116979
If \(\alpha \in\left[0, \frac{\pi}{2}\right)\),then \(\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}\) is always greater than or equal to
1 \(2 \tan \alpha\)
2 1
3 2
4 \(\sec ^2 \alpha\)
Explanation:
A Here, \(\alpha \in\left[0, \frac{\pi}{2}\right) \Rightarrow \tan \alpha\) is (+ve) As, we know, if \(\mathrm{a}>0 \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) i.e., \(\quad \mathrm{AM} \geq \mathrm{GM}\) \(\therefore \frac{\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}{2} \geq \sqrt{\sqrt{\mathrm{x}^2+\mathrm{x}} \frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}\) \((\because \quad \text { using } \geq \mathrm{GM})\) \(\Rightarrow \quad \sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}} \geq 2 \tan \alpha\)
116979
If \(\alpha \in\left[0, \frac{\pi}{2}\right)\),then \(\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}\) is always greater than or equal to
1 \(2 \tan \alpha\)
2 1
3 2
4 \(\sec ^2 \alpha\)
Explanation:
A Here, \(\alpha \in\left[0, \frac{\pi}{2}\right) \Rightarrow \tan \alpha\) is (+ve) As, we know, if \(\mathrm{a}>0 \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) i.e., \(\quad \mathrm{AM} \geq \mathrm{GM}\) \(\therefore \frac{\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}{2} \geq \sqrt{\sqrt{\mathrm{x}^2+\mathrm{x}} \frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}\) \((\because \quad \text { using } \geq \mathrm{GM})\) \(\Rightarrow \quad \sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}} \geq 2 \tan \alpha\)
116979
If \(\alpha \in\left[0, \frac{\pi}{2}\right)\),then \(\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}\) is always greater than or equal to
1 \(2 \tan \alpha\)
2 1
3 2
4 \(\sec ^2 \alpha\)
Explanation:
A Here, \(\alpha \in\left[0, \frac{\pi}{2}\right) \Rightarrow \tan \alpha\) is (+ve) As, we know, if \(\mathrm{a}>0 \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) i.e., \(\quad \mathrm{AM} \geq \mathrm{GM}\) \(\therefore \frac{\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}{2} \geq \sqrt{\sqrt{\mathrm{x}^2+\mathrm{x}} \frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}\) \((\because \quad \text { using } \geq \mathrm{GM})\) \(\Rightarrow \quad \sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}} \geq 2 \tan \alpha\)
116979
If \(\alpha \in\left[0, \frac{\pi}{2}\right)\),then \(\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}\) is always greater than or equal to
1 \(2 \tan \alpha\)
2 1
3 2
4 \(\sec ^2 \alpha\)
Explanation:
A Here, \(\alpha \in\left[0, \frac{\pi}{2}\right) \Rightarrow \tan \alpha\) is (+ve) As, we know, if \(\mathrm{a}>0 \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) i.e., \(\quad \mathrm{AM} \geq \mathrm{GM}\) \(\therefore \frac{\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}{2} \geq \sqrt{\sqrt{\mathrm{x}^2+\mathrm{x}} \frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}\) \((\because \quad \text { using } \geq \mathrm{GM})\) \(\Rightarrow \quad \sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}} \geq 2 \tan \alpha\)
116979
If \(\alpha \in\left[0, \frac{\pi}{2}\right)\),then \(\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}\) is always greater than or equal to
1 \(2 \tan \alpha\)
2 1
3 2
4 \(\sec ^2 \alpha\)
Explanation:
A Here, \(\alpha \in\left[0, \frac{\pi}{2}\right) \Rightarrow \tan \alpha\) is (+ve) As, we know, if \(\mathrm{a}>0 \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) i.e., \(\quad \mathrm{AM} \geq \mathrm{GM}\) \(\therefore \frac{\sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}{2} \geq \sqrt{\sqrt{\mathrm{x}^2+\mathrm{x}} \frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}}}\) \((\because \quad \text { using } \geq \mathrm{GM})\) \(\Rightarrow \quad \sqrt{\mathrm{x}^2+\mathrm{x}}+\frac{\tan ^2 \alpha}{\sqrt{\mathrm{x}^2+\mathrm{x}}} \geq 2 \tan \alpha\)
