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This is the second of our many quizzes on Composite functions. Click on this to access more such competitive quizzes on Algebra.

Composite functions Quiz 2

mathematics

    Find the constant term of the function $f\left( x+9 \right)$ if$\left\{ f\left( x \right)={{x}^{2}}-9 \right\}$ .

 Substitute $\left(x+9\right)$ in place of $x$ in $f\left(x\right)$ function and simplify to get the constant term.

  $ \text{Given}, f\left( x+9 \right) \text{if} \left\{ f\left( x \right)={{x}^{2}}-9 \right\} \\ f\left( x \right)={{x}^{2}}-9 \\   \Rightarrow f\left( x \right)={{\left( x+9 \right)}^{2}}-9 \\   \Rightarrow f\left( x \right)={{x}^{2}}+18x+81-9 \\   \therefore f\left( x \right)={{x}^{2}}+18x+72 \\  \text{The constant term of the function is }72. $

 Find the value of  $f\left( {{\left( x-1 \right)}^{2}} \right)\text{ if }\left\{ f\left( x \right)=4\sqrt{x}+4 \right\}$.

 Substitute ${{\left( x-9 \right)}^{2}}$ in place of $x$ in $f\left(x\right)$ function and simplify to get the constant term.

  $ \text{Given}, f\left( {{\left( x-1 \right)}^{2}} \right)\text{ if }\left\{ f\left( x \right)=4\sqrt{x}+4 \right\}\\ f\left( x \right)=4\sqrt{x}+4 \\   \Rightarrow f\left( x \right)=4\sqrt{{{\left( x-1 \right)}^{2}}}+4 \\ \therefore f\left( x \right)=4x $

   What is the coefficient of  ${{x}^{2}}$ in $g\left( f\left( x+1 \right) \right)$if given that $\left\{ f\left( x \right)=-{{x}^{3}}+4 \right\}$ and $\left\{ g\left( x \right)=\left( x+1 \right) \right\}$.

 Substitute $\left( x+1 \right)$ in $f\left( x \right)$ and then substitute the obtained answer in $g\left( x \right)$ to get the coefficient.

 $\left\{ f\left( x \right)=-{{x}^{3}}+4 \right\} \text{and } \left\{ g\left( x \right)=\left( x+1 \right) \right\}.\\ g\left( f\left( x+1 \right) \right) \\ =g\left( -{{\left( x+1 \right)}^{3}}+4 \right) \\   =g\left( -x^3-3x^2-3x+3 \right) \\ =-x^3-3x^2-3x+3+1 \\  =-x^3-3x^2-3x+4 \\ \text{Therefore, the required coefficient is “-3”.}$

    What will be the sum of the coefficient of $x$ and ${{x}^{2}}$ in the expansion of $g\circ f$ if $\left\{ f\left( x \right)=5x+2 \right\}$and $\left\{ g\left( x \right)=-{{x}^{2}}+2 \right\}$.

  Substitute  $f\left( x \right)$ in $g\left( x \right)$ in place of $x$ to get the coefficient values and then add them. 

 $ \text{Given}: \\ \left\{ f\left( x \right)=5x+2 \right\}\text{and }\left\{ g\left( x \right)=-{{x}^{2}}+2 \right\}.\\   g\circ f=g\left( f\left( x \right) \right) \\   \Rightarrow g\circ f=g\left( 5x+2 \right) \\   \Rightarrow g\circ f=-{{\left( 5x+2 \right)}^{2}}+2 \\   \therefore g\circ f=-25{{x}^{2}}-20x-4 \\   \text{The sum of coefficient of } x \text{ and }{{x}^{2} }\text{ is,} \\ = \left( -20-25 \right)=-45 $

 What is the difference between the values of functions $\left( f\circ g \right)\left( 3 \right)$ and $\left( g\circ f \right)\left( 3 \right)$ if given that $\left\{ f\left( x \right)={{x}^{2}}+4x,g\left( x \right)=6x+2, \right\}$.

 Substitute  $f\left( 3 \right)$ in $g\left( x \right)$ in place of $x$ to get the value and then substitute $g\left( 3 \right)$ in $f\left( x \right)$ in place of $x$ to get the value. Now find the difference of the two values.

$ \text{Given}: \\ \left\{ f\left( x \right)={{x}^{2}}+4x,g\left( x \right)=6x+2, \right\}\\   g\circ f(3)=g\left( f\left( 3 \right) \right) \\   \Rightarrow g\circ f(3)=g\left( {{3}^{2}}+4(3) \right) \\   \Rightarrow g\circ f(3)=g\left( 21 \right) \\  \therefore g\circ f(3)=6(21)+2=128 \\   f\circ g(3)=f\left( g\left( 3 \right) \right) \\   \Rightarrow f\circ g(3)=f\left( 6(3)+2 \right) \\   \Rightarrow f\circ g(3)=f\left( 20 \right) \\   \therefore f\circ g(3)={{20}^{2}}+4(20)=480 \\   \text{The difference between } f\circ g(3) \text{ and }g\circ f(3) \text{ is}, \\  =480-128=352   $