Question

The greatest among (sin1+cos1),$(\sqrt{sin1}+\sqrt{cos1}),(sin1-cos-1)$ and 1 is

• A. $sin1+cos1$
• B. $(\sqrt{sin1}+\sqrt{cos1})$
• C. $sin1-cos1$
• D. $1$

Answer 1

The correct option is A $sin1+cos1$

Take,

$$\begin{array}{c}\text{(Sint + cos1) multiply and divide it ly}\sqrt{2}\text{.}\\ \text{we get,}\\ \sqrt{2}[\frac{1}{\sqrt{2}}\sin 1+\frac{1}{\sqrt{2}}\cos 1]\end{array}$$

$\Rightarrow \sqrt{2}[\cos \frac{\pi }{4}\sin 1+\sin \frac{\pi }{4}\cos 1]$

$\Rightarrow \sqrt{2}\left[\sin (\frac{\pi }{4}+1)\right]$

$\Rightarrow$ Range $\sin \theta \in [-1,1]$

$\Rightarrow [\sqrt{2},\sqrt{2}]$ Range of $\sin 1+\cos 1$

maximum value will to $\sqrt{2}$.

$(\sin 1-\cos 1)$. multiply and divide it $\mathrm{ly}\sqrt{2}$

weget,

$$

\sqrt{2}\left[\dfrac{1}{\sqrt{2}} \sin 1-\dfrac{1}{\sqrt{2}} \cos 1\right]

$$

$\Rightarrow \sqrt{2}\left[\sin (\frac{\pi }{4}-1)\right]$

and $\sqrt{2}$ will le maximum value

$\sqrt{\sin 1}+\sqrt{\cos 1}$

. by. A.M and GM.

$\frac{\sqrt{\sin 1}+\sqrt{\cos 1}}{2}\ge \sqrt{\sqrt{\sin 1}\cos 1}$

$\frac{\sqrt{\sin 1}+\sqrt{\cos 1}}{2}⩾\sqrt{\frac{1}{\sqrt{2}}\sqrt{2\sin 1\cos 1}}$

$\sqrt{\sin 1}+\sqrt{\cos 1}\ge 2\sqrt{\frac{1}{\sqrt{2}}\sqrt{\sin 2}}[-1⩽\sin 2⩽1]$

$\sqrt{\sin 1}+\sqrt{\cos 1}⩾\frac{2}{\sqrt{2}}$

$\sqrt{\sin 1}+\sqrt{\cos 1}⩾\sqrt{2}$

$\therefore \sqrt{\sin 1}+\sqrt{\cos 1}$ will be greatest

among all. option $\left(b\right)=\sqrt{\sin 1}+\sqrt{\cos 1})$