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Functions Test 8

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Functions Test 8
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Weekly Quiz Competition
  • Question 1
    1 / -0
    If $$f(x)=\dfrac{x}{\sqrt{1+x^{2}}}$$ then $$fofof(x)=$$
    Solution
    $$fofof\left(x\right)=fof\left(\dfrac{x}{\sqrt{1+x^{2}}}\right)$$
    $$=f\left(\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+x^{2}}}}\right)$$
    $$=f\left(\dfrac{x}{\sqrt{1+2x^{2}}}\right)$$
    $$=\left(\dfrac{\dfrac{x}{\sqrt{1+2x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+2x^{2}}}}\right)$$
    $$=\dfrac{x}{\sqrt{1+3x^{2}}}$$
  • Question 2
    1 / -0
    The domain of $$\displaystyle f(x)=\sqrt{\frac{x+7}{x+5}}$$ is
    Solution
    $$f$$ is defined when $$\displaystyle \frac{x+7}{x+5}\ge 0$$  &  $$x+5\neq 0$$ i.e. $$x\neq -5$$
    $$\displaystyle \frac{(x+5)(x+7)}{(x+5)^{2}}\ge 0$$
    $$\Rightarrow (x+5)(x+7) \ge 0$$
    $$\Rightarrow x\in (-\infty ,-7]\cup [-5,\infty )$$
    As $$f$$ is not defined at -5
    $$x\in (-\infty ,-7]\cup (-5,\infty )$$

  • Question 3
    1 / -0

    The domain of $$f(x)=\sqrt{[x]^{2}-[x]-6}$$ is
    Solution
    $$f$$ is defined if term under root is positive.
    $$\Rightarrow [x]^{2}-[x]-6\geq 0$$
    $$([x]+2)([x]-3)\geq 0$$
    $$\Rightarrow [x]\ge 3 $$ or $$[x]\le -2$$
    $$\Rightarrow x\in (-\infty ,-1)\cup [3,\infty )$$

  • Question 4
    1 / -0
    The domain of $$f(x)=\sqrt{\sin^{-1}(\log_{9}x)}$$ is
    Solution
    $$\sin^{-1}x$$ is defined if $$x\in [-1,1]$$
    But as $$\sin^{-1}$$ is under square root, it has to be positive
    $$\Rightarrow 0\leq \log_{9} x\leq 1$$
    $$\Rightarrow 1\leq x\leq 9$$
  • Question 5
    1 / -0
    The domain of $$f(x)=\sin^{-1}\displaystyle \left\{\log_{3}\left(\frac{x^{2}}{3}\right)\right\}$$ is
    Solution
    $$\sin^{-1}x$$ is defined if $$x\in [-1,1]$$
    $$\Rightarrow -1\leq \log_{3}\dfrac{x^{2}}{3}\leq 1$$
    $$\Rightarrow \dfrac{1}{3}\leq \dfrac{x^{2}}{3}\leq 3$$
    $$1\leq x^{2}\leq 9$$
    $$\Rightarrow x\in [-1,-3]\cup [1,3]$$
    $$\therefore x\in [-3,-1]\cup [1,3]$$
  • Question 6
    1 / -0
    The domain of $${ f }({ x })=\tan ^{ -1 } (5x) $$ is
    Solution
    $$\tan^{-1}x$$ is defined for all $$x$$ as range of $$\tan x$$ is from $$(-\infty ,\infty )$$.
    $$\therefore$$ domain of $$\tan^{-1}5x$$ is $$(-\infty ,\infty )$$.
    Hence, option 'A' is correct.
  • Question 7
    1 / -0
    The domain of $$f(x)=\sqrt{\left | x \right |-x} $$ is
    Solution
    function is defined when $$|x|-x\geqslant 0$$
    $$\Rightarrow|x|\geqslant x$$
    $$|x|\geqslant x\  \forall \ x\in R$$
    $$\therefore  f$$ is defined every where in $$R$$
  • Question 8
    1 / -0
    The domain of $$ \displaystyle f(x)= { \sin }^{ -1 }\left(\dfrac { 2x-3 }{ 5 } \right)$$ is
    Solution
    $$\sin^{-1}x$$ is defined if $$x\in [-1,1]$$
    $$\Rightarrow -1\leq \dfrac{2x-3}{5}\leq 1$$
    $$-5\leq 2x-3\leq 5$$
    $$-2\leq 2x\leq 8$$
    $$-1\leq x\leq 4$$

  • Question 9
    1 / -0
    The domain of  $$f(x)=\log_{x}12$$ is
    Solution
    $$ x > 0 $$ and $$x\not=1$$
    Therefore, $$(0, 1) \cup(1,\infty)$$.
    Hence, the correct answer is 'B'.
  • Question 10
    1 / -0
    The domain of $${f}({x})=\log|x^{2}-9|$$ is
    Solution
    $$f$$ is defined for all $$[x^{2}-9]>0   \Rightarrow    x\neq \pm 3$$
    $$\therefore$$ domain of $$f$$ is $$R-\left \{ -3,3 \right \}$$
    Hence, option 'A' is correct.
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