Question

If $a\le {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x\le b$, then find the value of a and b.
A. $a=\dfrac{\pi }{4},b=\dfrac{3\pi }{4}$
B. $a=0,b=\dfrac{\pi }{2}$
C. $a=\dfrac{\pi }{2},b=\pi $
D. none of these

Answer 1

Hint: We try to find the identity for ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$. From the domain of both ${{\sin }^{-1}}x$ and ${{\cos }^{-1}}x$, we try to find the maximum and minimum value of $g\left( x \right)={{\tan }^{-1}}x$. Putting the end values of $x=-1$ and $x=1$, we get the values of a and b.

Complete step-by-step answer:
We have the identity formula of ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ when $x\in \left[ -1,1 \right]$.
We also know the domain of both ${{\sin }^{-1}}x$ and ${{\cos }^{-1}}x$ is $\left[ -1,1 \right]$.
So, for $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x$ to be defined we have to take the domain as $\left[ -1,1 \right]$.
Now ${{\tan }^{-1}}x$ is an increasing function as if $g\left( x \right)={{\tan }^{-1}}x$ and ${{g}^{'}}\left( x \right)=\dfrac{1}{1+{{x}^{2}}}>0$.
So, we can say at point $x=-1$, it has minimum value for $g\left( x \right)={{\tan }^{-1}}x$ and at point $x=1$, it has maximum value for $g\left( x \right)={{\tan }^{-1}}x$.
So, the maximum and minimum value of $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x$ depends on totally $g\left( x \right)={{\tan }^{-1}}x$. The main function $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x$ attains minimum value at $x=-1$ and maximum value at point $x=1$.
So, we can say at point $x=-1$, the value for $g\left( x \right)={{\tan }^{-1}}x$ is $g\left( -1 \right)={{\tan }^{-1}}\left( -1 \right)=\dfrac{-\pi }{4}$ and at point $x=1$, the value for $g\left( x \right)={{\tan }^{-1}}x$ is $g\left( 1 \right)={{\tan }^{-1}}1=\dfrac{\pi }{4}$.
Now we find the maximum and minimum value of $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x$.
At point $x=-1$, the minimum value for $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x$ is $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}-\dfrac{\pi }{4}=\dfrac{\pi }{4}$ and at point $x=1$, the maximum value for $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x$ is $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}+\dfrac{\pi }{4}=\dfrac{3\pi }{4}$.
So, for $a\le {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x\le b$, the value of a and b is $a=\dfrac{\pi }{4},b=\dfrac{3\pi }{4}$.

So, the correct answer is “Option A”.

Note: We can’t change the domain for the function $y\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x$. The $\left[ -1,1 \right]$ range is the key to find the maximum and minimum value. The principal value of $g\left( x \right)={{\tan }^{-1}}x$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.