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Relations and Functions Test - 35

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Relations and Functions Test - 35
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  • Question 1
    1 / -0
    The number of one-one functions that can be defined from A={4,8,12,16}A=\{4,8,12,16\} to BB is 5040,5040, then n(B)=n(B)=
    Solution
    The first element 44 of AA can be mapped to any of the nn elements in BB
    Similarly second element 88 of AA can be mapped to any of (n1)(n-1) elements in BB
    \Rightarrow Total no of one - one functions is
    n(n1)(n2)(n3)=5040n(n-1)(n-2)(n-3)=5040
    n=10\Rightarrow n=10
  • Question 2
    1 / -0
    If f:R+[1,)f:R^+\rightarrow [1,\infty), defined by f(x)=x2+1f\left( x \right) ={ x }^{ 2 }+1, then the value of f1(17){ f }^{ -1 }\left( 17 \right) and f1(3){ f }^{ -1 }\left( -3 \right)  respectively are
    Solution
    Given, f(x)=x2+1f(x)=x^{2}+1
    Put f(x)=yf(x)=y
    y=x2+1y=x^{2}+1
    x=±y1\Rightarrow x=\pm \sqrt{y-1}
    f1(y)=±y1\Rightarrow f^{-1}(y)=\pm \sqrt{y-1}
    So, f1(17)=±4f^{-1}(17)=\pm 4
    f1(17)=4f^-{1} (17) = 4 [ As 4-4 \notin domain R+R^+]

    and f1(3)=ϕf^{-1}(-3)=\phi
    f1(3)f^{-1}(-3) is not defined in R+R^+ as square root of negative number is not defined in R+R^+
    Option D is correct
  • Question 3
    1 / -0
    The number of injections that are possible from AA to itself is 720,720, then n(A)=n (A) =
    Solution
    first element of AA in domain can be mapped to
    any of the nn elements of AA in range.
    2nd2^{nd} element of AA in domain can be mapped to any of (n1)(n-1) elements of AA in range.
    \therefore Total no of injections possible is
    n×(n1)×(n2)...×3×2×1=n!n\times (n-1)\times (n-2)...\times 3\times 2\times 1=n!
    n!=720\therefore n!=720
    n=6\Rightarrow n=6


  • Question 4
    1 / -0
    The number of one-one functions that can be defined from A={1,2,3}A = \left \{ 1,2,3 \right \} to  B={a,e,i,o,u}  B = \left \{ a,e,i,o,u \right \} is 
    Solution
    You can correlate this to permutation and combination problems.

    We have to arrange 3 people on 5 places.

    So total no of permutations =5P3={_{}}^{5}P_{3}
  • Question 5
    1 / -0
    The number of non-surjective mappings that can be defined from A={1,4,9,16} A = \left \{ 1,4,9,16 \right \}   to B={2,8,16,32,64}  B=\left \{ 2,8,16,32,64 \right \} is
    Solution
    Here n(B)>n(A)n(B)>n(A), no function can be a surjective.
    No of functions == No of non-surjectives

    Any elements of AA can be mapped to any of 55 elements in BB
    So non surjective mapping from AA to BB
    =5×5×5×5=5\times5\times5\times5
    =625=625
  • Question 6
    1 / -0
    If f:AB  f:A\rightarrow B  is a constant function which is onto then BB is
    Solution
    f(x)f(x) is a constant fucntion \Rightarrow Range of f(x)f(x) is a singleton set. 
    For ff to be an onto function, Co domain BB should be equal to Range.
    B\therefore B is a singleton set.
  • Question 7
    1 / -0
    If  f:AB f:A\rightarrow B  is a bijection then f1of=  f^{-1} of = 
    Solution
    Since ff is bijective, its inverse will also be bijective.
    f(x)=yf(x)=y
    x=f1(y)x=f^{-1}(y)
    f(x)=f1(y)f(x)=f^{-1}(y)
    y=f(f)1(y)=f(x)=yy=f(f)^{-1}(y)=f(x)=y
    Hence, it is an identity.
  • Question 8
    1 / -0
    The number of bijection that can be defined from A={1,2,8,9} A = \left \{ 1,2,8,9 \right \}   to  B={3,4,5,10}  B = \left \{ 3,4,5,10 \right \} is
    Solution
    There are 4 inputs {1,2,8,9}\{1, 2, 8, 9\} and 4 outputs {3,4,5,10}\{3, 4, 5, 10\}. Hence function will be bijective if and only if each output is connected with only one input.
    Therefore number of bijective function is 4!=244! = 24
  • Question 9
    1 / -0
    The number of possible surjection from A={1,2,3,...n}A=\{1,2,3,...n\} to B={1,2}B = \{1,2\} (where n2)n \geq 2) is 6262, then n=n=
    Solution
    Each element in AA can be mapped onto any of two elements of BB
    \therefore Total possible functions are 2n2^{n}
    For the fnsf^{{n}'s} to be surjections , they shouldn't be mapped alone to any of the two elements.
    \therefore Total no of surjections =2n2= 2^{n}-2
    2n2=622^{n}-2=62
    n=6\Rightarrow n=6

  • Question 10
    1 / -0
    The number of injections possible from A={1,3,5,6}A=\{1,3,5,6\} to B={2,8,11}B =\{2,8,11\} is
    Solution
    co-domain n(B)<n(B) < domain n(A)n(A)
    There can't be any one-one function from AA to B.B.
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