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

Answer:

\displaystyle \text{Given } f (x) = x^2 - 3x +4

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

\displaystyle \text{Given } f(x) = f(2x+1)

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

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

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

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

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

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

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Question 2: \displaystyle \text{If } f(x) = (x-a)^2 (x-b)^2  \text{, find }  f(a+b) .

Answer:

\displaystyle \text{Given } f(x) = (x-a)^2 (x-b)^2

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

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

Answer:

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

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

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

\displaystyle \Rightarrow xyb -ya = ax - b

\displaystyle \Rightarrow xyb - ax = ay - b

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

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

\displaystyle \Rightarrow \therefore x = f(y) \text{. Hence proved.}\

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

Answer:

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

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

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

\displaystyle \therefore f(f(f(x))) = x \text{. Hence proved.}\

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

Answer:

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

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

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

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

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

\displaystyle \therefore f(f(x)) = x \text{. Hence proved.}\

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\displaystyle \text{Question 6: } \text{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}

\displaystyle \text{Find } \text{i) } f(\frac{1}{2}) \hspace{1.0cm}  \text{ii) } f(-2) \hspace{1.0cm} \text{iii) } f(1) \hspace{1.0cm} \text{ iv) } f(\sqrt{3}) \hspace{1.0cm}  \text{ v) } f(-\sqrt{3})

Answer:

\displaystyle \text{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}

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

\displaystyle \text{ii) } f(-2) = (-2)^2 = 4 \text{ [ Using } f(x) = x^2 , x < 0 ]

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

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

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

Answer:

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

\displaystyle \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

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

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

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

Answer:

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

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

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

\displaystyle = 2 \sin \theta \cos \theta = \sin 2 \theta

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

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

Answer:

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

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

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

Answer:

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

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

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

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

\displaystyle = x

\displaystyle \text{Hence } f(f(x)) =x for all \displaystyle x

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

Answer:

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

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

Adding i) and ii)

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

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

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

Subtracting ii) from 1

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

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

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

Adding iii) and iv) we get

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

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

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

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

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

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

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