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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
    If $$ n (A) = 5 $$ and $$ n (B) = 7 , $$ then the number of relations on $$ A \times B $$ is :
    Solution
    Given, $$ n (A) = 5 $$ and $$ n (B) = 7 $$
    Therefore, number of relations on $$ A \times B = 2^{[ n(A) \times n(B) ]} $$
    $$ = 2^{(5 \times 7)} = 2^{(35)} $$
  • Question 2
    1 / -0
    The functions $$f, g$$ and $$h$$ satisfy the relations $$f^{ ' }\left( x \right) =g\left( x+1 \right) $$ and $$g^{ ' }\left( x \right) =h\left( x-1 \right) $$. Then, $$f^{ '' }\left( 2x \right) $$ is equal to
    Solution
    We have $$f^{ ' }\left( x \right) =g\left( x+1 \right) $$
    $$\Rightarrow f^{ '' }\left( x \right) =g^{ ' }\left( x+1 \right) $$
    But $$g^{ ' }\left( x \right) =h\left( x-1 \right) $$
    $$\Rightarrow g^{ ' }\left( x+1 \right) =h\left( x+1-1 \right) $$
    $$=h\left( x \right) $$
    Therefore, $$ f^{ '' }\left( x \right) =h\left( x \right) $$
    $$\Rightarrow f^{ '' }\left( 2x \right) =h\left( 2x \right) $$
  • Question 3
    1 / -0
    Let $$ \phi (x) = \dfrac {b(x-a)}{b-a} + \dfrac {a(x-b)}{a-b} , $$ where $$ x \epsilon R $$ and $$a$$ and $$b$$  are fixed real numbers with $$ a \ne b. $$ Then $$ \phi ( a + b) $$ is equal to :
    Solution
    Given, $$ \phi (x) = \dfrac {b(x-a)}{b-a} + \dfrac {a(x-b)}{a-b}  $$
    Therefore, $$ \phi (a+b) = \dfrac {b(a+b-a)}{b-a} + \dfrac {a(a+b-b)}{a-b} $$
    $$ = \dfrac {b^2}{b-a} + \dfrac {a^2}{(a-b)} $$
    $$ = - \dfrac {b^2}{(a-b)} + \dfrac {a^2}{(a-b)} $$
    $$ = (a-b) \dfrac {(a+b)}{a-b} $$
    $$ = a+b $$
    $$= \phi (a) + \phi (b) $$
  • Question 4
    1 / -0
    If two sets $$A$$ and $$B$$ are having $$39$$ elements in common, then the number of elements common to each of the sets $$A\times B$$ and $$B\times A$$ are
    Solution
    If set $$A$$ and set $$B$$ have $$39$$ common elements, then the number of common elements in set $$A\times B$$ and set $$B\times A\,=39^2$$
  • Question 5
    1 / -0
    Let $$n$$ be a fixed positive integer. Define a relation $$R$$ in the set $$Z$$ of integers by $$aRb$$ if and only if $$\dfrac {n}{a - b}$$. The relation $$R$$ is
    Solution
    Given $$aRb$$ such that $$R:\dfrac{n}{a-b}\epsilon Z$$
    $$(A) aRa:$$ $$\dfrac{n}{a-a}=\infty$$ and $$\infty$$ does not belongs to  $$Z$$ 
    So not a reflexive relation and hence (A) and (D) options are ruled out.
    $$(B)aRb\epsilon Z $$ we need to check whether $$bRa\epsilon Z$$ or not.
    $$\dfrac{n}{a-b}\epsilon Z$$ Now, $$\dfrac{n}{b-a}\rightarrow -\dfrac{n}{a-b}$$
    As $$\dfrac{n}{a-b}$$ is integer so negative of it also integer 
    $$\therefore bRa\epsilon Z$$
    Hence relation $$R$$ is Symmetric.
    $$(C)$$ Now $$aRb\epsilon Z$$ and $$bRc\epsilon Z $$ but we can't say anything about $$aRc\epsilon Z.$$ Hence not a transitive relation.
  • Question 6
    1 / -0
    A function $$f$$ satisfies the relation $$f\left( { n }^{ 2 } \right) =f\left( n \right) +6$$ for $$n\ge 2$$ and $$f\left( 2 \right) =8$$. Then, the value of $$f\left( 256 \right) $$ is
    Solution
    Given that, $$f\left( { n }^{ 2 } \right) =f\left( n \right) +6$$

    On putting $$n=2$$, we get
    $$f\left( 4 \right) =f\left( { 2 }^{ 2 } \right) =f\left( 2 \right) +6=8+6=14$$         $$\left[ \because f\left( 2 \right) =8 \right] $$

    On putting $$n=4$$, we get
    $$f\left( 16 \right) =f\left( { 4 }^{ 2 } \right) =f\left( 4 \right) +6=14+6=20$$

    On putting $$n=16$$, we get
    $$f\left( 256 \right) =f\left( { 16 }^{ 2 } \right) =f\left( 16 \right) +6=20+6=26$$
  • Question 7
    1 / -0

    Directions For Questions

    Let a real valued polynomial function $$f(x)$$ with leading coefficient $$'x'$$ satisfied the relation $$(x^{2} - 9x + 18)f(x) - (x^{2} + 3x)f(x - 3) = 0$$.

    ...view full instructions

    Number of distinct real roots of the equation $$f(x) = 0$$, is
  • Question 8
    1 / -0
    Let $$ A = \{ 1,2,3,4 \} $$ and $$R$$ be a relation in $$A$$ given by $$ R = \{ (1,1) , (2,2) (3,3) , (4,4) , (1,2) , (3,1) , (1,3) \} $$ then $$R$$ is :
    Solution
    $$ A = \{ 1, 2, 3, 4\} $$
    $$ R = \{ (1,1) , (2,2),(3,3),(4,4),(1,2),(3,1)(1,3) \} $$ 
    clearly $$ (1,1) \epsilon R, (1,2) \epsilon (2,2) \epsilon R , (3,3) \epsilon R, (4,4) \epsilon R $$
    $$ \Rightarrow R $$ is Reflective.
    also $$ (1,1) \epsilon R, (1,2) \epsilon R \Rightarrow (1,2) R $$
    and $$(1,2) \epsilon R, (1,3) \epsilon R \Rightarrow (1,3) \epsilon R \Rightarrow R $$ is transitive
    But for $$ (1,2) \epsilon R (2,1) \notin R, \Rightarrow R $$ is not symmetric 
    $$ \Rightarrow R $$ is reflective and transitive only.
  • Question 9
    1 / -0
    The graph of a constant function $$f(x)=k$$ is?
    Solution
    A graph of a constant function is always a horizontal line.
    Horizontal line is a line which is parallel to $$x$$-axis.

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