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Relations and Functions Test - 54

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Relations and Functions Test - 54
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  • Question 1
    1 / -0
    Find the number of binary operations on the set $$\left \{a, b\right \}$$ 

    Solution
    Considering 'n' elements in a set, the no. of binary operations on that set will be
    $$=n^{n^{2}}$$

    Applying this concept to the above set gives us

    $$=2^{2^{2}}$$  ...................  $$\because n=2$$

    $$=2^{4}$$
    $$=16$$.
  • Question 2
    1 / -0
    If $$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$$, then the set of real numbers $$x$$ such that $$f\left\{ g(x) \right\} =x$$ is
    Solution
    Given $$f(x)={2}^{100}x+1, g(x)={3}^{100}x+1$$

    Now $$fo{g(x)}=x$$

    $$\Rightarrow$$ $$f({3}^{100}.x+1)=x\\$$
    $$\Rightarrow$$ $${2}^{100}({3}^{100}.x+1)+1=x\\$$
    $$\Rightarrow$$ $${6}^{100}.x+{2}^{100}+1=x\\$$

    $$\Rightarrow$$ $$x(1-{6}^{100})=(1+{2}^{100})$$

    $$\Rightarrow$$ $$x=\cfrac{1+{2}^{100}}{1-{6}^{100}}$$

    Hence $$fog(x)=x$$ represent a singleton set.
  • Question 3
    1 / -0
    In three element group $$\{e, a, b\}$$ where $$e$$ is the identity, $$a^5b^4$$ is equal to 
    Solution
    Given group $$G\equiv\{e,a,b\}$$
    It is given that $$e$$ is an identity.
    Therefore only possibilities we have are
    Either $$a\cdot a = e$$ and $$b\cdot b = e$$ .... (i)
    Or $$a\cdot b = b\cdot a = e$$    ...(ii)

    Case I:
    $$a\cdot a = e$$ and $$b\cdot b = e$$
    $$a^5b^4 = a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b\cdot b$$
    $$\Rightarrow a^5b^4 = e\cdot a\cdot a\cdot a\cdot b\cdot b\cdot e$$
    $$\Rightarrow a^5b^4 = e\cdot a\cdot e$$
    $$\Rightarrow a^5b^4 = a$$

    Case II:
    $$a\cdot b = b\cdot a = e$$
    $$a^5b^4 = a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b\cdot b$$
    $$\Rightarrow a^5b^4 = a\cdot a\cdot a\cdot a\cdot e\cdot b\cdot b\cdot b$$
    $$\Rightarrow a^5b^4 = a\cdot a\cdot a\cdot e\cdot b\cdot b$$
    $$\Rightarrow a^5b^4 = a\cdot a\cdot e\cdot b$$
    $$\Rightarrow a^5b^4 = a\cdot e$$
    $$\Rightarrow a^5b^4 = a$$

    Hence, both cases resulted in $$a$$.
  • Question 4
    1 / -0
    Find the value of $${f}^{-1}(1.5)$$ if $$f(x)=\sqrt [ 3 ]{ { x }^{ 3 }+1 } $$.
    Solution
    As per the condition, $$1.5 = \sqrt[3]{x^3+1}$$
    Taking cube roots on both the sides,
    $$1.5^3=x^3+1$$
    $$3.375 - 1 = x^3$$
    $$2.375 = x^3$$
    $$x \approx 1.3$$
  • Question 5
    1 / -0
    If $$f(x)\, =\, (p\, -\, x^n)^{1/n},\, p\, >\, 0$$ and $$n$$ is a positive integer, then $$f(f(x)) =$$
    Solution
    Given, $$f(x)=(p-x^{n})^{\dfrac{1}{n}}$$

