Q72 / 170 · Manipal (MET) 2020 · Mathematics

According to Newton-Raphson method, the value of $\sqrt{12}$ upto three places of decimal will be

  • A
    $3463$
  • B
    $3.462$
  • C
    $3.467$
  • Correct answer
    D
    None of these

Solution

Let $x=\sqrt{12}$ $\Rightarrow \quad x^2=12$ $\Rightarrow \quad x^2-12=0$ Let $\quad f(x)=x^2-12$ The first approximation in the Newton-Raphson method is given by $x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime}\left(x_0\right)}$ $=x_0-\frac{x_0^2-12}{2 x_0}=\frac{x_0^2+12}{2 x_0}$ $\because \quad 3 \lt \sqrt{12} \lt 3.5$ We can take $x_0=3.5$ $\therefore \quad x_1=\frac{(3.5)^2+12}{2 \times 3.5}=\frac{12.25+12}{7}$ $=\frac{24.25}{7}=3.464$

Hence, the correct answer is option: 'D'.

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