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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

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

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Grade:11

2 Answers

Susmita
425 Points
5 years ago
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 x
dy/dx=2cosx-6cosxsinx=2cosx(1-3sinx)
d2y/dx2=-2sinx(1-3sinx)+2cosx(-3cosx)
=-2sinx+6sin2x-6cos2x
for max or min dy/dx=0.
Either cosx=0 or sinx=1/3
Take 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)=2root2/3
d2y/dx2=-2/3+6/9-24/9=-2/3-18/9=-2/3-2=-8/2=-4
y=1+2/3+24/9=5/3+8/3=14/3
Hope you can do the rest.
 
Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago
for all parts who have to find first of all differentiate the function and then find maxima and minima
im solving part b here
y=1+2sinx+3cos^2x
dy/dx=2cosx-6sinx cosx=2 cosx(1-3sinx)
dy/dx=0 for cosx=0 or sinx=1/3
when cosx=0 , sinx can be -1 or 1
so find now
y=1+2=3 when sinx=1
y=1-2=-1 when sinx=-1 minimum
y=14/3 when sinx=1/3 maximum

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