    $$\therefore f(f(x))=(p-((p-x^{n})^{\dfrac{1}{n}})^{n})^{\dfrac{1}{n}}$$

    $$=(p-(p-x^{n}))^{\dfrac{1}{n}}$$

    $$=(x^{n})^{\dfrac{1}{n}}$$

    $$=x$$
  • Question 6
    1 / -0
    If $$\displaystyle { Q }_{ 1 }$$ is the set of all relations other than $$1$$ with the binary operation $$\displaystyle \ast $$ defined by $$\displaystyle a\ast b=a+b-ab$$ for all $$a, b \in \displaystyle Q_1$$, then the identity in $$\displaystyle { Q }_{ 1 }$$ with respect to $$\displaystyle { Q }_{ 1 }$$ is
    Solution
    Let $$b$$ be the identity
    $$a*b=a$$
    $$a+b-ab=a$$
    $$b-ab=0$$
    $$b(1-a)=0$$
    $$b=0$$ and $$a=1$$
    However, $$1$$ is excluded.
    Hence
    $$b=0$$
    Hence $$0$$ is the identity element
  • Question 7
    1 / -0
    Which of the following is a subgroup of the group $$G = \left \{1, 2, 3, 4, 5, 6\right \}$$ under $$\otimes_{7}$$.
    ($$\otimes_7$$: under multiplication modulo 7)
    Solution
    (c) can not be a subgroup as identity $$'1'$$ is not present
    (d) can not be a subgroup of $$2\otimes_{7} 3 = 6\not {\epsilon} \left \{1, 2, 3\right \}$$
    (a) can not be a subgroup as $$2\otimes_{7} 6 = 5\not {\epsilon} \left \{2, 6, 1\right \}$$
    $$\therefore$$ (b) is a subgroup
  • Question 8
    1 / -0
    If $$f: R\rightarrow R^{+}$$ and $$g: R^{+} \rightarrow R$$ are such that $$g(f(x)) = |\sin x|$$ and $$f(g(x)) = (\sin \sqrt {x})^{2}$$, then a possible choice for f and g is
    Solution
    As f returns only positive values and takes any real number it can be mod or square hence option 2 is eliminated 
    For g only positive values has to be given and it will return any real number
    so on trying the option only option $$A$$ and $$C$$ results in the given equation ie $$g(f(x)) = |sin x|$$

    Now function g should take positive values and return real values which is satisfied by only option $$A$$ as in option $$C$$ square root of positive number will always result in positive number while $$sin\sqrt{x}$$ will give -ve real numbers as well.
  • Question 9
    1 / -0
    In $$Z$$, the set of all integers, the inverse of $$-7$$ w.r.t. defined by $$a\times b=a+b+7$$ for all $$a, b, \in Z$$ is : 
    Solution
    Consider the identity element of 'b' for any element $$a\in Z$$
    $$a\times b=a$$
    $$a+b+7=a$$
    $$b+7=0$$
    $$b=-7$$.
    Hence identity element is -7.
    Hence $$e=-7$$.
    Now for the inverse element
    $$a\times a^{-1}=e$$
    $$a+a^{-1}+7=-7$$       ...$$(e=-7)$$
    $$a^{-1}=-14-a$$
    Hence
    $$7^{-1}=-14-(-7)$$
    $$=-7$$.
  • Question 10
    1 / -0
    Let Q be the set of all rational numbers in [0, 1] and $$f : [0, 1]\rightarrow [0, 1]$$ be defined by $$f(x)=\begin{cases}x&for&x\in Q\\ 1-x&for&x\notin Q\end{cases}$$
    Then the set $$S=\{x\in [0, 1]: (f\, o \, f)(x)=x\}$$ is equal to
    Solution
    Let $$x\in Q$$, then,
    $$f(x)=x$$ where $$x\in Q$$
    So, $$fof(x)=f(f(x))=f(x)=x$$   as  $$x\in Q$$
    $$\therefore fof(x)=x$$   when $$x \in Q$$
    Now, 
    Let $$x\notin Q$$ then
    $$f(x) =1-x$$
    $$\therefore fof(x)=1-(1-x) = x$$
    as $$1-x\notin Q$$ as $$x\notin Q$$
    where $$x\notin Q$$
    $$fof(x)=\begin{cases}x \,where\, x\in Q \,\&\, x\in [0, 1] \\ x\, where x\notin Q\, \&\, x\in [0, 1] \end{cases}$$
    $$\therefore$$ the set $$S = [0, 1]$$
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