Question
If $$\alpha $$ satisfies the inequation $${x^2} - x - 2 > 0$$ then a value exists for
A.
$${\sin ^{ - 1}}\alpha $$
B.
$${\sec ^{ - 1}}\alpha $$
C.
$${\cos ^{ - 1}}\alpha $$
D.
None of these
Answer :
$${\sec ^{ - 1}}\alpha $$
Solution :
$$\eqalign{
& {x^2} - x - 2 > 0 \cr
& \Rightarrow \,\,\left( {x - 2} \right)\left( {x + 1} \right) > 0 \cr} $$
$$ \Rightarrow \,\,x < - 1\,\,{\text{or, }}x > 2,$$ using sign scheme.
We know that $${\sin^{ - 1}}x$$ is defined for $$\left| x \right| \leqslant 1,{\cos ^{ - 1}}x$$ is defined for $$\left| x \right| \leqslant 1$$ and $${\sec^{ - 1}}x$$ is defined for $$\left| x \right| \geqslant 1,\,{\text{i}}{\text{.e}}{\text{., }}x \leqslant - 1\,\,{\text{or, }}x \geqslant 1.$$