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Relations and Functions Test - 34

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Relations and Functions Test - 34
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  • Question 1
    1 / -0
    The number of ordered pairs $$(x, y)$$ of real numbers that satisfy the simultaneous equations.
    $$x + y^{2} = x^{2} + y = 12$$ is.
    Solution
    $$x + y^{2} = x^{2} + y = 12$$
    $$x + y^{2} = x^{2} + y$$
    $$x - y = x^{2} - y^{2} \Rightarrow x = y, x + y = 1$$
    when $$x = y, x^{2} + x = 12$$
    $$x^{2} + x - 12 = 0$$
    $$x^{2} + 4x - 3x - 12 = 0$$
    $$(x + 4)(x - 3) = 0$$
    $$x = -4, 3$$
    $$(3, 3),(-4, -4)$$
    When $$y = 1 - x$$
    $$x + (1 - x)^{2} = 12$$
    $$x^{2} - x - 11 = 0$$
    $$x = \dfrac {1\pm \sqrt {1 + 44}}{2}$$ for two value of $$x$$ there are two value of $$y$$.
    So four pair.
  • Question 2
    1 / -0
    If $$e^{x} = y + \sqrt {1 + y^{2}}$$, then the value of $$y$$ is
    Solution
    Given $$e^{x} = y + \sqrt {1 + y^{2}}$$

    $$\implies e^{x} - y = \sqrt {1 + y^{2}}$$

    Squaring both side, we have

    $$e^{2x} + y^{2} - 2e^{x}y = 1 + y^{2}$$
    $$\implies 2e^{x}y = e^{2x} - 1$$

    $$\implies y = \dfrac {e^{2x} - 1}{2e^{x}} \implies y = \dfrac {1}{2} [e^{x} - e^{-x}]$$

    Hence, option B is correct.
  • Question 3
    1 / -0
    The number of ordered pairs $$\left( m,n \right)$$, where $$ m, n \in \left\{ 1,2,3,\dots ,50 \right\}$$, such that $$ { 6 }^{ m }+{ 9 }^{ n }$$ is a multiple of $$5$$ is
    Solution
    For a number to be divisible by $$5$$ the last digit shall be a multiple of $$5$$
    Any power of 6 has its last digit(or unit's digit) equal to 6.
    The power of $$9$$ of the form
    $$4p+1$$ has it's last digit as 9
    $$4p+2$$ has it's last digit as 1
    $$4p+3$$ has it's last digit as 9
    $$4p$$ has it's last digit as 1
    Thus $$6^{m}+9^{n}$$ will be a multiple of $$5$$ only if $$m$$ is any number from the given set and $$n$$ is a number of the form $$4p+1$$ or $$4p+3$$.
    Therefore the number of ways to select $$m$$ is $$\binom{50}{1}$$ and the number of ways to select $$n$$ from the given set is $$\binom{25}{1}$$
    hence the total number of ordered pairs $$(m,n)$$ is $$\binom{50}{1} \times \binom{25}{1}=50 \times 25=1250$$

