frac{a}{x} + ax - 10a - \frac{b}{x}-bx+10b+\frac{a}{x}-ax+\frac{b}{x}-bx}{a^2-b^2}$

$=$ $\frac{\frac{2a}{x} - 10a + 10b - 2bx}{a^2-b^2}$

$\Rightarrow f(x) =$ $\frac{\frac{a}{x} - 5a + 5b - bx}{a^2-b^2}$

$=$ $\frac{\frac{a}{x} - bx}{a^2-b^2}$ $+$ $\frac{- 5a + 5b }{a^2-b^2}$

$=$ $\frac{\frac{a}{x} - bx}{a^2-b^2}$ $-$ $\frac{5 }{a+b}$

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