Q196 / 270 · NDA 2023 (Phase 2) · Mathematics

A function is defined by $f(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \\ 2 & x+4 & 6 \\ 3 & 6 & x+9 \end{array}\right|$ The function attains local minimum value at:

  • A
    $x=-\frac{28}{3}$
  • B
    $x=-1$
  • Correct answer
    C
    $x=0$
  • D
    $x=\frac{28}{3}$

Solution

For local minima $f^{\prime}(x)=0 \Rightarrow 3 x^2+28 x=0 \Rightarrow x=\frac{-28}{3}, 0$ Now, $\quad f^{\prime \prime}(x)=6 x+28$ Now, $\quad f^{\prime \prime}(0)=28 \gt 0 \quad \& \quad f^{\prime \prime}\left(\frac{-28}{3}\right) \gt 0$ So, $\mathrm{f}(\mathrm{x})$ has local minimum value at $\mathrm{x}=0$

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

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