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This is one of our many quizzes on Domain of a function. Click on this to access more such competitive quizzes on Domain and range relation.

Domain of a function Quiz 2

mathematics

 Find the domain of $\cos \left( \log \left( x \right) \right)$.

  We will first substitute x = 0. Thus we get, $\\  y=\cos \left( \log \left( x \right) \right) \\ \Rightarrow y=\cos \left( \log \left( 0 \right) \right) \\ \Rightarrow y=\cos \left( -\infty  \right) \\ $ Since, the value of cosine oscillated between -1 and 1 so it will never gives that value at infinity. So, we will discard a and b options. $ \\ $ Now, we will take one value which will be less than 0, $x = -1$. Thus we get $y=\cos \left( \log \left( -1 \right) \right)$. As log is not defined on negative terms therefore option d is discarded as well. $ \\ $ Here, domain is all the values of $x$ which are greater than 0 that satisfy the above equation. Since, all positive numbers greater than 0 for $x$ define the equation therefore, the domain of the given function is $\left\{ x\in \mathbb{R}:x > 0 \right\}$.

 Consider the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)$. If the domain of the function $f\left( x \right)$ is $\left[ a,b \right]$ then what will be the domain of the inverse function?

 The domain of a function is the range of its inverse function.

  As it is given that the domain of the function i.e., x is $\left[ a,b \right]$ so, $\\ a\le x\le b \\ \Rightarrow f\left( a \right)\le f\left( x \right)\le f\left( b \right) \\ $ Since, the domain of the inverse function is basically the range of $f\left( x \right)$ so, $\left[ f\left( a \right),f\left( b \right) \right]$ will be the domain of the inverse function.      

 What do you think will be the domain for $y=\dfrac{1}{\sqrt{{{x}^{2}}+\left[ x \right]}}$

 Since, the function is inside the root so, we will try to figure out if the function inside it is greater than or less than or equal to 0 or not.

 We will consider $\sqrt{{{x}^{2}}+\left[ x \right]}$. Since, this is in the denominator so the given function to be defined we need it to be not equal to 0. Also, if we take values for $x$ less than 0 then we will get, $y=\dfrac{1}{\sqrt{{{\left( -1 \right)}^{2}}+\left[ -1 \right]}}=\dfrac{1}{\sqrt{1-1}}$ which is not defined. 

 If we are given a function $\left| \sqrt{{{x}^{\dfrac{1}{3}}}}+6 \right|$. What do you think will be the range of it?

 The number in modulus is greater than and equal to 0.

 As we know that any number inside modulus is always greater than 0 so, we have $\left| \sqrt{{{x}^{\dfrac{1}{3}}}}+6 \right|\ge 0 \\ \Rightarrow 0\le \left| \sqrt{{{x}^{\dfrac{1}{3}}}}+6 \right|< \infty  \\ \Rightarrow 0\le \left| \sqrt{{{x}^{\dfrac{1}{3}}}}+6 \right|< \infty  \\ $ Clearly, the range is $\left[ 0,\infty  \right)$.  

 The range of $f\left( x \right)={{\left( \dfrac{1}{1-x} \right)}^{\dfrac{1}{2}}}$ is 

 The square root function is defined only for non-negative arguments also it is important to have a defined function.

 $f\left( x \right)={{\left( \dfrac{1}{1-x} \right)}^{\dfrac{1}{2}}}=y \\ \Rightarrow y=\sqrt{\dfrac{1}{1-x}} \\ f\left( x \right)$ is defined when $1-x\ge 0 \\ \Rightarrow x < 1 \\ i.e., x\in \left( -\infty ,1 \right)$