The absolute minimum value of $x^4-x^2-2 x+5$ is

The absolute minimum value of $x^4-x^2-2 x+5$ is
  1. equal to 5
  2. equal to 3
  3. equal to 7
  4. Does not exist

Solution

$x^4-x^2-2 x+5=f(x)$ (say) Then, $f^{\prime}(x)=4 x^3-2 x-2$ Equate $f^{\prime}(x)=0 \Rightarrow 4 x^3-2 x-2=0$ $\Rightarrow \quad 2 x^3-x-1=0$ $\Rightarrow \quad 2 x\left(x^2-1\right)+(x-1)=0$ $\Rightarrow \quad 2 x(x-1)(x+1)+(x-1)=0$ $\Rightarrow \quad(x-1)[2 x(x+1)+1]=0$ $\Rightarrow \quad(x-1)\left(2 x^2+2 x+1\right)=0$ $\Rightarrow \quad x-1=0$ and $2 x^2+2 x+1=0$ $\Rightarrow x=1$ and $x=-\frac{2 \pm \sqrt{4-8}}{4}$ (Imaginary) $\Rightarrow x=1$ and $x=(-1 \pm 1 i) / 2$ $f^{\prime \prime}(x)=12 x^2-2$ $f^{\prime \prime}(x) / x=1=10>0$ $\Rightarrow f(x)$ is minimum at $x=1$ Now, $f(\mathrm{l})=(\mathrm{l})^4-(\mathrm{l})^2-2+5=3$ $\therefore$ Absolute minimum value is 3 .

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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