Question
If $$\alpha \in \left( {0,\frac{\pi }{2}} \right){\text{then }}\sqrt {{x^2} + x} + \frac{{{{\tan }^2}\alpha }}{{\sqrt {{x^2} + x} }}$$ is always greater than or equal to
A.
$$2\tan \alpha $$
B.
1
C.
2
D.
$${\sec ^2}\alpha $$
Answer :
$$2\tan \alpha $$
Solution :
Let $$a = \sqrt {{x^2} + x} \,\,{\text{and}}\,\,b = \frac{{{{\tan }^2}\alpha }}{{\sqrt {{x^2} + x} }}$$
then using A.M. $$ \geqslant $$ G.M., we get $$\frac{{a + b}}{2} \geqslant \sqrt {ab} $$
$$\eqalign{
& \Rightarrow \,\,a + b \geqslant 2\sqrt {ab} \cr
& \Rightarrow \,\,\sqrt {{x^2} + x} + \frac{{{{\tan }^2}\alpha }}{{\sqrt {{x^2} + x} }} \geqslant 2\sqrt {{{\tan }^2}\alpha } = 2\tan \alpha \cr
& \left[ {\because \,\,\alpha \in \left( {0,\frac{\pi }{2}} \right)} \right]. \cr} $$