A) 1
B) 2
C) 3
D) None of these
Correct Answer: A
Solution :
| [a] Let \[f(x)=\frac{\sin 2x}{\sin \left( x+\frac{\pi }{4} \right)}=\] |
| \[\sqrt{2}\left\{ \frac{{{(\sin x+\cos )}^{2}}-1}{\sin x+\cos x} \right\}\] |
| \[=\sqrt{2}\left( \frac{{{y}^{2}}-1}{y} \right)\], where \[y=\sin x+\cos x\] |
| Let \[\phi (y)=\sqrt{2}\left( \frac{{{y}^{2}}-1}{y} \right),\] and \[g(x)\] |
| \[=\sin x+\cos x\] |
| We have, \[g'(x)=\cos x-\sin x\]. |
| For max or min. \[g'(x)=0\Rightarrow \tan x=1\] |
| \[\Rightarrow x=\pi /4\]. For this value of x. |
| \[g''(x)<0\]. Thus, \[g(x)\] is max at \[x=\pi /4\] and hence the domain of \[g(x)\] is \[[1,\sqrt{2}]\] i.e. y lies between 1 and \[\sqrt{2}\] |
| Now, \[\phi '(y)=\sqrt{2}\left( 1+\frac{1}{{{y}^{2}}} \right)>0\] for all \[y\in [1,\sqrt{2}]\]. |
| That is \[\phi (y)\] is increasing for all \[y\in [1,\sqrt{2}]\] |
| Thus it attains the greatest value at \[\sqrt{2}\] and is equal to \[\sqrt{2}\left( \frac{{{(\sqrt{2})}^{2}}-1}{\sqrt{2}} \right)=1\] |
| Hence, greatest value of f(x) on \[[0,\pi /2]\,\,=\] greatest value of \[\phi (y)\] on \[[1,\,\,\sqrt{2}]=1\]. |
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