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`        Please answer the question nd thanks in advance.the question number is 14 and has 3 parts .plese answer all the parts`
11 months ago

## Answers : (2)

```							Let me do any one for you.to learn,please do the rest yourself.if you face any problem then ask me again.y=1+2sinx+3cos2 xdy/dx=2cosx-6cosxsinx=2cosx(1-3sinx)d2y/dx2=-2sinx(1-3sinx)+2cosx(-3cosx)=-2sinx+6sin2x-6cos2xfor max or min dy/dx=0.Either cosx=0 or sinx=1/3Take cosx=0,then x=90,sinx=1.d2y/dx2=-2+6=8>0 this is the condition for minima.y=1+2=3(min)Take sinx=1/3,cosx=root(1-sin2x)=2root2/3d2y/dx2=-2/3+6/9-24/9=-2/3-18/9=-2/3-2=-8/2=-4y=1+2/3+24/9=5/3+8/3=14/3Hope you can do the rest.
```
11 months ago Deepak Kumar Shringi
askIITians Faculty
4391 Points
```							for all parts who have to find first of all differentiate the function and then find maxima and minimaim solving part b herey=1+2sinx+3cos^2xdy/dx=2cosx-6sinx cosx=2 cosx(1-3sinx)dy/dx=0 for cosx=0 or sinx=1/3when cosx=0 , sinx can be -1 or 1so find nowy=1+2=3 when sinx=1y=1-2=-1 when sinx=-1 minimumy=14/3 when sinx=1/3 maximum
```
11 months ago
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