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Functions Test 42

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Functions Test 42
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  • Question 1
    1 / -0
    The domain of the function $$f(x) =  \dfrac{1} {\sqrt{\left \{ \sin {x} \right \} + \left \{ \sin (\pi + x) \right \}}}$$, where {.} denotes the fractional part, is
    Solution
    d. $$f(x) = \dfrac{1} {\sqrt{\left \{ \sin {x} \right \} + \left \{ \sin (\pi + x) \right \}}} = \dfrac{1} {\sqrt{\left \{ \sin {x} \right \} + \left \{- \sin {x} \right \}}}$$

    Now $$\left \{ \sin {x} \right \} + \left \{- \sin {x} \right \}$$ = $$\begin {cases}
                                                                                             0, & \sin x \space is \space an \space integer\\
                                                                                             1, & \sin x \space is \space not \space an \space integer
    \end {cases}$$
    For $$f(x)$$ to get defined $$\left \{ \sin {x} \right \} + \left \{- \sin {x} \right \} \neq 0$$
    $$\Rightarrow \sin x \neq integer$$
    $$\Rightarrow \sin x \neq \pm 1, 0$$
    $$\Rightarrow x \neq \dfrac {n \pi} {2}, n \space \epsilon  \space 1$$
    Hence, the domain is $$R - \left \{\dfrac {n \pi} {2}\ \text{where}\ n \space \epsilon \space I  \right \}$$



  • Question 2
    1 / -0
    Domain (D) and range (R) of $$f(x) = \sin^{-1}\left (\cos^{-1} [x]  \right )$$ where [.] denotes the greatest integer function is 
    Solution

  • Question 3
    1 / -0

    Directions For Questions

    If $$a_{o} = x, a_{n+1} = f(a_n)$$, where n = 0, 1, 2,.....then answer the following question

    ...view full instructions

    If f: $$R\rightarrow R$$ be given by $$f(x) = 3 + 4x$$ and $$a_n = A + Bx$$, then which of the following is not true?
    Solution
    Since $$a_1 = g(x) = 3 + 4x$$
    $$\therefore a_2 = g\{g^2(x)\} = g(3+4x) = 3 + 4(3+4x) = (4^2 - 1) + 4^x$$

    $$a_3 = g\{g^(x)\} = g(15 + 4^2x) = 3 + 4 (15 + 4^2x) = 63 + 4^3x = (4^3 - 1) + 4^3 x$$

    Similarly, we get $$a_n = (4^n - 1) + 4^n x$$
    $$\Rightarrow A = 4^n - 1 \space and B = 4^n$$

    $$\Rightarrow A + B + 1 = 2^{2n + 1}, |a - b| = 1\space and\space lim_{n \to \infty}\dfrac{4^n - 1}{4^n}$$

    $$= lim_{n \to \infty}\left(1 - \dfrac{1}{4^n}\right) = 1$$
  • Question 4
    1 / -0

    Directions For Questions

    $$f(x) = \begin{cases} 2x + a,   x \geq -1\\ bx^2 + 3, x < -1 \end{cases}$$

    and      $$g(x) = \begin{cases} x + 4,    0 \leq x \leq 4\\  -3x -2,   -2 < x < 0 \end{cases}$$

    ...view full instructions

    g(f(x)) is not defined if
    Solution

  • Question 5
    1 / -0
    The domain of $$f(x) = \sin^{-1} \left [ 2x^{2} - 3 \right ]$$, where [.]denotes the greatest integer function, is
    Solution
    We must have $$ -1 \leq \left [2x^{2} - 3  \right ] \leq 1$$

    $$\Rightarrow -1 \leq 2x^{2} - 3 \leq 2 \Rightarrow 1 \leq x^{2} < \frac{5} {2}$$

    $$\Rightarrow x \space \epsilon \left ( - \frac{5} {2}, -1 \right ] \cup \left [ 1, \frac{5} {2} \right )$$

  • Question 6
    1 / -0
    Let $$f(x)$$ and $$g(x)$$ be differentiable for $$0\times  < 1$$ such that $$f(0)=0, g(0), f(1)=6$$. Let there exist a real number $$c$$ in $$(0,1)$$ such that $$f'(c)=2g'(c)$$, then the value of $$g(1)$$ must be 
    Solution

  • Question 7
    1 / -0
    If g is the inverse of function $$f$$ and $$f'(x) = \frac{1}{1 + x}$$, then the value of g'(x) is equal to:
    Solution
    Since g is the inverse of $$f, f^{-1}(x) = g(x)$$
    $$\therefore f[f^{-1}(x)] = f [g(x)] = x$$
    $$\therefore f'[g(x)] \cdot \frac{d}{dx} [g(x)] = 1$$
    $$\therefore f'[g(x)] \times g'(x) = 1$$
    $$\therefore g'(x) = \frac{1}{f'[g(x)]}$$, where $$f'(x) = \frac{1}{1 + x^7}$$
    $$\therefore g(x) = 1 + [g(x)]^7$$
  • Question 8
    1 / -0
    Which one of the following is onto function define R to R .
    Solution

  • Question 9
    1 / -0
    Which of the following in one -one function defined from R to R
  • Question 10
    1 / -0
    Which of the following is onto function-
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