  • Question 4
    1 / -0
    If $$A$$ is a finite set having $$n$$ elements, then the number of relations which can be defined in $$A$$ is
    Solution
    A relation is simply a subset of cartesian product $$ A\times A$$. 
    If $$ {A\times A}= [(a_1,a_1), (a_1,a_2),.....(a_1,a_n), $$
    $$ (a_2,a_1),(a_2,a_2),........(a_2,a_n) $$
    $$ ...... $$
    $$ (a_n,a_1), (a_n,a_2).........(a_n,a_n)]$$
    We can select first element of ordered pair in $$n$$ ways and second element in $$n$$ ways.
    So, clearly this set of ordered pairs contain $$n^2 $$ pairs.
    Now, each of these $$n^{2}$$ ordered pairs can be present in the relation or can't be. So, there are $$2$$ possibilities for each of the $$n^{2}$$ ordered pairs.
    Thus, the total no. of relations is $$2^{n^{2}}$$.
    Hence, option C is correct.
  • Question 5
    1 / -0
    Let R be a relation from A = {1, 2, 3, 4} to B = {1, 3, 5} such that
    R = [(a, b) : a < b, where a $$\varepsilon$$ A and b $$\varepsilon$$ B].
    What is RoR$$^{-1}$$ equal to?
    Solution
    $$R$$ gives the set $$\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$$
    $$R^{-1}$$ gives the set $$\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4),(3,3)\}$$
    To compute $$RoR^{-1}$$, we pick 1 element from $$R^{-1}$$ and its corresponding relation from $$R$$.
    eg: $$(3,1)\in R^{-1}$$ and $$(1,3)\in R\implies (3,3)\in RoR^{-1}$$
    Similarly, computing for all such pairs, we have
    $$RoR^{-1}=\{(3,3),(3,5),(5,3),(5,5)\}$$
  • Question 6
    1 / -0
    The relation R define on the set of natural numbers as {(a, b) : a differs from b by 3} is given.
    Solution
    Let $$R=\{(a,b): a, b \in N, a-b=3\}$$
    $$=\{(n+3,n):n \in N\}$$
    $$=\{(4,1), (5, 2), (6, 3).......\}$$
    Hence, option B is correct.
  • Question 7
    1 / -0
    Let $$f(x) = \dfrac {2025^{x}}{45 + 2025^{x}}$$, then the value of $$f\left (\dfrac {1}{2025}\right ) + f\left (\dfrac {2}{2025}\right ) + f\left (\dfrac {3}{2025}\right ) + ... + f\left (\dfrac {2024}{2025}\right )$$ is equal to
    Solution
    We know $$(45)^2=2025$$
    Let $$a=45$$
    $$\rightarrow f(x)=\dfrac{a^{2x}}{a+a^{2x}}$$
    $$Val=f(\dfrac{1}{2025})+f(\dfrac{2}{2025})+......+f(\dfrac{2024}{2025})$$
    Now by seeing carefully we can observe that we are asked something of form $$f(y)+f(1-y)$$ where 
    y=$$\dfrac{1}{2025},\dfrac{2}{2025},.......,\dfrac{1012}{2025}$$
    $$f(y)+f(1-y)=\dfrac{a^{2y}}{a+a^{2y}}+\dfrac{a^{2(1-y)}}{a+a^{2(1-y)}}$$
    $$\dfrac{a^{2y}}{a+a^{2y}}+\dfrac{a^2a^{-2y}}{a+a^2a^{-2y}}=\dfrac{a^{2y}}{a+a^{2y}}+\dfrac{a}{a^{2y}+a}=1$$
    $$\therefore Val=1+1+1.....1012$$ times$$=1012$$
  • Question 8
    1 / -0
    If $$f(x) = \dfrac {a^{x} + a^{-x}}{2}$$ and $$f(x + y) + f(x - y) = k f(x) f(y)$$, then $$k$$ is equal to
    Solution
    Given, $$f(x)=\dfrac {a^x+a^{-x}}{2}$$
    On putting $$y = 0$$, in the given relation, we $$f(x) + f(x) = k\cdot f(x)\cdot f(0)$$
    $$\Rightarrow 2f(x) = k\cdot f(x)$$     .....$$ [\because f(0)=1]$$
    $$\Rightarrow  k = 2$$
  • Question 9
    1 / -0
    $$R$$ is a relation on $$N$$ given by $$R = \left \{(x, y)|4x + 3y = 20\right \}$$. Which of the following doesnot belong to $$R$$?
    Solution
    $$R$$ is given by $$\{(x,y)|4x+3y=20\}$$
    $$4 \times (-4)+3 \times 12=-16+36=20$$
    $$4 \times 5+3 \times 0=20+0=20$$
    $$4 \times 3+3 \times 4=12+12=24\neq 20$$
    $$4 \times 2+3 \times 4=8+12=20$$
    So, among the given options $$C$$ does not satisfy the given condition.
    All except $$C$$, does not belong to the set $$R.$$
    Hence, option C is correct.
  • Question 10
    1 / -0
    If $$f\left( 1 \right) =1$$, $$f\left( 2n \right) =f\left( n \right) $$ and $$f\left( 2n+1 \right) ={ \left\{ f\left( n \right)  \right\}  }^{ 2 }-2$$ for $$n=1,2,3,\dots $$, then the value of $$f\left( 1 \right) +f\left( 2 \right) +\cdots +f\left( 25 \right) $$ is
    Solution
    Given that, $$f\left( 1 \right) =1$$, $$f\left( 2n \right) =f\left( n \right) $$, $$f\left( 2n+1 \right) ={ \left\{ f\left( n \right)  \right\}  }^{ 2 }-2$$

    Now, $$f\left( 2 \right) =f\left( 2\times 1 \right) =f\left( 1 \right) =1$$,

    $$f\left( 3 \right) =f\left( 2\times 1+1 \right) $$
    $$={ \left\{ f\left( 1 \right)  \right\}  }^{ 2 }-2=1-2=-1$$,
    $$f\left( 4 \right) =f\left( 2\times 2 \right) =f\left( 2 \right) =1$$
    $$f\left( 5 \right) =f\left( 2\times 2+1 \right) $$

    $$={ \left\{ f\left( 2 \right)  \right\}  }^{ 2 }-2=1-2=-1$$
    and so on

    Now, $$f\left( 1 \right) +f\left( 2 \right) +\cdots +f\left( 25 \right) =1+1-1+1-1+\cdots +\left( -1 \right) =1$$
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