p(x) is a polynomial of degree 5 and has extremum at -1 and 1 and lim x -0(p(x)/x^3 -2)=4 if m and M are maximum and minimum value of y=f'(x) on the set A={x|x^2+6x<5x} then m/M is

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p(x) is a polynomial of degree 5 and has extremum at -1 and 1 and lim x -0(p(x)/x^3 -2)=4 if m and M are maximum and minimum value of y=f'(x) on the set A={x|x^2+6x<5x} then m/M is

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consider p(x)=ax^5+bx^4+cx^3+dx^2+ex+f

Now d=e=f=0 as lim x -0(p(x)/x^3 -2)=4 so power of x<3 causing the destabilisation of finite value on right side.

so our polynomial p(x) becomes p(x)=ax^5+bx^4+cx^3

Now, putting all values in limit and we get c-2=4 implies c=6.

Also p(x) has extremum at -1 and 1 so putting the value firstly x=-1 and equating dp(x)/dx=0.

We get equation (1)

Using same for x=1 and we get equation (2)

Solving equation (1) and (2) we get "a" and "b"

Putting value of "a", "b", "c" in p(x) we get our desired polynomial.

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