Question
The set of values of $$\lambda \in R$$ such that $${\tan ^2}\theta + \sec \theta = \lambda $$ holds for some $$\theta $$ is
A.
$$\left( { - \infty , 1} \right]$$
B.
$$\left( { - \infty , - 1} \right]$$
C.
$$\phi $$
D.
$$\left[ { - 1, + \infty } \right)$$
Answer :
$$\left[ { - 1, + \infty } \right)$$
Solution :
$$\eqalign{
& {\sec ^2}\theta + \sec \theta - \left( {\lambda + 1} \right) = 0 \cr
& \therefore \,\,\sec \theta = \frac{{ - 1 \pm \sqrt {1 + 4\left( {\lambda + 1} \right)} }}{2} = \frac{{ - 1 \pm \sqrt {4\lambda + 5} }}{2}. \cr} $$
For real, $$\sec \theta ,4\lambda + 5 \geqslant 0,\,{\text{i}}{\text{.e}}{\text{., }}\lambda \geqslant - \frac{5}{4}.$$
Also, $$\sec\theta \geqslant 1\,\,{\text{or, sec}}\theta \leqslant - {\text{1}}{\text{.}}$$
$$\eqalign{
& \therefore \,\,\frac{{ - 1 \pm \sqrt {4\lambda + 5} }}{2} \geqslant 1\,\,{\text{or, }}\frac{{ - 1 \pm \sqrt {4\lambda + 5} }}{2} \leqslant - 1 \cr
& \Rightarrow \,\, - 1 + \sqrt {4\lambda + 5} \geqslant 2\,\,\,{\text{or, }} - 1 - \sqrt {4\lambda + 5} \leqslant - 2 \cr
& \Rightarrow \,\,4\lambda + 5 \geqslant 9\,\,{\text{or, }}4\lambda + 5 \geqslant 1 \cr
& \Rightarrow \,\,\lambda \geqslant 1\,\,\,{\text{or, }}\lambda \geqslant - 1 \cr
& \therefore \,\,\lambda \geqslant - \frac{5}{4}\,{\text{and }}\lambda \geqslant 1\,\,\,{\text{or, }}\lambda \geqslant - \frac{5}{4}\,{\text{and }}\lambda \geqslant - 1 \cr
& \therefore \,\,\lambda \geqslant 1\,\,{\text{or, }}\lambda \geqslant - 1 \cr
& \therefore \,\,\lambda \in \left[ { - 1, + \infty } \right). \cr} $$