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Relations and Functions Test - 84

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Relations and Functions Test - 84
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  • Question 1
    1 / -0
    If $$f(x)= sin^{2}x$$ and the composite functions g{f(x)}=|sin x|, then the function g(x)=
  • Question 2
    1 / -0
    If $$f:N\rightarrow N,f(x)=x+3$$, then $$\quad { f }^{ -1 }(x)=.....$$
  • Question 3
    1 / -0
    If $$f(x)=8x^3$$ and $$g(x)=x^{1/3}$$ then $$(g o f)(x)=?$$
    Solution
    $$(g o f)(x)=g[f(x)]=g(8x^3)=(8x^3)^{1/3}=2x$$.
  • Question 4
    1 / -0
    If $$f(x)=\dfrac{1}{(1-x)}$$ then $$(f o f o f)(x)=?$$
    Solution
    $$(f o f)(x)=f\{f(x)\}=f\left(\dfrac{1}{1-x}\right)=\dfrac{1}{\left(1-\dfrac{1}{1-x}\right)}=\dfrac{1-x}{-x}=\dfrac{x-1}{x}$$

    $$\Rightarrow \{f o (f o f)\}(x)=f\{(f o f)(x)\}$$

    $$=f\left(\dfrac{x-1}{x}\right)$$

    $$(f o f)(x)=\dfrac{1}{1-\dfrac{x-1}{x}}=x$$.
  • Question 5
    1 / -0
    If $$f(x)=x^2, g(x)=\tan x$$ and $$h(x)=log x$$ then $$\{h o (g o f)\}\left(\sqrt{\dfrac{\pi}{4}}\right)=?$$
    Solution
    $$\{h o ( g o f)\}(x)=(h o g)\{f(x)\}=(h o g)(x^2)$$

    $$=h\{g(x^2)\}=h(\tan x^2)=log (\tan x^2)$$.

    $$\therefore \{h o (g o f)\}\sqrt{\dfrac{\pi}{4}}=\log\left(\tan\dfrac{\pi}{4}\right)=\log 1=0$$.
  • Question 6
    1 / -0
    If $$f=\{(1, 2), (3, 5), (4, 1)\}$$ and $$g=\{(2, 3), (5, 1), (1, 3)\}$$ then $$(g o f)=?$$
    Solution
    $$Dom(g o f)=dom(f)=\{1, 3, 4\}$$.

    $$(g o f)(1)=g\{f(1)\}=g(2)=3,$$

    $$ (g o f)(3)=g\{f(3)\}=g(5)=1$$

    $$(g o f)(4)=g\{f(4)\}=g(1)=3$$

    $$\therefore g o f=\{(1, 3), (3, 1), (4, 3)\}$$.
  • Question 7
    1 / -0
    If $$f(x)=(x^2-1)$$ and $$g(x)=(2x+3)$$ then $$(g o f)(x)=?$$
    Solution
    $$(g o f)(x)=g[f(x)]=g(x^2-1)=2(x^2-1)+3=(2x^2+1)$$.
  • Question 8
    1 / -0
    If $$f(x)=x^2-3x+2$$ then $$(f o f)(x)=?$$
    Solution

  • Question 9
    1 / -0
    If $$ f(x)=\dfrac{2}{x-3}, g(x)=\dfrac{x-3}{x+4} $$ and $$ h(x)=-\dfrac{2(2 x+1)}{x^{2}+x-12} $$ then $$ \lim _{x \rightarrow 3}[f(x)+g(x)+h(x)] $$ is
    Solution
    c. We have $$ f(x)+g(x)+h(x)=\dfrac{x^{2}-4 x+17-4 x-2}{x^{2}+x-12} $$
    $$ =\dfrac{x^{2}-8 x+15}{x^{2}+x-12}=\dfrac{(x-3)(x-5)}{(x-3)(x+4)} $$
    $$ \therefore \lim _{x \rightarrow 3}[f(x)+g(x)+h(x)]=\lim _{x \rightarrow 3} \dfrac{(x-3)(x-5)}{(x-3)(x+4)}=-\dfrac{2}{7} $$.
  • Question 10
    1 / -0

    Directions For Questions

    Consider the function $$f(x) = f\left(\dfrac{x - 1}{x}\right) = 1 + x \space \forall \space x \space \epsilon \space R - {0, 1}\space and \space g(x) = 2f(x) - x + 1$$

    ...view full instructions

    The number of roots of the equation g(x) = 1 is
    Solution

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