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Functions Test 18

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Functions Test 18
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  • Question 1
    1 / -0
    Which of the following is an onto function
    Solution
    A function is called onto iff,  Range$$=$$ co-domain.
    A. In $$[0,\pi]$$, Range of $$\sin x$$ is $$[0,1]\Rightarrow$$ Not onto. 
    B. In $$[0,\pi]$$, Range of $$\cos x$$ is $$[-1,1] =$$co-domain  $$\Rightarrow $$onto function.                          
    C.
    Range of $$e^x$$ is $$(0, \infty)\neq$$ co-domain.
    D.
    Range of  $$x^3$$  is not R, since $$x\in Q$$
  • Question 2
    1 / -0
    Let f : {x,y,z} $$\rightarrow$$ {a,b,c} be a one-one function. It is known that only one of the following statment is true, and only one such function exists :

    find the function f (as ordered pair).(i) f(x) $$\neq$$ b
    (i) f(y) = b

    (ii) f(z) $$\neq$$ a
    Solution
    When (i) is true, then $$f(x) \neq b, f(y) \neq b , f(z) = a,b,c $$

    $$\Rightarrow$$ Two ordered pair function is possible $$ f(x) = a, f(y) = c, f(z) = b$$ or $$f(x) = c, f(y) = a, f(z) = b$$

    But given $$f$$ is one-one and only one such function is possible. Hence (i) can't be true.

    When (ii) is true, then $$f(y) = b, f(z) =c , f(x) = a\Rightarrow f(x) = a, f(y) = b, f(z) = c$$

    Clearly if (ii) is true then it is  satisfying every condition.

    Hence ordered pair of $$f$$ is $$\{(x,a), (y,b), (z,c)\}$$
  • Question 3
    1 / -0
    The function $$f(x)$$ is defined on the interval $$[0, 1]$$. Find the domain of the function:
    $$f(\sin x)$$
    Solution
    Given $$f(x)$$ defined in interval $$[0,1]$$

    $$\therefore f(\sin x)$$ will be defined if $$\sin x \in [0,1]$$

    $$i.e  \sin x \geq 0\Rightarrow 0 \leq x \leq \pi$$ in one period. We know 

    $$\sin x$$ is periodic with period $$2\pi$$

    Hence $$x \in [2k\pi, 2k\pi+\pi]$$ where $$ k \in I$$
  • Question 4
    1 / -0
    Write explicitly, functions of $$y$$ defined by the following equations and also find the domains of the given implicit functions :
    $$10^{x} + 10^{y} = 10$$
    Solution
    $$10^{x} + 10^{y} = 10$$
    $$10^{y} = 10 - 10^{x}$$
    $$y = \log_{10}(10 - 10^{x})$$

    Domain :
    For $$y$$ to be defined 
    $$10 - 10^{x} > 0 $$
    $$\Rightarrow 10 > 10^{x}$$
    $$\Rightarrow x < 1 $$
    $$\Rightarrow x \in (-\infty, 1)$$
  • Question 5
    1 / -0
    If $$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},g(x)=\begin{cases} x\quad \quad if\quad x\leq 1 \\ 2-x\quad if\quad x>1 \end{cases}$$

    and $$x\, \in\, (1, 2)$$, then $$g(f(x))$$ is equal to
    Solution
    $$x\, \epsilon\, (1, 2)$$

    Hence for x >1, $$f(x) = 5-x^2$$and $$f(x) \epsilon (1,4) $$

    $$f(x) >1$$

    $$g(f(x)) = 2 - f(x)=  2-5+x^2 = x^2-3$$ 
  • Question 6
    1 / -0
    If $$g(x) = 2x + 1$$ and $$h(x) = 4x^{2} + 4x + 7$$, find a function $$f$$ such that $$f o g = h$$
    Solution
    $$f(g(x)) = h = 4x^2+4x+7=(2x+1)^2+6=(g(x))^2+6$$

    $$\Rightarrow f(x) =x^2+6$$
  • Question 7
    1 / -0
    Which of the following are two distinct linear functions which map the interval $$[-1, 1]$$ onto $$[0, 2]$$
    Solution
    Out of two linear functions one will be increasing and other will be decreasing.

    Let $$f(x) = ax+b$$

    For increasing: $$f(-1)=0=> b-a=0\Rightarrow b=a$$ and $$f(1) = 2\Rightarrow a+b=2\Rightarrow a=b=1\therefore f(x) = 1+x$$

    For decreasing: $$f(-1)=2=> b-a=2$$ and $$f(1) = 0\Rightarrow a+b=2\Rightarrow b=1, a=-1\therefore f(x) = 1-x$$
  • Question 8
    1 / -0
    Find domain of the given function:
    $$f(x)\,=\, \displaystyle \frac{\sqrt{\cos\, x\, -\,\frac{1}{2}}}{\sqrt{6\, +\, 35\, x\, -\, 6x^{2}}}$$
    Solution
    $$f(x)\, =\, \displaystyle \dfrac{\sqrt{\cos x\, -\, \dfrac{1}{2}}}{\sqrt{6\, +\, 35x\, -\, 6x^{2}}}$$
    For $$f(x)$$ to be defined $$ \cos\, x\, -\, \displaystyle \dfrac{1}{2}\, \geq\, 0$$
    $$\Rightarrow\,

    x\, \in\, \left[ \displaystyle - \dfrac{\pi}{3},\, \displaystyle

    \dfrac{\pi}{3}\, \right]\, \cup\, \displaystyle \left[\dfrac{5 \pi}{3},\,

    \displaystyle \dfrac{7 \pi}{3} \right]$$
    Also $$ 6x^{2}\, -\, 35 x\, -\, 6\, <\, 0\, \Rightarrow\, -\, \dfrac{1}{6}\, <\, x\, <\, 6$$
    Thus combining above two we get required domain $$\Rightarrow\,

    x\, \in\, \left[-\, \displaystyle \dfrac{\pi}{3},\, \displaystyle

    \dfrac{\pi}{3} \right]\, \cup\, \left[ \displaystyle \dfrac{5 \pi}{3},\, 6

    \right)$$
  • Question 9
    1 / -0
    The functions whose domain is $$R$$ are
    Solution
    From the graphs we can clearly see that, the domain of cosech x and coth x is [R-{0}]
    so, Option B,C,D are wrong.
    And the domain of sinh x, cosh x, tanh x, sech x is R.
    So, option A is correct.

  • Question 10
    1 / -0
    The domain of the following function $$f(x)\, =\, \sqrt {\displaystyle \frac{1\, -\, 5^{x}}{7^{-x}\, -\, 7}}$$ is
    Solution
    $$f(x)\, =\, \sqrt {\displaystyle \frac{1\, -\, 5^{x}}{7^{-x}\, -\, 7}}$$
    For $$f(x)$$ to be defined (i) $$1\, -\, 5^{x}\, \geq\, 0\, \Rightarrow\, 1\, \geq\, 5^{x}\, \Rightarrow\, x\, \leq\, 0$$
    And $$7^{-x}\, -\, 7\, >\, 0$$
    $$\Rightarrow\, x\, <\, -1$$
    $$\Rightarrow\, x\, \in\, (-\infty,\, -1)$$
    or (ii) $$1\, -\, 5^{x}\, \leq\, 0\, \Rightarrow\, x\, \geq\, 0$$
     & $$7^{-x}\, -\, 7\, <\, \, 0\, \Rightarrow\, x\, >\, -1$$
    $$\Rightarrow\, x\, \in\, [0,\, \infty)$$
    Thus domain is $$x\, \in\, (\,\infty,\, -1)\, \cup [0,\, \infty)$$
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