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Functions Test 61

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Functions Test 61
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Let $$f(x)=\cos^{-1}(3x-1)$$. Then, dom $$(f)=?$$
    Solution
    $$\cos^{-1}(3x-1)$$ is defined only when $$(3x-1)\in [-1, 1]$$
    $$\Rightarrow 3x\in[0, 2]\Rightarrow x\in\left[0, \dfrac{2}{3}\right]\Rightarrow dom(f)=\left[0, \dfrac{2}{3}\right]$$.
  • Question 2
    1 / -0
    Let $$f(x)=\sqrt{\cos x}$$. Then, dom $$(f)=?$$
    Solution
    $$f(x)$$ is defined only when $$\cos x \geq 0$$

    $$\Rightarrow x$$ lies in $$1$$st or $$4$$th quadrant

    $$\Rightarrow dom(f)=\left[0, \dfrac{\pi}{2}\right]\cup\left[\dfrac{3\pi}{2}, 2\pi\right]$$.
  • Question 3
    1 / -0
    Let $$f(x)=log(1-x)+\sqrt{x^2-1}$$. Then, dom$$(f)=?$$
    Solution
    Let $$f(x)=g(x)+h(x)$$, where $$g(x)=\log (1-x)$$ and $$h(x)=\sqrt{x^2-1}$$.

    $$g(x)$$ is defined only when $$1-x > 0\Rightarrow x < 1$$. So, dom$$(g)=(-\infty, 1)$$.

    $$h(x)$$ is defined only when $$x^2-1\geq 0\Rightarrow x\geq 1$$ or $$x\leq -1$$.

    $$\therefore dom(h)=(-\infty, -1]\cup[1, \infty)$$.

    $$\therefore dom(f)=dom(g)\cap dom(h)=(-\infty, -1]$$.
  • Question 4
    1 / -0

    Directions For Questions

    $$f(x) = \begin{cases} x-1, -1 \leq x \leq 0\\x^2, 0\leq x\leq 1 \end{cases}$$ and g(x) = sin x, consider the functions.
    $$h_1(x) = f(|g(x)|) \space and \space h_2(x) = |f(g(x))|$$.

    ...view full instructions

    Which of the following is not true about $$h_2(x)$$
    Solution

  • Question 5
    1 / -0
    If $$ f(x)=\dfrac{2}{x-3}, g(x)=\dfrac{x-3}{x+4} $$ and $$ h(x)=-\dfrac{2(2 x+1)}{x^{2}+x-12} $$ then $$ \lim _{x \rightarrow 3}[f(x)+g(x)+h(x)] $$ is
    Solution
    c. We have $$ f(x)+g(x)+h(x)=\dfrac{x^{2}-4 x+17-4 x-2}{x^{2}+x-12} $$
    $$ =\dfrac{x^{2}-8 x+15}{x^{2}+x-12}=\dfrac{(x-3)(x-5)}{(x-3)(x+4)} $$
    $$ \therefore \lim _{x \rightarrow 3}[f(x)+g(x)+h(x)]=\lim _{x \rightarrow 3} \dfrac{(x-3)(x-5)}{(x-3)(x+4)}=-\dfrac{2}{7} $$.
  • Question 6
    1 / -0
    For  a real number y, let [y] denotes the greatest integer less than or equal to y. Then the function
    $$ f(x)= \dfrac{tan(\pi \left[ x-\pi  \right ])}{1+[x]^2} $$ is
    Solution

  • Question 7
    1 / -0

    Directions For Questions

    Consider the function $$f(x) = f\left(\dfrac{x - 1}{x}\right) = 1 + x \space \forall \space x \space \epsilon \space R - {0, 1}\space and \space g(x) = 2f(x) - x + 1$$

    ...view full instructions

    The number of roots of the equation g(x) = 1 is
    Solution

  • Question 8
    1 / -0
    The domain of the function $$f(x) = \sqrt{\ln_{(|x| - 1)} \left (x^{2} + 4x +4 \right )} $$ is 
    Solution

  • Question 9
    1 / -0
    If $$f : X \rightarrow Y$$, where  $$X$$ and $$Y$$ are sets containing natural numbers, $$f(x) = \frac{x + 5} {x + 2}$$ then the number of elements in the domain and range of $$f(x) are respectively
    Solution

  • Question 10
    1 / -0
    The domain of definition of the function $$f(x)$$ given by the equation $$2^x + 2^y = 2$$ is                                                                  (IIT-JEE, 2000)
    Solution

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