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Question: \[{\sec ^{ - 1}}x + {\tan ^{ - 1}}x\] is real if \[x \in ( - \infty , - a] \cup [b,\infty ).\] Find ...

sec⁡−1x+tan⁡−1x{\sec ^{ - 1}}x + {\tan ^{ - 1}}x is real if x∈(−∞,−a]∪[b,∞).x \in ( - \infty , - a] \cup [b,\infty ). Find the value of a+ba + b.
A. −1 - 1
B. 0
C. 1
D. 2

Explanation

Solution

In this question, we will proceed by converting the inverse function of secant into inverse function of tangent by using the formula sec⁡−1x=tan⁡−1x2−1{\sec ^{ - 1}}x = {\tan ^{ - 1}}\sqrt {{x^2} - 1} . Further compare the values for xx to be real to get the required value of the solution.

Complete step-by-step answer:
Given that sec⁡−1x+tan⁡−1x{\sec ^{ - 1}}x + {\tan ^{ - 1}}x is real if x∈(−∞,−a]∪[b,∞).x \in ( - \infty , - a] \cup [b,\infty ).
We know that sec⁡−1x=tan⁡−1x2−1{\sec ^{ - 1}}x = {\tan ^{ - 1}}\sqrt {{x^2} - 1}
Substituting sec⁡−1x=tan⁡−1x2−1{\sec ^{ - 1}}x = {\tan ^{ - 1}}\sqrt {{x^2} - 1} in sec⁡−1x+tan⁡−1x{\sec ^{ - 1}}x + {\tan ^{ - 1}}x, we get
⇒sec⁡−1x+tan⁡−1x=tan⁡−1x+tan⁡−1x2−1\Rightarrow {\sec ^{ - 1}}x + {\tan ^{ - 1}}x = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\sqrt {{x^2} - 1}
Since the value of sec⁡−1x+tan⁡−1x{\sec ^{ - 1}}x + {\tan ^{ - 1}}x is real, the value of tan⁡−1x+tan⁡−1x2−1{\tan ^{ - 1}}x + {\tan ^{ - 1}}\sqrt {{x^2} - 1} is also real.
We get real values of tan⁡−1x+tan⁡−1x2−1{\tan ^{ - 1}}x + {\tan ^{ - 1}}\sqrt {{x^2} - 1} if and only if x2−1⩾0\sqrt {{x^2} - 1} \geqslant 0.
So, consider the values of x2−1⩾0\sqrt {{x^2} - 1} \geqslant 0
⇒x2−1⩾0\Rightarrow \sqrt {{x^2} - 1} \geqslant 0
Squaring on both sides, we get

⇒(x2−1)2⩾02 ⇒x2−1⩾0 ⇒x2⩾1  \Rightarrow {\left( {\sqrt {{x^2} - 1} } \right)^2} \geqslant {0^2} \\\ \Rightarrow {x^2} - 1 \geqslant 0 \\\ \Rightarrow {x^2} \geqslant 1 \\\

Rooting on both sides, we get

⇒x2⩾1 ⇒∣x∣⩾1  \Rightarrow \sqrt {{x^2}} \geqslant \sqrt 1 \\\ \Rightarrow \left| x \right| \geqslant 1 \\\

We know if ∣x∣⩾a\left| x \right| \geqslant a then x∈(−∞,−a]∪[a,∞)x \in ( - \infty , - a] \cup [a,\infty ).
So, we have ∣x∣⩾1\left| x \right| \geqslant 1 as x∈(−∞,−1]∪[1,∞)x \in ( - \infty , - 1] \cup [1,\infty ).
By comparing x∈(−∞,−1]∪[1,∞)x \in ( - \infty , - 1] \cup [1,\infty ) and x∈(−∞,−a]∪[b,∞)x \in ( - \infty , - a] \cup [b,\infty ), we have a=1a = 1 and b=1b = 1.
Now, consider the value of a+ba + b i.e., a+b=1+1=2a + b = 1 + 1 = 2.
Therefore, the value of a+ba + b is 2.

Thus, the correct option is D. 2

Note:
To solve these kinds of problems, remember the formulae of inverse trigonometry and the conversions of inequalities. Some of the important inequality conversions are
1. If ∣x∣⩾1\left| x \right| \geqslant 1 then x∈(−∞,−1]∪[1,∞)x \in ( - \infty , - 1] \cup [1,\infty )
2. If a⩽x⩽ba \leqslant x \leqslant b then x∈[a,b]x \in \left[ {a,b} \right]
3. If a<x⩽ba < x \leqslant b then x∈(a,b]x \in (a,b]
4. If a⩽x<ba \leqslant x < b then x∈[a,b)x \in [a,b)
5. If a<x<ba < x < b then x∈(a,b)x \in \left( {a,b} \right)