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

Answer:

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) .

Answer:

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) .

Answer:

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

Answer:

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 .

Answer:

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})

Answer:

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

Answer:

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

Answer:

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)}

Answer:

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

Answer:

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) .

Answer:

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)

Adding i) and 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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