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Relations and Functions Test - 43

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Relations and Functions Test - 43
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  • Question 1
    1 / -0
    Let f:(,1](,1]\displaystyle f:(-\infty,1]\rightarrow (-\infty,1] such that f(x)=x(2x).\displaystyle f(x)=x(2-x). Then f1(x)f^{-1}(x) is
    Solution
    We have, f(x)=x(2x)f(x)=x(2-x)

    y=x(2x)y=x(2-x)

    y=x(x2)-y=x(x-2)

    y=x22x-y=x^{2}-2x

    1y=x22x+11-y=x^{2}-2x+1

    1y=(x1)21-y=(x-1)^{2}

    ±1y=x1\pm\sqrt{1-y}=x-1

    Now f:(,1](,1]f:(-\infty,1] \rightarrow (-\infty,1]

    x1=1yx-1=-\sqrt{1-y}

    x=11yx=1-\sqrt{1-y}

    Hence f1(x)=11xf^{-1}(x)=1-\sqrt{1-x}
  • Question 2
    1 / -0
    If f(x)=11x,x0,1\displaystyle f(x)=\frac{1}{1-x},x\neq 0,1 then the graph of the function y=f{f(f(x))},x>1,\displaystyle y=f\left \{ f(f(x)) \right \},x> 1, is
    Solution
    f(f(x))=1111xf(f(x))=\dfrac{1}{1-\dfrac{1}{1-x}}

    =1xx =g(x)=-\dfrac{1-x}{x} =g(x)

    f(g(x))=11+1xxf(g(x))=\dfrac{1}{1+\dfrac{1-x}{x}}

    =xx+1x=\dfrac{x}{x+1-x}

    =x=x

    y=xy=x is an equation of straight line.

    Hence, 'C' is correct.
  • Question 3
    1 / -0
    Let f:{x,y,z}{a,b,c}\displaystyle f:\left \{ x,y,z \right \}\rightarrow \left \{ a,b,c \right \} be a one-one function and only one of the conditions (i)f(x)b,(ii)f(y)=b,(iii)f(z)a(i)f(x)\neq b, (ii)f(y)=b,(iii)f(z)\neq a is true then the function ff  is given by the set 
    Solution
    f:{x,y,z}{a,b,c}f:\left\{ x,y,z \right\} \rightarrow \left\{ a,b,c \right\}  is  a  one-one  function 

    \Rightarrow   each  element  in {x,y,z} \left\{ x,y,z \right\}  will  have  exactly  one  image  in  {a,b,c}\left\{ a,b,c \right\}

    and  no  two  elements  of {x,y,z} \left\{ x,y,z \right\}   will  have  same  image  in  {a,b,c}\left\{ a,b,c \right\}

     Coming to the given  3  conditions, only one is true.

     1)  if  f(x)bf\left( x \right) \neq b  is  true  then  f(y)=bf\left( y \right) =b  is  false  which  makes  f(z)af\left( z \right) \neq a  true   f(x)b\Longrightarrow   f\left( x \right) \neq b  is  false.

     2)  if  f(y)=bf\left( y \right) =b  is  true  then  f(x)bf\left( x \right) \neq b  will  also  be  true   f(y)=b\Longrightarrow   f\left( y \right) =b  is  false

     f(z)a \therefore   f\left( z \right) \neq a  is  the  true  condition  and  remainig two  are  false  conditions.

     f(x)=b, f(y)=a, f(z)=c\therefore   f\left( x \right) =b,  f\left( y \right) =a,  f\left( z \right) =c

     hence  f={(x,b),(y,a),(z,c) }f=\left\{ \left( x,b \right) ,\left( y,a \right) ,\left( z,c \right)  \right\}
  • Question 4
    1 / -0
    If f(x)=3x5\displaystyle f(x)=3x-5 then f1(x)=f^{-1}(x)=
    Solution
    f(x)=3x5f(x)=3x-5

    Since f(x)f(x) is a purely linear function (straight line), it is therefore a one-one function.

