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Functions Test 24

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Functions Test 24
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  • Question 1
    1 / -0
    $$f(x) = x^{2} + d$$ and $$g(x) = 2x^{2}$$, where d is a constant. If $$\dfrac {f(g(2))}{f(2)} = 4$$, find the value of $$d$$.
    Solution
    Given, $$f(x)=x^2+d$$ and $$g(x)=2x^2$$
    $$g(2)=2(2^{2})=8$$
    Hence, $$f(g(2))=f(8)=64+d$$
    Therefore $$\dfrac{f(g(2))}{f(2)}=\dfrac{64+d}{4+d}=4$$
    Hence, $$64+d=16+4d$$
    $$\Rightarrow 48=3d$$ 
    $$\Rightarrow d=16$$
  • Question 2
    1 / -0
    If $$f(x) =$$ $$x^2$$ and $$g(x) = 2x$$, calculate the value of $$f(g(-3))-g(f(-3))$$.
    Solution
    Given $$f(x)=x^2$$ and $$g(x)=2x$$.
    Then $$f(-3)={(-3)}^2=9$$ and $$g(-3)=2(-3)=-6$$
    So, $$f(g(-3))={[g(-3)]^2}={[-6]}^2=36$$
    and $$g(f(-3))=2*f(-3)=2*9=18$$
    The value of $$f(g(-3))-g(f(-3))$$ is $$36-18=18$$
  • Question 3
    1 / -0
    Find the correct expression for $$\displaystyle f\left( g\left( x \right)  \right) $$ if $$\displaystyle f(x)=4x+1$$ and $$\displaystyle g\left( x \right) ={ x }^{ 2 }-2$$  
    Solution
    $$f(x)=4x+1$$ and $$g(x)=x^2-2$$
    $$\Rightarrow f(g(x))=f(x^2-2)$$
    $$\Rightarrow (4(x^2-2)+1$$
    $$\Rightarrow 4x^2-8+1$$
    $$\Rightarrow 4x^2-7$$
  • Question 4
    1 / -0
    If $$p(x) = \dfrac{x}{x-2}$$ and $$q(x) = \sqrt{9-x}$$, find the value of $$(p\circ  q)(5)$$
    Solution
    Given, $$p = \dfrac{x}{x-2}$$ $$q=\sqrt{9-x}$$
    For $$(p\circ q)$$, substitute the value $$p$$ in $$q$$
    ie., $$\dfrac{\sqrt{9-x}}{\sqrt{9-x}-2}(5)$$ is a undefined.
  • Question 5
    1 / -0
    Consider the functions $$\displaystyle f\left( x \right) =\sqrt { x } $$ and $$\displaystyle g\left( x \right) =7x+b$$. Find the value of $$b$$, if the composite function, $$\displaystyle y=f\left( g\left( x \right)  \right) $$ passes through $$(4, 6)$$. 
    Solution
    Given, $$f(x)=\sqrt{x}$$ and $$g(x)=7x+b$$
    Therefore, $$f(g(x))=\sqrt{7x+b}$$
    Now it passes through $$(4,6)$$. 
    Hence, $$y_{4,6}=\sqrt{7x+b}_{4,6}$$ 
    $$\Rightarrow 6=\sqrt{28+b}$$ 
    $$\Rightarrow 36=28+b$$ 
    $$\Rightarrow b=8$$
  • Question 6
    1 / -0
    Find $$g(x)$$, if $$f(x) = 7x + 12$$ and $$f(g(x) = 21x^{2} + 40$$
    Solution
    Given, $$f(x) = 7x+12, f(g(x))=21x^2+40$$ 
    $$\therefore f(g(x)) = 7g(x)+12 = 21 {x}^{2} + 40$$
    $$\therefore 7 g(x) = 21 {x}^{2}+28$$
    $$\therefore g(x) = 3 {x}^{2} + 4$$
  • Question 7
    1 / -0
    Given a function $$f(x) = \dfrac {1}{2}x - 4$$ and the composite function $$f(g(x)) = g(f(x))$$, determine which among the following can be $$g(x)$$:
    I. $$2x - \dfrac {1}{4}$$
    II. $$2x + 8$$
    III. $$\dfrac {1}{2}x - 4$$
    Solution
    If $$g(x)=2x-\dfrac14$$

    $$g(f(x))=x-8-\dfrac14$$

    $$f(g(x))=x-\dfrac18-4$$

    Case II

    $$g(f(x))=2(\dfrac{x}{2}-4)+8=x$$

    $$f(g(x))=\dfrac12(2x+8)-4=x+4-4=x$$

    Case III

    $$g(f(x))=\dfrac12(\dfrac{x-8}{2})-4=\dfrac x4-6$$

    $$f(g(x))=\dfrac12(\dfrac{x-8}{2})-4=\dfrac x4-6$$
    Option D is correct
  • Question 8
    1 / -0
    If $$f(x) = 2x$$ and $$f(f(x)) = x + 1$$, then the value of $$x $$ is
    Solution
    Given $$f(x)=2x$$ and $$f(f(x))=x+1$$
    Then $$ f(f(x))=x+1$$
    $$\Rightarrow 2[f(x)]=x+1$$
    $$\Rightarrow  2(2x)=x+1$$
    $$\Rightarrow 4x=x+1$$
    $$\Rightarrow x=\dfrac {1}{3}$$
  • Question 9
    1 / -0
    Find the value of  $$g(f(2))$$, if $$f(x) = e^{x}$$ and $$g(x) = \dfrac {x}{2}$$
    Solution
    Given, $$f(x)=e^x, g(x)=\dfrac {x}{2}$$
    $$\Rightarrow f(2) = { e }^{ 2 }=7.389$$
    $$\Rightarrow g(f(2)) = g(7.389) =\dfrac { 7.389}{2} = 3.7$$
  • Question 10
    1 / -0
    The above figure shows the graph of the function $$f(x)$$, the value of $$f(f(3))$$ is:

    Solution
    From the graph shown, we need to find $$f(f(3))$$
    The value of the function at $$x = 3$$ is given by $$f(3) = -2$$
    Next, $$f(f(3)) = f(-2) = 0$$
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