One root of the equation $e^{x}-3x^{2}=0$ lies in the interval $(3,4)$, the least number of iterations of the bisection method, so that $|\text{Error}|<10^{-3}$ is
(a) $10$
(b) $6$
(c) $8$
(d) $4$
One root of the equation $e^{x}-3x^{2}=0$ lies in the interval $(3,4)$, the least number of iterations of the bisection method, so that $|\text{Error}|<10^{-3}$ is
(a) $10$
(b) $6$
(c) $8$
(d) $4$
It's very easy. For Bisection method we always have $$n\ge \frac{\log{(b-a)}-\log{\epsilon}}{\log2}$$ Here we have $\epsilon=10^{-3}$, $a=3$, $b=4$ and $n$ is the number of iterations $$n\ge \frac{\log{(1)}-\log{10^{-3}}}{\log2}\approx 9.9658$$ Then $n=10$.
The number of bisection steps is simply equal to the number of binary digits you gain from the initial interval (you are dividing by 2). Then it's a simple conversion from decimal digits to binary digits.