A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) \[2\]
D) \[-2\]
Correct Answer: B
Solution :
| [b] The given eq. can be written as |
| \[({{y}^{2}}+2)\left( y-\frac{\cos x}{1+\sin x} \right)=0\] |
| \[\Rightarrow y=\frac{\cos x}{1+\sin x}[\because {{y}^{2}}+2\ne 0]\] |
| \[=\frac{\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}{1+\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}=\frac{1-{{t}^{2}}}{1+{{t}^{2}}+2t}=\frac{(1-t)(1+t)}{{{(1+t)}^{2}}}\] |
| \[=\frac{1-t}{1+t}=\frac{2}{1+t}-1,\] where \[t=\tan \frac{x}{2}\] |
| \[\Rightarrow y'(t)=\frac{-2}{{{(1+t)}^{2}}}.At\,x=\frac{\pi }{2},t=1\] |
| \[\therefore \,\,y'(1)=\frac{-2}{{{(1+1)}^{2}}}=-\frac{1}{2}\] |
You need to login to perform this action.
You will be redirected in
3 sec
