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Question: If \(a\le {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x\le b\), then find the value of a and b. ...

If a≤sin⁡−1x+cos⁡−1x+tan⁡−1x≤ba\le {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x\le b, then find the value of a and b.
A. a=π4,b=3π4a=\dfrac{\pi }{4},b=\dfrac{3\pi }{4}
B. a=0,b=π2a=0,b=\dfrac{\pi }{2}
C. a=π2,b=πa=\dfrac{\pi }{2},b=\pi
D. none of these

Explanation

Solution

We try to find the identity for sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}. From the domain of both sin⁡−1x{{\sin }^{-1}}x and cos⁡−1x{{\cos }^{-1}}x, we try to find the maximum and minimum value of g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x. Putting the end values of x=−1x=-1 and x=1x=1, we get the values of a and b.

Complete step-by-step answer:
We have the identity formula of sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} when x∈[−1,1]x\in \left[ -1,1 \right].
We also know the domain of both sin⁡−1x{{\sin }^{-1}}x and cos⁡−1x{{\cos }^{-1}}x is [−1,1]\left[ -1,1 \right].
So, for y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1xy\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x to be defined we have to take the domain as [−1,1]\left[ -1,1 \right].
Now tan⁡−1x{{\tan }^{-1}}x is an increasing function as if g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x and g′(x)=11+x2>0{{g}^{'}}\left( x \right)=\dfrac{1}{1+{{x}^{2}}}>0.
So, we can say at point x=−1x=-1, it has minimum value for g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x and at point x=1x=1, it has maximum value for g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x.
So, the maximum and minimum value of y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1xy\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x depends on totally g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x. The main function y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1xy\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x attains minimum value at x=−1x=-1 and maximum value at point x=1x=1.
So, we can say at point x=−1x=-1, the value for g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x is g(−1)=tan⁡−1(−1)=−π4g\left( -1 \right)={{\tan }^{-1}}\left( -1 \right)=\dfrac{-\pi }{4} and at point x=1x=1, the value for g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x is g(1)=tan⁡−11=π4g\left( 1 \right)={{\tan }^{-1}}1=\dfrac{\pi }{4}.
Now we find the maximum and minimum value of y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1xy\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x.
At point x=−1x=-1, the minimum value for y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1xy\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x is y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1x=π2−π4=π4y\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=1x=1, the maximum value for y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1xy\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x is y(x)=sin⁡−1x+cos⁡−1x+tan⁡−1x=π2+π4=3π4y\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≤sin⁡−1x+cos⁡−1x+tan⁡−1x≤ba\le {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x\le b, the value of a and b is a=π4,b=3π4a=\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(x)=sin⁡−1x+cos⁡−1x+tan⁡−1xy\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x. The [−1,1]\left[ -1,1 \right] range is the key to find the maximum and minimum value. The principal value of g(x)=tan⁡−1xg\left( x \right)={{\tan }^{-1}}x is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].