Question

Let $ g\left( x \right)=ax+b $ , where $ a<0 $ and $ g $ is defined from $ \left[ 1,3 \right] $ onto $ \left[ 0,2 \right] $ then the value of $ {{\cot }^{-1}}\left( {{\cos }^{-1}}\left( \left| \sin x \right|+\left| \cos x \right| \right)+{{\sin }^{-1}}\left( -\left| \sin x \right|-\left| \cos x \right| \right) \right) $ is equal to \[\]
A. $ g\left( 1 \right) $ \[\]
B. $ g\left( 2 \right) $ \[\]
C. $ g\left( 3 \right) $ \[\]
D. $ g\left( 1 \right)+g\left( 3 \right) $ \[\]

Answer 1

Hint: We find the range of the function in the innermost bracket $ \left| \sin x \right|+\left| \cos x \right| $ using by taking $ y=\left| \sin x \right|+\left| \cos x \right| $ and then squaring and simplifying. We use symmetry to find the range of $ -\left| \sin x \right|-\left| \cos x \right|=-\left( \left| \sin x \right|+\left| \cos x \right| \right) $ . We find for what values of that range $ {{\cos }^{-1}}x,{{\sin }^{-1}}x $ are well defined and then find $ {{\cot }^{-1}}x $ at those values. We find the function $ g\left( x \right)=ax+b $ g to be linear decreasing function and compare for what values of $ x $ $ g\left( x \right)={{\cot }^{-1}}x $ .\[\]

