Question

Which of the following is possible for real values of $\theta$ and x.
A. $\cos \theta =x+\dfrac{1}{x}$
B. $sec\theta =\dfrac{{{x}^{2}}}{1+{{x}^{2}}}$
C. $\operatorname{cosec}\theta =\dfrac{x}{1+{{x}^{2}}}$
D. $\tan \theta =\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}-3x+5}$

Answer 1

Hint: We know the domain and range of cosine functions. The domain of a cosine function is
$(-\infty ,\infty )$ and its range is [-1,1]. Here, we have, $f(x)=cos\theta$ where $f(x)=x+\dfrac{1}{x}$ . Find the range of the function $f(x)$ and also, we have the range of cosine function. If their ranges overlap, then we have possible real solutions of $\theta$ and x. For part (B) and part (C), we have $f(x)=\dfrac{{{x}^{2}}}{1+{{x}^{2}}}$ and $f(x)=\dfrac{x}{1+{{x}^{2}}}$ respectively. The domain of a sec function and cosec function is $\operatorname{R}-\left( \dfrac{\pi }{2}+n\pi \right)$ and its range is $(-\infty ,-1]\cup [1,\infty )$ . For part (D), we have $f(x)=\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}-3x+5}$ . The range of tan function is the real number that is $(-\infty ,\infty )$ . Now, if the range of trigonometric function and function $f(x)$ have real numbers as its intersection, then we will have possible real values of $\theta$ and x.

Complete step-by-step answer:
A. $\cos \theta =x+\dfrac{1}{x}$
$f(x)=cos\theta$ ,where $f(x)=x+\dfrac{1}{x}$
Now, we have
$f(x)=x+\dfrac{1}{x}$
We can simplify it into simpler form.
$\Rightarrow f(x)=x+\dfrac{1}{x}$

