Question

The domain of the function \[f(x) = {\sin ^{ - 1}}\left[ {\dfrac{{\left( {\left| x \right| + 5} \right)}}{{{x^2} + 1}}} \right]\] is \[( -\infty , - a] \cup [a,\infty )\]. Find the value of a.
A.\[\dfrac{{\sqrt {17} - 1}}{2}\]
B.\[\dfrac{{\sqrt {17} }}{2}\]
C. \[\dfrac{{1 + \sqrt {17} }}{2}\]
D. \[\dfrac{{\sqrt {17} }}{2} + 1\]

Answer 1

Hints The value of sine inverse varies between -1 to 1 so obtain the inequality \[ - 1 \le \dfrac{{\left( {\left| x \right| + 5} \right)}}{{{x^2} + 1}} \le 1\] then solve. First solve the inequality \[{x^2} + 1 \ge \left| x \right| + 5\] then solve the inequality \[ - ({x^2} + 1) \le \left| x \right| + 5\] to obtain the value of a.

Formula used
The quadratic formula for a quadratic equation \[a{x^2} + bx + c = 0\] is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step by step solution
The value of sine inverse varies between -1 to 1, so we can write \[ - 1 \le \dfrac{{\left( {\left| x \right| + 5} \right)}}{{{x^2} + 1}} \le 1\].
Case 1:
Take the right-hand side of the inequality for evaluation.
\[{x^2} + 1 \ge \left| x \right| + 5\]
Hence,
\[{x^2} - \left| x \right| - 4 \ge 0 - - - (1)\]
Use quadratic formula to obtain the roots,
\[x = \dfrac{{1 \pm \sqrt {1 - 4.1( - 4)} }}{{2.1}}\]
\[ = \dfrac{{1 \pm \sqrt {17} }}{2}\]
Therefore, two roots are \[\left( {\dfrac{{1 + \sqrt {17} }}{2}} \right),\left( {\dfrac{{1 - \sqrt {17} }}{2}} \right)\] .
Now, (1) can be written as,
 \[\left( {x - \dfrac{{1 + \sqrt {17} }}{2}} \right)\left( {x - \dfrac{{1 - \sqrt {17} }}{2}} \right) \ge 0\]
So, \[x - \dfrac{{1 + \sqrt {17} }}{2} \ge 0\] or \[x - \dfrac{{1 - \sqrt {17} }}{2} \le 0\]
The value of \[\dfrac{{1 + \sqrt {17} }}{2}\]is positive and \[\dfrac{{1 - \sqrt {17} }}{2}\] is negative.
Hence a modulus function cannot be less than a negative number.
So, \[\left| x \right| \le \dfrac{{1 - \sqrt {17} }}{2}\] is not possible, so \[\left| x \right| \ge \dfrac{{1 + \sqrt {17} }}{2}\] .
Now, as \[\left| x \right| \ge \dfrac{{1 + \sqrt {17} }}{2}\] hence, the required interval is \[x \ge \dfrac{{1 + \sqrt {17} }}{2}\] or \[x \le - \dfrac{{1 + \sqrt {17} }}{2}\] or, \[( - \infty , - \dfrac{{1 + \sqrt {17} }}{2}] \cup [\dfrac{{1 + \sqrt {17} }}{2},\infty )\] .
Case 2:
Take the left-hand side of the inequality for evaluation.
\[ - ({x^2} + 1) \le \left| x \right| + 5\]
Hence,
\[{x^2} + \left| x \right| + 6 \ge 0 - - - (2)\]
Use quadratic formula to obtain the roots,
\[x = \dfrac{{ - 1 \pm \sqrt {1 - 4.1.6} }}{{2.1}}\]
\[ = \dfrac{{1 \pm \sqrt { - 23} }}{2}\]
The roots are imaginary therefore, we cannot proceed with case 2.
Therefore, \[a = \dfrac{{1 + \sqrt {17} }}{2}\] .

Therefore the correct option is C.

Note Sometime students want to find the roots by middle term factor method, but in this problem, we cannot find the roots by middle term factor method, we need to use quadratic formula to find the roots.