Question

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

Answer 1

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

Complete step-by-step answer:
Given that \[{\sec ^{ - 1}}x + {\tan ^{ - 1}}x\] is real if \[x \in ( - \infty , - a] \cup [b,\infty ).\]
We know that \[{\sec ^{ - 1}}x = {\tan ^{ - 1}}\sqrt {{x^2} - 1} \]
Substituting \[{\sec ^{ - 1}}x = {\tan ^{ - 1}}\sqrt {{x^2} - 1} \] in \[{\sec ^{ - 1}}x + {\tan ^{ - 1}}x\], we get
\[ \Rightarrow {\sec ^{ - 1}}x + {\tan ^{ - 1}}x = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\sqrt {{x^2} - 1} \]
Since the value of \[{\sec ^{ - 1}}x + {\tan ^{ - 1}}x\] is real, the value of \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}\sqrt {{x^2} - 1} \] is also real.
We get real values of \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}\sqrt {{x^2} - 1} \] if and only if \[\sqrt {{x^2} - 1} \geqslant 0\].
So, consider the values of \[\sqrt {{x^2} - 1} \geqslant 0\]
\[ \Rightarrow \sqrt {{x^2} - 1} \geqslant 0\]
Squaring on both sides, we get
\[
   \Rightarrow {\left( {\sqrt {{x^2} - 1} } \right)^2} \geqslant {0^2} \\
   \Rightarrow {x^2} - 1 \geqslant 0 \\
   \Rightarrow {x^2} \geqslant 1 \\
\]
Rooting on both sides, we get
\[
   \Rightarrow \sqrt {{x^2}} \geqslant \sqrt 1 \\
   \Rightarrow \left| x \right| \geqslant 1 \\
\]
We know if \[\left| x \right| \geqslant a\] then \[x \in ( - \infty , - a] \cup [a,\infty )\].
So, we have \[\left| x \right| \geqslant 1\] as \[x \in ( - \infty , - 1] \cup [1,\infty )\].
By comparing \[x \in ( - \infty , - 1] \cup [1,\infty )\] and \[x \in ( - \infty , - a] \cup [b,\infty )\], we have \[a = 1\] and \[b = 1\].
Now, consider the value of \[a + b\] i.e., \[a + b = 1 + 1 = 2\].
Therefore, the value of \[a + b\] 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 \[\left| x \right| \geqslant 1\] then \[x \in ( - \infty , - 1] \cup [1,\infty )\]
2. If \[a \leqslant x \leqslant b\] then \[x \in \left[ {a,b} \right]\]
3. If \[a < x \leqslant b\] then \[x \in (a,b]\]
4. If \[a \leqslant x < b\] then \[x \in [a,b)\]
5. If \[a < x < b\] then \[x \in \left( {a,b} \right)\]