Weird Roots
If f ( x ) = ( x − a ) ( x − b ) for a , b ∈ R , then the minimum number of roots of equation π ( f ′ ( x ) ) 2 cos ( π ( f ( x ) ) ) + sin ( π ( f ( x ) ) ) f ′ ′ ( x ) = 0 in ( α , β ) , where f ( α ) = + 3 = f ( β ) ,and α < a < b < β will be:
The answer is 8.
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1 solution
As it is a quadratic f'(x) has exactly one root between a and b.
Given a<b f(x) has two roots between alpha and beta.
For sin(pi f(x) ) to be zero in (alpha,beta) f(x) has to be a integer.
Also given f(alpha) = f(beta) = 3.
There for f(x) must possess 1 and 2 as their integral values for two x each, which you totally neglected.
Moreover the roots were asked in (alpha,beta) and not in [alpha,beta] ( understand the brackets ) therefore you cannot take alpha and beta to be your solutions, though you did take them
For minimum roots assuming min(f(x)) is not less than -1 we get minimum roots of sin(pi f(x) )f'(x) to be 7.
Hence minimum roots of given equation is 6 ( From Rolle's Theorem ).
So answer should be 46656.
I hope you understand it and make the required corrections.
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I just now worked on the problem.You could solve the differential equation and we get that cos(pi*f(x))=-1 ( Use f(alpha)=f(beta)=3 to find the values of the two arbitrary constants).So f(x) must be odd integer.Between alpha and beta ,the odd integers that f(x) can attain are 1,-1,-3,-5,.... and so on.For the case of minimum roots,the vertex of the parabola can be made to lie above the line y=-1.Hence,this ensures that f(x)=-1,-3,-5.... do not have any solution.Hence,we get that f(x)=1 will have two real solutions between alpha and beta.So,the minimum number of roots must be 2.
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I am not getting cos(pi f(x)) as -1
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@Parv Mor – I think you have made an error in solving the differential equation.You do get cos (pi*f (x))=-1.
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@Manish Maharaj – Okay got it. But still then answer comes out to be 4 right?
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@Parv Mor – No.So,now you see that cos(pi*f(x))=-1.So,f(x) must be an odd integer.Between alpha and beta the odd integers that f(x) can attain are 1,-1,-3,-5.....However,if we want minimum number of roots,we can make the vertex of the parabola to lie between the lines y=0 and y=-1 so that f(x)=-1,-3,-5... will have no real solutions.In any case f(x)=1 will always have two real and distinct real roots.Hence the minimum number of roots between alpha and beta must be 2.
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@Indraneel Mukhopadhyaya – Yes,the answer must be 4.I thought you meant that the minimum number of roots are 4.
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@Indraneel Mukhopadhyaya – Wait but putting cos(pi f(x) ) in the original equation we get f'(x) to be zero.
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@Parv Mor – Yes,that's correct.So,we must have that f (x) is an odd integer as well as f'(x)=0.f'(x)=0 implies that f (x) must be constant.Since it is given that f (alpha)=f (beta)=3,the constant is equal to 3.So we must have f (x)=3 whose two roots are alpha and beta.Hence between alpha and beta the number of roots of f (x)=3 is 0.So,the answer to this question is undefined.
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@Indraneel Mukhopadhyaya – But dont you think its contrary to the rolles theorem. As for given interval sin(pif(x)) is zero at seven x's. Therefore its differentiation must have atleast 6 roots in respective intervals??
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@Parv Mor – Why are you applying Rolles theorem to sin(pi*f(x))?The question asks about minimum number of roots for f(x).
@Parv Mor – We have d/dx(sin(pi (f(x)) (d/dx(f(x)))=0 (which is essentially the condition given in the question.)Hence,integrating both sides,we get sin(pi f(x)) d/dx(f(x))=c1 (first arbitrary constant).Rearranging the differential equation in variable separable form ,we get sin(pi f(x)) d(f(x))=c1dx.Integrating both sides,we get (-1/pi)cos(pi (f(x)))=c1x+c2 (c2 is second arbitrary constant).Multiply by -pi on both sides,we get cos(pi f(x))=c1x+c2 (the constants are arbitrary constants,so it does not matter if we multiply any fixed constant to it).First substitute alpha in the equation,then beta and using f(alpha)=f(beta)=3,we get c1(alpha)+c2=c1(beta)+c2=cos(3pi)=-1.From first two equality,we get c1(alpha)=c1(beta).Since it is given that alpha<beta,we must have c1=0.Now,using c1(alpha)+c2=-1,we get c2=-1.Hence,we get the particular solution of our differential equation as cos(pi*f(x))=-1.
you are conceptually wrong.since f(x) is a quadratic so f'(x) has exactly one root. and also you didn't use the other conditions given in question.
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then minimum number of roots will come out to be 5, a , b , α , β and one of f ′ ( x ) where α < β
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yes there are atleast five roots of sin(pi(f(x)).f'(x) as correctly mention by you, then by applying rolle's theorem its derivative has atleast 4 roots which is the answer of the given question.
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@Ojas Dhiman – That's what I found. The number of roots I found is at least 8. In fact any root of the equation is a zero of the second derivative of g ( x ) = − π 1 cos ( π f ( x ) ) Since in [ α , β ] the argument of cosine ranges monotonically from − 3 π to − ϵ and again monotonically from − ϵ to 3 π (where ϵ is a positive number and namely ϵ = − π 4 ( b − a ) 2 , the function g has in [ α , β ] at least 9 stationary points (that are altrnatively local minima and maxima). This means the second derivatives of g has at least 8 zeros.
The equation can be rewritten as
d x d ( s i n ( π ( f ( x ) ) ) × f ′ ( x ) ) = 0
In the closed interval [ α , β ] , we can verify that sin ( π f ( x ) ) has 8 roots (corresponding to f ( x ) = 0 , 1 , 2 , 3 ) and f ′ ( x ) has one root. These roots are of multiplicity 1, and are distinct from each other.
So s i n ( π ( f ( x ) ) ) × f ′ ( x ) has 9 roots, and it's derivative has 8 roots in the open interval ( α , β ) .