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The equation $f(x)$ is given as $x^{3}+4 x+1=0$. Considering the initial approximation at $x=1$ then the value of $\mathrm{x}_{1}$ is given as
a) $1.67$
b) $1.87$
c) $1.86$
d) $1.85$
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$\mathrm{C}$

Explanation:

Iterative formula for Newton Raphson method is given by $\mathrm{x}(1)=\mathrm{x}(0)+\frac{f(x(0))}{f^{\prime} x(x(0))}$
Hence $x_{0}=1$ (initial guess), $f\left(x_{0}\right)=6$ and $f^{\prime}\left(x_{0}\right)=7$. Substituting the values in the equation we get $x_{1}=1.857$ which is $1.86$ rounded to 2 decimal places.

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