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This is one of our many quizzes on Combining and composing functions. Click on this to access more such competitive quizzes on Functions.

Combining and composing functions Quiz 2

mathematics

 If $f\left( x \right) = \cos x$ and $g\left( x \right) = {\sin ^{ - 1}}\left( x \right)$ then find $g\left( {f\left( {\dfrac{\pi }{3}} \right)} \right)?$ 

 Simplify composite function from argument to parent function step by step.

$f\left( x \right) = \cos x$ and $g\left( x \right) = {\sin ^{ - 1}}\left( x \right) \\ $ Then $ \\ g\left( {f\left( {\dfrac{\pi }{3}} \right)} \right) = {\sin ^{ - 1}}\left( {\cos \left( {\dfrac{\pi }{3}} \right)} \right) \\ \Rightarrow  {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\ \Rightarrow  \dfrac{\pi }{6} \\  $

 If $g\left( x \right) = \sqrt {x - 7} ,$ then $g\left( {{x^2} + 7} \right) = $ 

 Put the given argument in the place of $x$ in the function and simplify it.

We have $g\left( x \right) = \sqrt {x - 7} $ So, $ \\ g\left( {{x^2} + 7} \right) = \sqrt {\left( {{x^2} + 7} \right) - 7}  \\ \Rightarrow  \sqrt {{x^2} + 7 - 7}  \\ \Rightarrow  \sqrt {{x^2}}  \\ \Rightarrow  x  $

 If $p:R \to R$ is defined as $p\left( x \right) = {\left( {x - 1} \right)^2}$ then find $p\left( {p\left( x \right)} \right)$ 

 Take the function as the argument of the function itself and simplify it according to the options given.

We have $p\left( x \right) = {\left( {x - 1} \right)^2}$ Then $ \\ p\left( {p\left( x \right)} \right) = {\left( {{{\left( {x - 1} \right)}^2} - 1} \right)^2} \\ \Rightarrow  {\left( {{x^2} - 2x + 1 - 1} \right)^2} \\ \Rightarrow  {\left( {{x^2} - 2x} \right)^2} \\ \Rightarrow  {x^2}{\left( {x - 2} \right)^2} \\  $

 Let two functions be $p\left( x \right) = \left\{ {\left( {4,\;5} \right),\;\left( {5,\;6} \right),\;\left( {6,\;7} \right),\;\left( {7,\;8} \right)} \right\}\;{\text{and}}\;q\left( x \right) = \left\{ {\left( {1,\;5} \right),\;\left( {2,\;4} \right),\;\left( {3,\;6} \right),\;\left( {4,\;8} \right)} \right\}$ find $p\left( {q\left( 3 \right)} \right)$

 Firstly find the value of the argument function and then find the value of the parent function taking the outcome of the argument function as the argument.

We have to find $p\left( {q\left( 3 \right)} \right)$ $ \\ $ At first find $q\left( 3 \right)$ $ \\ $ From the function $q\left( x \right),\;q\left( 3 \right) = 6$  $ \\ $ Now, $p\left( {q\left( 3 \right)} \right) = p\left( 6 \right)$ $ \\ $ And in the function $p\left( x \right),\;p\left( 6 \right) = 7$ $ \\ $ Therefore $p\left( {q\left( 3 \right)} \right) = 7$

 If $p\left( x \right) = \left[ x \right] - x + 1\;{\text{and}}\;q\left( x \right) = \left\{    - 1,\;\;x < 0 \\  0,\;\;\;\;x = 0 \\  1,\;\;\;\;x > 0 \\   \right.$ then $q\left( {p\left( x \right)} \right)$ for all $x$ equals to

 Simplify the function $p\left( x \right)$ by finding its range with help of properties of greatest integer function, i.e. $x = \left[ x \right] + \left\{ x \right\}$ and range of fractional part function is $\left( {0,\;1} \right)$

We have $p\left( x \right) = \left[ x \right] - x + 1\;{\text{and}}\;q\left( x \right) = \left\{    - 1,\;\;x < 0 \\  0,\;\;\;\;x = 0 \\  1,\;\;\;\;x > 0 \\   \right. \\ $ Write $p\left( x \right) = \left[ x \right] - x + 1 > 0$  $ \\ $ Since $\left[ x \right] - x =  - \left\{ x \right\}$ $ \\ $ And $ - 1 <  - \left\{ x \right\} < 0 \Leftrightarrow  - 1 < \left[ x \right] - x < 0$  $ \\ $ Then $0 < \left[ x \right] - x + 1 < 1 \Leftrightarrow 0 < p\left( x \right) < 1$ $ \\ $ Therefore $q\left( {p\left( x \right)} \right) = 1\;\;\;\left[ {\because p\left( x \right) > 0} \right]$