Q67 / 125 · BITSAT 2009 · Mathematics

The minimum value of the function \(y=x^4-2 x^2+1\) in the interval \(\left[\frac{1}{2}, 2\right]\) is

  • Correct answer
    A
    0
  • B
    2
  • C
    8
  • D
    9

Solution

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^4-2 \mathrm{x}^2+1\right)=4 \mathrm{x}\left(\mathrm{x}^2-1\right)\) For \(\max\). or \(\min , \frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(4 \mathrm{x}\left(\mathrm{x}^2-1\right)=0 ;\), either \(\mathrm{x}=0\) or \(\mathrm{x}= \pm 1\) \(x=0\) and \(x=-1\) does not belong to \(\left[\frac{1}{2}, 2\right]\) \(\frac{d^2 y}{d x^2}=12 x^2-4 \therefore\left(\frac{d^2 y}{d x^2}\right)_{x=1}\) \(=12(1)^2-4=8 > 0\) \(\therefore \quad\) there is minimum value of function at \(\mathrm{x}=1\) \(\therefore \quad\) minimum value is \(y(1)=1^4-2(1)^2+1=1-2+1=0\)

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

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