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Functions Test 28

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Functions Test 28
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  • Question 1
    1 / -0
    If A={1,3,5,7}A = \left \{ 1 , 3 , 5 , 7 \right \} and B={1,2,3,4,5,6,7,8} B = \left \{ 1 , 2 , 3, 4 , 5 , 6 , 7 , 8 \right \} then the number of one-to-one functions from AA into BB is 
    Solution
    Given, A={1,3,5,7}A = \{1, 3, 5, 7\}
    and B={1,2,3,4,5,6,7,8}B =\{1, 2, 3, 4, 5, 6, 7, 8\}
    Here, n(A)=4n(A) = 4 and n(B)=8n(B) = 8
    Thus number of one-one function from AA into BB
    $$= \,^8 P_4 = 8.7.6.5 = 1680$$
  • Question 2
    1 / -0
    If f(x)=3x+5f(x)=3x+5 and g(x)=x21g(x)={ x }^{ 2 }-1, then (fg)\left( f\circ g \right) (x21)({ x }^{ 2 }-1) is equal to
    Solution
    f(x)=3x+5f(x)=3x+5
    and g(x)=x21g(x)={x}^{2}-1
    fg(x21)=f[g(x21)]=f[(x21) 21]\therefore f\circ g\left( { x }^{ 2 }-1 \right) =f[g\left( { x }^{ 2 }-1 \right) ]=f\left[ { \left( { x }^{ 2 }-1 \right)  }^{ 2 }-1 \right]
    =3(x42x2)+5=3x46x2+5=3\left( { x }^{ 4 }-2{ x }^{ 2 } \right) +5=3{ x }^{ 4 }-6{ x }^{ 2 }+5
  • Question 3
    1 / -0
    If g(x)=x2+x2g(x)={ x }^{ 2 }+x-2 and 12(gf)(x)=2x25x+2\cfrac { 1 }{ 2 } (g\circ f)(x)=2{ x }^{ 2 }-5x+2, then f(x)f(x) is
    Solution
    12(gf)(x)=2x25x+2\cfrac { 1 }{ 2 } \left( g\circ f \right) (x)=2{ x }^{ 2 }-5x+2

    12g[f(x)]=2x25x+2\Rightarrow \cfrac { 1 }{ 2 } g\left[ f(x) \right] =2{ x }^{ 2 }-5x+2\quad

    [{f(x)} 2+{f(x)}2]=2[2x25x+2][g(x)=x2+x2]\left[ { \left\{ f(x) \right\}  }^{ 2 }+\left\{ f(x) \right\} -2 \right] =2\left[ 2{ x }^{ 2 }-5x+2 \right] \quad \left[ \because g(x)={ x }^{ 2 }+x-2 \right]

    f(x) 2+f(x)(4x210x+6)=0\Rightarrow f{ \left( x \right)  }^{ 2 }+f(x)-\left( 4{ x }^{ 2 }-10x+6 \right) =0  ....represents quadratic equation

    We have a formula for solving quadratic equation ax2+bx+c=0ax^2+bx+c=0 is

    x=b±b24ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

    f(x)=1±1+4(4x210x+6)  2\therefore f(x)=\cfrac { -1\pm \sqrt { 1+4\left( 4{ x }^{ 2 }-10x+6 \right)  }  }{ 2 }

    =1±(4x5)2=2x3=\cfrac { -1\pm(4x-5) }{ 2 } =2x-3 or 22x2-2x
  • Question 4
    1 / -0
    If (ax2+bx+c)y+ax2+bx+c=0(ax^2 + bx + c)y +a'x^2+b'x+c=0, then the condition that x may be a rational function of y is
    Solution

  • Question 5
    1 / -0
    Let f(x)=x2f(x) = |x - 2|, where xx is a real number. Which one of the following is true?
    Solution
    From the graph of the function it can be observed clearly that ff is not one-one function.
    So option DD is correct.
  • Question 6
    1 / -0
    If f(x)f\left( x \right) and g(x)g\left( x \right) are two functions with g(x)=x1xg\left( x \right) =x-\dfrac { 1 }{ x } and fg(x)=x31x3f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } , then f(x)f^{ ' }\left( x \right) is equal to
    Solution
    fg(x)=x31x3f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }
    Writing x31x3{ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } using (ab) 3=a3b33ab(ab){ \left( a-b \right)  }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right) , we have
    (x1x ) 3=x31x33x1x(x1x ){ \left( x-\dfrac { 1 }{ x }  \right)  }^{ 3 }={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } -3x\cdot \dfrac { 1 }{ x } \left( x-\dfrac { 1 }{ x }  \right)
    x31x3=(x1x ) 3+3(x1x )\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x }  \right)  }^{ 3 }+3\left( x-\dfrac { 1 }{ x }  \right)
    We have,
    f(g(x) )=x31x3=(x1x ) 3+3(x1x )f\left( g\left( x \right)  \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x }  \right)  }^{ 3 }+3\left( x-\dfrac { 1 }{ x }  \right)
    As g(x)=x1xg\left( x \right) =x-\dfrac { 1 }{ x } , this yields
    f(x1x )=(x1x ) 3+3(x1x )f\left( x-\dfrac { 1 }{ x }  \right) ={ \left( x-\dfrac { 1 }{ x }  \right)  }^{ 3 }+3\left( x-\dfrac { 1 }{ x }  \right)
    On putting x1x=tx-\dfrac { 1 }{ x } =t, we get
    f(t)=t3+3tf\left( t \right) ={ t }^{ 3 }+3t
    Thus, f(x)=x3+3xf\left( x \right) ={ x }^{ 3 }+3x
    and f(x)=3x2+3f^{ ' }\left( x \right) =3{ x }^{ 2 }+3
  • Question 7
    1 / -0
    If f(x)=x,xRf(x)=\left| x \right| ,x\in R, then
    Solution

  • Question 8
    1 / -0
    f,g:RRf,g:R\rightarrow R are functions such that f(x)=3xsin(πx2 ) ,g(x)=x3+2xsin(πx2 ) f(x)=3x-\sin { \left( \cfrac { \pi x }{ 2 }  \right)  } ,g(x)={ x }^{ 3 }+2x-\sin { \left( \cfrac { \pi x }{ 2 }  \right)  }
    The value of ddxf1(g1(x)) x=12\cfrac { d }{ dx } { f }^{ -1 }{ \left( { g }^{ -1 }(x) \right)  }_{ x=12 } is equal to
    Solution
    Given that,
    g:RRg:R \to R

    f(x)=3xsin(πx2)f\left( x \right) = 3x - \sin \left( {\dfrac{{\pi x}}{2}} \right)

    g(x)=x3+2xsin(πx2)g\left( x \right) = {x^3} + 2x - \sin \left( {\dfrac{{\pi x}}{2}} \right)

    df1dx(g(x))x=12=230+x\therefore \dfrac{{d{f^{ - 1}}}}{{dx}}{\left( {g'\left( x \right)} \right)_{x = 12}} = \dfrac{2}{{30 + x}}

    Hence, the option (A)(A) is correct.
  • Question 9
    1 / -0
    The graph of a constant function f(x)=kf(x)=k is?
    Solution
    A graph of a constant function is always a horizontal line.
    Horizontal line is a line which is parallel to xx-axis.

  • Question 10
    1 / -0
    If f,g,hf,g,h are three functions from a set of positive real numbers into itself satisfying the condition,
    f(x)g(x)=hx2+y2f(x) \cdot g(x)=h \sqrt{x^2 + y^2} such that x,yϵ(0,)x,y \epsilon (0,\infty).then, f(x)g(x)\dfrac{f(x)}{g(x)} is a?
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