Class 11: Functions – Exercise 3.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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