Complete step by step answer:
Let us first find the value of inverse cotangent function in the given in the question as a composites function. We have;
\[{{\cot }^{-1}}\left( {{\cos }^{-1}}\left( \left| \sin x \right|+\left| \cos x \right| \right)+{{\sin }^{-1}}\left( -\left| \sin x \right|-\left| \cos x \right| \right) \right)\]
We know that inverse cosine function $ {{\cos }^{-1}}x $ and inverse sine function $ {{\sin }^{-1}}x $ can take values only from the interval $ \left[ -1,1 \right] $ since they are defined as $ {{\cos }^{-1}}x:\left[ -1,1 \right]\to \left[ 0,\pi \right] $ and $ {{\sin }^{-1}}x:\left[ -1,1 \right]\to \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] $ .\[\]
We see that $ {{\cos }^{-1}}x $ has the argument $ \left| \sin x \right|+\left| \cos x \right| $ and $ {{\sin }^{-1}}x $ has the argument $ -\left| \sin x \right|-\left| \cos x \right|=-\left( \left| \sin x \right|+\left| \cos x \right| \right) $ . Let us find the range of $ \left| \sin x \right|+\left| \cos x \right| $ . Let us assume.
\[y=\left| \sin x \right|+\left| \cos x \right|\]
We square both side and use the algebraic identity $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab $ to have;
\[\Rightarrow {{y}^{2}}={{\left| \sin x \right|}^{2}}+{{\left| \cos x \right|}^{2}}+2\left| \sin x \right|\left| \cos x \right|\]
We use the property of modulus $ \left| a \right|\left| b \right|=\left| ab \right| $ for $ a=\sin x,b=\cos x $ in the above step to have
\[\Rightarrow {{y}^{2}}={{\sin }^{2}}x+{{\cos }^{2}}x+2\left| \sin x\cos x \right|\]
We use Pythagorean trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ and the sine double angle formula $ \sin 2\theta =2\sin \theta \cos \theta $ for $ \theta =x $ in the above step to have;
\[\Rightarrow {{y}^{2}}=1+\left| \sin 2x \right|\]
We take square root both side and have ;
\[\Rightarrow y=\sqrt{1+\left| \sin 2x \right|}\]
We know the $ \sin \theta $ lies in between $ -1 $ to 1 for all real values of $ \theta . $ So for $ \theta =2x $ we have;
\[\begin{align}
  & -1\le \sin 2x\le 1 \\
 & \Rightarrow 0\le \left| \sin 2x \right|\le 1 \\
 & \Rightarrow 1\le 1+\left| \sin 2x \right|\le 1+1 \\
 & \Rightarrow 1\le 1+\left| \sin 2x \right|\le 2 \\
\end{align}\]
We take square roots in all the terms to have;
\[\begin{align}
  & \Rightarrow 1\le \sqrt{1+\left| \sin 2x \right|}\le \sqrt{2} \\
 & \Rightarrow 1\le y\le \sqrt{2} \\
 & \Rightarrow 1\le \left| \sin x \right|+\left| \cos x \right|\le \sqrt{2} \\
\end{align}\]
So we have $ \left| \sin x \right|+\left| \cos x \right|\in \left[ 1,\sqrt{2} \right] $ . We use the property of symmetric property of functions about $ x- $ axis and conclude that $ -\left( \left| \sin x \right|+\left| \cos x \right| \right)\in \left[ -\sqrt{2},-1 \right] $ . Now see that only value $ {{\cos }^{-1}}x $ can take from $ \left| \sin x \right|+\left| \cos x \right|\in \left[ 1,\sqrt{2} \right] $ is 1 since the maximum value that $ {{\cos }^{-1}}x $ c can take is 1 and the only value $ {{\sin }^{-1}}x $ can take from $ -\left( \left| \sin x \right|+\left| \cos x \right| \right)\in \left[ -\sqrt{2},-1 \right] $ is $ -1 $ since the minimum value $ {{\sin }^{-1}}x $ can take is $ -1. $ So we have
\[\begin{align}
  & \Rightarrow {{\cot }^{-1}}\left( {{\cos }^{-1}}\left( \left| \sin x \right|+\left| \cos x \right| \right)+{{\sin }^{-1}}\left( -\left| \sin x \right|-\left| \cos x \right| \right) \right) \\
 & \Rightarrow {{\cot }^{-1}}\left( {{\cos }^{-1}}\left( 1 \right)+{{\sin }^{-1}}\left( -1 \right) \right) \\
 & \Rightarrow {{\cot }^{-1}}\left( 0-\dfrac{\pi }{2} \right)={{\cot }^{-1}}\left( \dfrac{-\pi }{2} \right)=0....\left( 1 \right) \\
\end{align}\]
We are given t $ g\left( x \right)=ax+b $ , where $ a<0 $ and $ g $ is defined from $ \left[ 1,3 \right] $ onto $ \left[ 0,2 \right] $ and are asked for what values of $ x $ ,
\[\Rightarrow {{\cot }^{-1}}\left( {{\cos }^{-1}}\left( \left| \sin x \right|+\left| \cos x \right| \right)+{{\sin }^{-1}}\left( -\left| \sin x \right|-\left| \cos x \right| \right) \right)=g\left( x \right)\]
Since the value of inverse cotangent function is 0 for all $ x $ as obtained in (1) we have;
\[\begin{align}
  & \Rightarrow 0=g\left( x \right) \\
 & \Rightarrow g\left( x \right)=0 \\
\end{align}\]
We now need to find the $ x $ for which $ g\left( x \right)=0 $ . We see that $ g\left( x \right)=ax+b $ is a linear function with an equation of line with negative slope $ \left( \because a<0 \right) $ and hence the function $ g\left( x \right) $ is decreasing. So $ g $ maps the endpoints of the intervals $ \left[ 1,3 \right] $ oppositely to the endpoints of $ \left[ 0,2 \right] $ which means $ g\left( 1 \right)=2 $ and $ g\left( 3 \right)=0 $ . So the correct option is C. \[\]

Note:
 We note that the range and domain of inverse co-tangent function is given respectively as $ \mathsf{\mathbb{R}} $ and $ \left( 0,\pi \right) $ and hence here $ {{\cot }^{-1}}\left( \dfrac{-\pi }{2} \right) $ is well defined.\[\]

We also note that the plot of two functions $ y=f\left( x \right) $ and $ y=-f\left( x \right) $ will always be symmetrical about $ x- $ axis and the plot of the functions $ y=f\left( x \right),y=f\left( -x \right) $ will always be symmetrical about $ y- $ axis provided $ f\left( x \right) $ is not even.