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This is one of our many quizzes on Functions. Click on this to access more such competitive quizzes on Algebra.

Functions Quiz 2

mathematics

 Determine $f\left( {f\left( x \right)} \right)$ if $f\left( x \right) = \dfrac{{x + 1}}{{x - 1}}$.

 Substitute the value of $f\left( x \right)$ in place of $x$ and simplify to get the required answer.

 Given: $ f\left( x \right) = \dfrac{{x + 1}}{{x - 1}}$ $ \\ \Rightarrow f\left( {f\left( x \right)} \right) = \dfrac{{f\left( x \right) + 1}}{{f\left( x \right) - 1}} \\   \Rightarrow f\left( {f\left( x \right)} \right) = \dfrac{{\dfrac{{x + 1}}{{x - 1}} + 1}}{{\dfrac{{x + 1}}{{x - 1}} - 1}} = \dfrac{{\dfrac{{x + 1 + x - 1}}{{x - 1}}}}{{\dfrac{{x + 1 - x + 1}}{{x - 1}}}} = \dfrac{{2x}}{2} = x \\  $

 Consider a function $g\left( x \right) = x\left| x \right|$. Its domain is:

 Consider both possible values as the mode is given , identify the function values to check the possibility and determine the domain.

 Given the function $g\left( x \right) = x\left| x \right|$. $ \\ $ W.K.T $\left| x \right| = x$ or $\left| x \right| =  - x$. $ \\ $ Considering the first case: $x \geqslant 0$: $ \\ \Rightarrow g\left( x \right) = x \times x = {x^2}$. $ \\ $ Thus, for this case all values of $x \geqslant 0$ are possible. $ \\ $ Considering the second case where $x < 0$:<br>$ \Rightarrow g\left( x \right) = x \times  - x =  - {x^2}$<br>Hence, for this case also, all values of $x < 0$are possible.<br>Therefore, the domain of the given function is $R$.

 Determine the range of the given function. $ \\ f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}  {x + 1,1 \leqslant x < 3} \\   {3x - 1,3 \leqslant x < 5} \\   {4x - 7,5 \leqslant x < 7} \end{array}} \right.$

 Identify the domain and then function values at each point of the domain to get the required answer.

 From the given function, it is clear that the domain of the function is: $\left\{ {1,2,3,4,5,6} \right\} \\ \Rightarrow R = \left\{ {f\left( 1 \right),f\left( 2 \right),f\left( 3 \right),f\left( 4 \right),f\left( 5 \right),f\left( 6 \right)} \right\} \\ $ First case: $ \\  f\left( x \right) = x + 1,1 \leqslant x < 3 \\   \Rightarrow f\left( 1 \right) = 1 + 1 = 2 \\   \Rightarrow f\left( 2 \right) = 2 + 1 = 3 \\  $ Second case: $ \\   f\left( x \right) = 3x - 1,3 \leqslant x < 5 \\   \Rightarrow f\left( 3 \right) = 9 - 1 = 8 \\   \Rightarrow f\left( 4 \right) = 12 - 1 = 11 \\  $ Third case: $ \\ f\left( x \right) = 4x - 7,5 \leqslant x < 7 \\   \Rightarrow f\left( 5 \right) = 20 - 7 = 13 \\   \Rightarrow f\left( 6 \right) = 24 - 7 = 17 \\  $ Thus, $\left\{ {2,3,8,11,13,17} \right\}$ is the range of the given function.

 $f\left( x \right) = {\left( {x - 1} \right)^3} - {\left( {x + 1} \right)^3}$ is a:

 Use the cube formulas ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$ and ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)$ for simplification.

 Expand the given function to identify its type. $ \\   {\left( {x - 1} \right)^3} - {\left( {x + 1} \right)^3} \\   = {x^3} - 3{x^2} + 3x - 1 - {x^3} - 3{x^2} - 3x - 1 \\   =  - 6{x^2} - 2 \\  $ As the degree of the simplified function is 2, it is a quadratic function.

 Let $f\left( x \right) = mx + n$ and $f = \left\{ {\left( {2,2} \right),\left( {3,4} \right),\left( {4,5} \right),\left( {5,6} \right)} \right\}$. What are the values of $m$ and $n$ respectively?

 Use any two points to form two linear equations and then solve them.

 The given points of the function can be considered as $\left( {x,f\left( x \right)} \right)$. $ \\ $ Let us consider the first point $\left( {2,2} \right)$. $ \\ f\left( x \right) = mx + n \Rightarrow 2 = 2m + n \\ $ Similarly for the second point $\left( {3,4} \right)$, $ \\ f\left( x \right) = mx + n \Rightarrow 4 = 3m + n \\ $ Subtract the first equation from the second equation. $ \\  \Rightarrow 3m + n - 2m - n = 4 - 2 \\   \Rightarrow m = 2 \\  $ Substituting this value in the first equation, $ \\   2m + n = 2 \\   \Rightarrow 4 + n = 2 \\   \Rightarrow n =  - 2 \\  $