    Hence f(x)f(x) is revertible for all real x.x.

    y=3x5y=3x-5

    y+5=3xy+5=3x

    x=y+53x=\dfrac{y+5}{3}

    Replacing xx by yy, gives us 

    f1(x)=x+53f^{-1}(x)=\dfrac{x+5}{3}
  • Question 5
    1 / -0
    Which of the following function is one-one?
    Solution
    Option A - Consider x=2x = 2 and x=0 x =0 , the value of f(x)f(x) is same. Hence it is not one-one
    Option B -  If we replace xx by x-x, then the value of g(x)g(x) remains the same. Hence it is not one-one
    Option C - h(x)h(x) is an increasing function for the given values of xx. Hence it is one-one function
    Option D -f(x)f(x) is an even function. So it is not one-one for the given values of xx

  • Question 6
    1 / -0
    If f(x)=xn,nN\displaystyle f(x)=x^{n},n\in N and (gof)(x)=ng(x)(gof)(x)=ng(x) then g(x)g(x) can be 
    Solution
    f(x)=xnf(x)=x^{n}

    g(f(x))=ng(x)g(f(x))=ng(x) ...(i)(i)

    log(f(x))=nlog(x)\log(f(x))=n\log(x) ...(ii)(ii)

    Taking log(x)\log(|x|) as g(x)g(x) the above expression is reduced to eq ii.

    Hence

    g(x)=log(x)g(x)=\log(|x|).
  • Question 7
    1 / -0
    If f:R+Af:R^+\rightarrow A, Where A={x:5<x<}A=\{x:-5<x<\infty\} be defined by f(x)=x25f(x)=x^2-5. Then f1(7)=f^{-1}(7)=
    Solution
    The value f1(7)f^{-1}(7) will be the roots of the equation 

    x25=7x2=12x=±23x^2-5= 7\Rightarrow x^2 = 12\Rightarrow x = \pm 2\sqrt{3}

    But only positive.  x= 23\therefore  x =  2\sqrt{3}
  • Question 8
    1 / -0
    If the function f:[1,)[1,)f:[1,\infty)\rightarrow [1,\infty) is defined by f(x)=3x(x1)\displaystyle f(x)=3^{x(x-1)} then f1(x)f^{-1}(x) is 
    Solution
    For the given domain, the function is one one.
    Thus,
    f(x)=2x(x1)f(x)=2^{x(x-1)}

    y=2x2xy=2^{x^{2}-x}

    lny=(x2x)ln2lny=(x^{2}-x)ln2

    log2y=x2xlog_{2}y=x^{2}-x

    log2y=(x12)214log_{2}y=(x-\dfrac{1}{2})^{2}-\dfrac{1}{4}

    log2y+14=(x12)2log_{2}y+\dfrac{1}{4}=(x-\dfrac{1}{2})^{2}

    ±12(1+4log2y)=x12\dfrac{\pm 1}{2}(\sqrt{1+4log_{2}y})=x-\dfrac{1}{2}

    Since f:[1,)[1,)f:[1,\infty)\rightarrow[1,\infty)
    Thus domain is [1,)[1,\infty)
    Hence domain is positive so negative sign will be ignored.   SoSo

    x12=12(1+4log2y)x-\dfrac{1}{2}=\dfrac{1}{2}(\sqrt{1+4log_{2}y})

    x=12[1+1+4log2y]x=\dfrac{1}{2}[1+\sqrt{1+4log_{2}y}]
    Replacing x and y, gives us 

    f1(x)=12[1+1+4log2x]f^{-1}(x)=\dfrac{1}{2}[1+\sqrt{1+4log_{2}x}]

  • Question 9
    1 / -0
    The composite mapping fogfog of the map f:RR,f(x)=sinxf: R\rightarrow R,f(x)=\sin x and g:RR,g(x)=x2g: R\rightarrow R, g(x)=x^2 is
    Solution
    Composite mapping will be f(g(x))f(g(x))

    =f(x2)=f(x^2)

    =sin(x2)=\sin(x^2)

    Hence option C'C' is the answer.
  • Question 10
    1 / -0
    If the function f:RRf:R\rightarrow R be such that f(x)=x[x],\displaystyle f(x)=x-[x], where [y][y] denotes the greatest integer less than or equal to yy, then f1(x)f^{-1}(x) is
    Solution
    f(x)=x[x]={x}f\left( x \right) = x - \left[ x \right] = \left\{ x \right\} is not one-one function from R to R. Hence f1(x){f^{ - 1}}\left( x \right) can not be defined.
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