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Relations and Functions Test - 33

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Relations and Functions Test - 33
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  • Question 1
    1 / -0
    If $$f(t)=\dfrac{1+t}{1-t}, then f(-t)$$ is
    Solution
    Given, $$f(t) = \dfrac{1+t}{1-t}$$
    $$\therefore f(-t) = \dfrac{1-t}{1+t} = \dfrac{1}{f(t)}$$
  • Question 2
    1 / -0
    If $$\left\{ \left( 7,11 \right) ,\left( 5,a \right)  \right\} $$ represents a constant function, then the value of '$$a$$' is :
    Solution
    Since we know that in constant function image of every element is same.
    Here, image of 7 is 11, so image of any element will be 11.
    Therefore, image of 5 is too 11.
    Hence, $$a=11$$
  • Question 3
    1 / -0
    Let $$f : N\rightarrow R$$ be such that $$f(1) = 1$$ and $$f(1) = 2f(2) + 3f(3) + .... + nf(n) = n(n + 1)f(n)$$, for all $$n\epsilon N, n\geq 2$$, where $$N$$ is the set of natural numbers and $$R$$ is the set of real numbers. Then the value of $$f(500)$$ is
    Solution
    $$f(1) + 2f(2) + 3f(3) + ..... + nf(n) = n(n + 1) f(n)$$
    $$f(1) + 2f(2) + 3f(3) + ..... + (n - 1)f(n - 1) = (n - 1)nf(n - 1)$$
    Subtracting, $$nf(n) = n(n + 1) f(n) - n(n - 1)f(n - 1)$$
    $$\Rightarrow nf(n) = (n - 1)f(n - 1)$$
    Clearly, $$f(n) = \dfrac {1}{2n}$$. So, $$f(500) = \dfrac {1}{1000}$$
  • Question 4
    1 / -0
    Let the number of elements of the sets $$A$$ and $$B$$ be $$p$$ and $$q$$ respectively. Then the number of relations from the set $$A$$ to the set $$B$$ is
    Solution
    Given $$A$$ and $$B$$ are two sets with number of elements $$p$$ and $$q$$ respectively.

    The cartesian product of $$A$$ and $$B = A \times B = \{(a,b): (a\in A)$$ and $$(b\in B)\}$$
    Number of elements in $$A \times B = |A\times B| =|A|.|B| = pq$$

    Any relation from $$A$$ to $$B$$ is a subset of $$A\times B$$. 
    Hence number of relations from $$A$$ to $$B$$ is the number of subsets of $$A\times B$$
    $$= 2^{|A\times B|} = 2^{pq}$$
  • Question 5
    1 / -0
    If a function $$f(x)=x+\dfrac{1}{x}$$ is shifted along $$y-axis$$ as $$g(x)=f(x)+2$$. Then calculate the number of solution for $$g(x)=3$$.
    Solution
    $$g(x)=3\Rightarrow x+\dfrac1x +2=3 \Rightarrow x^2-x+1=0$$

     Now $$b^2-4ac=(-1)^2-4\times 1\times1=1-4=-3$$

    Hence as discriminant is negative there is no real solution to the equation but 1 complex pair as the solutions .
  • Question 6
    1 / -0
    The function $$f$$ is a differentiable function and satisfies the functional equation $$f(x) + f(y) = f(x + y)-  xy-  1$$ for every pair $$x, y$$ of real numbers. If $$f(1) = 1$$, then the number of integers $$n\neq  1$$ for which $$f(n) = n$$ is
    Solution

  • Question 7
    1 / -0
    In $$[0, 1]$$ Lagranges Mean Value theorem is NOT applicable to
    Solution
    For Lagrange's mean value theorem to be applicable to a function in the in interval $$[a,b]$$, the function must be continuous and differentiable in the $$[a,b]$$.

    A) $$f(x)=x|x|$$
              $$=x^2$$     ..$$x>0$$
    This is a quadratic function and so is continuous and differentiable in $$R$$

    B) $$f(x)=|x|$$
              $$=x$$     ..$$x>0$$

    C) $$f(x) =x$$ 
    Both these functions are linear and so they are continuous and differentiable in $$R$$

    So, Lagrange's mean value theorem is applicable to all the given functions.

    So, The answer is option (D).
  • Question 8
    1 / -0
    Let $$f(x) = x\left(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right), x > 1$$ Then
    Solution
    $$f(x)=x(\cfrac{1}{x-1} + \cfrac{1}{x} + \cfrac{1}{x+1})$$
    $$f(x)=x(\cfrac{2x}{x^2-1} + \cfrac{1}{x})$$
    $$f(x)=(\cfrac{2x^2}{x^2-1} +1)$$
    $$f(x)=\cfrac{2(x^2-1)+2}{x^2-1}+1 $$
    $$f(x)=2+\cfrac{2}{x^2-1} +1$$
    $$f(x)=3+\cfrac{2}{x^2-1} $$
    for $$x>1$$
    $$x^2>1$$
    $$x^2-1>0$$
    so,$$0<\cfrac{2}{x^2-1}<\infty$$
    thus $$f(x)>3$$

  • Question 9
    1 / -0
    The relation $$R$$ defined on set $$A = \left \{x :|x| < 3, x\epsilon I\right \}$$ by $$R = \left \{(x, y) : y = |x|\right \}$$ is
    Solution
    $$R$$ defined on set $$A = \left \{x :|x| < 3, x\epsilon I\right \}$$ by $$R = \left \{(x, y) : y = |x|\right \}$$
    $$A = \left \{x :|x| < 3, x\epsilon I\right \}$$
        $$=\left\{x:-3<x<3, x\epsilon I\right\}$$
    $$\therefore$$  $$R = \left \{(x, y) : y = |x|\right \}$$
             $$=\left\{(-2,2),(-1,1),(0,0),(1,1),(2,2)\right\}$$
    Hence, option A is correct.
  • Question 10
    1 / -0
    How many ordered pairs of (m, n) integers satisfy $$\displaystyle\frac{m}{12}=\frac{12}{n}$$?
    Solution
    $$m \times n = 144$$
    144 can be written in multiplication of two integers in the following ways:

    $$1 \times 144 = 2 \times 72 = 3 \times 48 = 4 \times 36 = 6 \times 24 = 8 \times 18 = 9 \times 16 = 12 \times 12$$

    For all the factors except $$(12, 12)$$ there would be 14 ordered pairs.
    Adding one more, yields 15 possibilities
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