& \Rightarrow f(x)={{(\sqrt{x})}^{2}}+\dfrac{1}{{{(\sqrt{x})}^{2}}}+2.\sqrt{x}.\dfrac{1}{\sqrt{x}}-2 \\\ & \Rightarrow f(x)={{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}-2 \\\ \end{aligned}$$ The maximum value of $${{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}$$ can be infinite while its minimum value is 0. We have square term in the function $$f(x)$$ , and when square term will be maximum then our function $$f(x)$$ will be maximum. Also, when square term will be minimum then the function $$f(x)$$ will have minimum value. The maximum value of $$f(x)$$ can be infinite while its minimum value is -2. Range of $$f(x)=\left[ -2,\infty \right]$$ . Range of $$\cos \theta =\left[ -1,1 \right]$$ . To satisfy the equality, we have to take the intersection of both ranges. That is when the range of $$f(x)=\left[ -1,1 \right]$$ , then we have possible real values of $$\theta $$ and x. B. $$sec\theta =\dfrac{{{x}^{2}}}{1+{{x}^{2}}}$$ $$f(x)=sec\theta $$ ,where $$f(x)=\dfrac{{{x}^{2}}}{1+{{x}^{2}}}$$ Now, we have $$f(x)=\dfrac{{{x}^{2}}}{1+{{x}^{2}}}$$ We can simplify it into simpler form. $$\Rightarrow f(x)=\dfrac{{{x}^{2}}}{1+{{x}^{2}}}$$ Dividing by $${{x}^{2}}$$ in numerator and denominator we get, $$\begin{aligned} & \Rightarrow f(x)=\dfrac{{{x}^{2}}}{1+{{x}^{2}}} \\\ & \Rightarrow f(x)=\dfrac{1}{\dfrac{1}{{{x}^{2}}}+{{x}^{2}}} \\\ & \Rightarrow f(x)=\dfrac{1}{{{\left( \dfrac{1}{x} \right)}^{2}}+{{\left( x \right)}^{2}}+2-2} \\\ & \Rightarrow f(x)=\dfrac{1}{{{\left( x+\dfrac{1}{x} \right)}^{2}}-2} \\\ \end{aligned}$$ The maximum value of $${{\left( x+\dfrac{1}{x} \right)}^{2}}$$ can be infinite while its minimum value is 0. We have a square term in the function $$f(x)$$ , and when the value of the square term will be 2 then our function $$f(x)$$ will be maximum. Also, when the square term will be minimum then the function $$f(x)$$ will have minimum value. Maximum value of $$f(x)=\dfrac{1}{{{\left( x+\dfrac{1}{x} \right)}^{2}}-2}=\dfrac{1}{2-2}=\dfrac{1}{0}=\infty $$ . Minimum value of $$f(x)=\dfrac{1}{{{\left( x+\dfrac{1}{x} \right)}^{2}}-2}=\dfrac{1}{0-2}=\dfrac{1}{-2}=\dfrac{-1}{2}$$ . Range of $$f(x)=\left[ -\dfrac{1}{2},\infty \right]$$ . Range of $$\sec \theta =(-\infty ,-1]\cup [1,\infty )$$ . To satisfy the equality, we have to take the intersection of both ranges. We get $$[1,\infty )$$ when we have intersection of $$\left[ -\dfrac{1}{2},0 \right]$$ and $$(-\infty ,-1]\cup [1,\infty )$$ . Thus, we have possible real values of $$\theta $$ and x. C. $$\operatorname{cosec}\theta =\dfrac{x}{1+{{x}^{2}}}$$ $$f(x)=cosec\theta $$ ,where $$f(x)=\dfrac{x}{1+{{x}^{2}}}$$ Now, we have $$f(x)=\dfrac{x}{1+{{x}^{2}}}$$ We can simplify it into simpler form. $$\Rightarrow f(x)=\dfrac{x}{1+{{x}^{2}}}$$ Dividing by x in numerator and denominator we get, $$\begin{aligned} & \Rightarrow f(x)=\dfrac{1}{\dfrac{1}{x}+x} \\\ & \Rightarrow f(x)=\dfrac{1}{{{(\sqrt{x})}^{2}}+\dfrac{1}{{{(\sqrt{x})}^{2}}}+2.\sqrt{x}.\dfrac{1}{\sqrt{x}}-2} \\\ & \Rightarrow f(x)=\dfrac{1}{{{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}-2} \\\ \end{aligned}$$ The maximum value of $${{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}$$ can be infinite while its minimum value is 0. We have a square term in the function $$f(x)$$ , and when the value of the square term will be 2 then our function $$f(x)$$ will be maximum. Also, when the square term will be minimum then the function $$f(x)$$ will have minimum value . Maximum value of $$f(x)=\dfrac{1}{{{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}-2}=\dfrac{1}{2-2}=\dfrac{1}{0}=\infty $$ . Minimum value of $$f(x)=\dfrac{1}{{{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}-2}=\dfrac{1}{0-2}=\dfrac{1}{-2}=\dfrac{-1}{2}$$ . Range of $$f(x)=\left[ -\dfrac{1}{2},\infty \right)$$ . Range of $$cosec\theta =(-\infty ,-1]\cup [1,\infty )$$ . To satisfy the equality, we have to take the intersection of both ranges. We get $$[1,\infty )$$ when we have intersection of $$\left[ -\dfrac{1}{2},0 \right]$$ and $$(-\infty ,-1]\cup [1,\infty )$$ . So, we have possible real values of $$\theta $$ and x. D. $$\tan \theta =\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}-3x+5}$$ $$\tan \theta =f(x)$$ where, $$f(x)=\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}-3x+5}$$ . Range of $$\tan \theta $$ can be any real number. Also, when we put any real value of x in the function $$f(x)$$ , we get only the real number. So, we have possible real values of $$\theta $$ and x. Note: We know that if the term in the denominator has minimum value then the whole fraction part switches to its maximum value. In part (B), if we follow this then, The maximum value of $${{\left( x+\dfrac{1}{x} \right)}^{2}}$$ can be infinite while its minimum value is 0. Maximum value of $$f(x)=\dfrac{1}{{{\left( x+\dfrac{1}{x} \right)}^{2}}-2}=\dfrac{1}{0-2}=-\dfrac{1}{2}$$ . Minimum value of $$f(x)=\dfrac{1}{{{\left( x+\dfrac{1}{x} \right)}^{2}}-2}=\dfrac{1}{\infty -2}=\dfrac{1}{\infty }=0$$ . This is a contradiction. We don’t have to think in this way for this question.