Q.

If the polynomial equation anxn+an1xn1+.........+a2x2+a1x+a0=0, n positive integer, has two different real roots α and β, then between  α and β, the equation nanxn1+(n1)an1xn2+.......+a1=0 has

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a

exactly one root

b

no root

c

atmost one root

d

atleast one root

answer is C.

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Detailed Solution

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Let fx=anxn+an1xn1+.........+a2x2+a1x+a0=0, which is a polynomial function in x of degree n of degree n. Hence f(x) is continuous and differentiable for all x

Let α<β we are given, f(α)=0=f(β)

By Rolle's theorem, f'(c)=0 for some value c, α<c<β

Hence the equation f'(x)=nanxn1+(n1)an1xn2+......+a1=0

has atleast one root between α and β

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If the polynomial equation anxn+an−1xn−1+.........+a2x2+a1x+a0=0, n positive integer, has two different real roots α and β, then between  α and β, the equation nanxn−1+(n−1)an−1xn−2+.......+a1=0 has