If a0n+1+a1n+a2n−1+…+an−12+an=0. Then the function f(x)=a0xn+a1xn−1+a2xn−2+…+an has in (0,1)

If a0n+1+a1n+a2n1++an12+an=0. Then the function f(x)=a0xn+a1xn1+a2xn2++an has in (0,1)

  1. A

    at least one zero 

  2. B

    at most one zero

  3. C

    only 3 zeros

  4. D

    only 2 zeros

    Register to Get Free Mock Test and Study Material

    +91

    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    Consider the function 

    ϕ(x)=a0xn+1n+1+a1xnn+a2xn1n1++an1x22+anx.

    Clearly ϕ(x), being a polynomial, is continuous on [0, 1] and differentiable on (0, 1)

    Also, ϕ(0)=0

    and, ϕ(1)=a0n+1+a1n+a2n1++an=0          [Given]

     ϕ(0)=ϕ(1)

    Thus ϕ(x) satisfies conditions of Rolle's theorem on [0, 1]

    Consequently, there exists c(0,1) such that ϕ(c)=0 i.e.

    c(0,1) is a zero of ϕ(x)=a0xn+a1xn1++an=f(x).

    Chat on WhatsApp Call Infinity Learn

      Register to Get Free Mock Test and Study Material

      +91

      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.