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Question

# If a0n+1+a1n+a2n−1+⋯+an−12+an=0, then the equation a0xn+a1xn−1+⋯+an−1x+an=0 has, in the interval (0,1)

A
Exactly one root
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B
Atleast one root
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C
Atmost one root
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D
No root
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Solution

## The correct option is B Atleast one rootConsider the functionf(x)=a0xn+1n+1+a1xnn+...an−1x22+anxThen f(0)=0 ...(i)And f(1)=a0n+1+a1n+...an−12+an=0 ...(given).Hence f(0)=f(1).Then applying Rolle's theorem, there exists atleast one 'c' 0<c<1 such thatf′(c)=0Or [a0xn+a1xn−1+a2xn−2+...an−1x+an]x=c=0Where 0<c<1.

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