GATE 2023 MA – Question 51
A twice continuously differentiable f on [a,b] has f′<0,f″<0 on (a,b), and f(ζ)=0 at ζ∈(a,b). Newton iteration is x_(k+1)=x_k−f(x_k)/f′(x_k). If x_k∈(ζ,b), which hold?
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Correct answer: (A) x_(k+1)>ζ; (C) x_(k+1)<x_k
Explanation
f(x_k)<0 and f′(x_k)<0, so a positive quantity is subtracted and the next iterate is below x_k. By the mean-value theorem f(x_k)=f′(η)(x_k−ζ). Strict concavity gives f′(x_k)<f′(η)<0, so 0<f′(η)/f′(x_k)<1. Consequently x_(k+1)−ζ=(x_k−ζ)[1−f′(η)/f′(x_k)]>0. **A,C**.