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Relations and Functions Test - 82

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Relations and Functions Test - 82
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  • Question 1
    1 / -0
    l qt f(x) be a function satisfying f'(x)=f(x) with f(0)=1 and g be the function satisfying f(x)+g(x)=$$X^{2}$$, the value of the integral $$\int _{ 0 }^{ 1 }{ f(x)g(x)\quad dx\quad is } $$
    Solution

  • Question 2
    1 / -0
    $$f : R \rightarrow R$$  defined by  $$f ( x ) = \dfrac { x } { x ^ { 2 } + 1 } , \forall x \in R$$  is
    Solution
    $$\begin{aligned} f(x) &=\dfrac{x}{1+x^{2}} \\-f(4)=& \dfrac{4}{1+16}=\dfrac{4}{17} \\ f\left(\dfrac{1}{4}\right) &=\frac{\dfrac{1}{4}}{1+\dfrac{1}{16}}=\dfrac{4}{17} \end{aligned}$$
    Hence two different input has Same output $$\cdot$$ so not one-one
    $$f(x)=y$$
    $$y=\dfrac{x}{1+x^{2}}=1 \quad y+y x^{2}-x=0$$
    $$x=\dfrac{1 \pm \sqrt{1-4 y^{2}}}{2 y}$$
    $$1-4 y^{2} \geq 0$$
    $$(1+2 y)(1-x y) \geq 0$$
    $$\dfrac{1}{-3} \leq y<\dfrac{1}{2}$$
    Range of $$f(x)$$ is $$\left[\dfrac{-1}{2}, \dfrac{1}{2}\right]$$
    Range of $$f(x) \neq$$ codomain $$f(x)$$ Hence f(x) is neither one-one nor onto option D is correct.
  • Question 3
    1 / -0
    If f (x) = cosx and g (x) = x$$^2$$ then (gof) (x) is ....
    Solution

  • Question 4
    1 / -0
    Suppose that $$g(x)=1+\sqrt { x } and\quad f(g(x))=3+2\sqrt { x } +x\quad then\quad f(x)\quad is$$
    Solution

  • Question 5
    1 / -0
    If  $$f ( x ) = \sqrt { x ^ { 2 } + 1 } , g ( x ) = \dfrac { x + 1 } { x ^ { 2 } + 1 }$$  and  $$h ( x ) = 2 x - 3 ,$$  then  $$f ^ { \prime } \left( h ^ { \prime } \left( g ^ { \prime } ( x ) \right) =\right.$$
    Solution

  • Question 6
    1 / -0
    If the operation * is defined as $$\left( a\times b \right) ={ a }^{ 2 }+{ b }^{ 2 }$$ then $$\left( 3\times 4 \right) =5$$ is
  • Question 7
    1 / -0
    Consider the binary operation $$*$$ on $$Q$$ defined by $$x*y=1+12x+xy,x,y\rightarrow Q$$ then $$2*3$$ equals
    Solution

  • Question 8
    1 / -0
    Let A=(5,6); how many binary operations can be defined on this set?
    Solution

  • Question 9
    1 / -0
    If =$$f=\left\{ (-2,4),(0,6),(2,8) \right\} $$ and 
    $$g=\left\{ (-2,-1),(0,3),(2,5) \right\} $$, then 
    $$\left( \frac { 2f }{ 3g } +\frac { 3g }{ 2f }  \right) (0)=\quad $$
    Solution

  • Question 10
    1 / -0
    Number of solution of the equation  $$f ( x ) = g ( x )$$  are same as number of point of intersection of the curves $$y = f ( x )$$  and  $$y = g ( x )$$  hence answer the following question.
    Number of the solution of the equation  $$x ^ { 2 } = | x - 2 | + | x + 2 | - 1$$  is
    Solution

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