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

kartikay kala Grade: 11

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

one year ago

Answers : (2)

Susmita

425 Points

							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+3cos²x
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.

one year ago

Disapproved

Deepak Kumar Shringi

askIITians Faculty

4399 Points

							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

one year ago

Approved

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