Q111 / 180 · KCET 2015 · Mathematics

If \( x \) is real, then the minimum value of \( x^{2}-8 x+17 \) is

  • Correct answer
    A
    \( 11 \)
  • B
    \( 02 \)
  • C
    \( 03 \)
  • D
    \( 04 \)

Solution

Given that, $f(x)=x^{2}-8 x+17$ So, $f^{\prime}(x)=2 x-8$ Now, $f^{\prime}(x)=0$ $2 x-8=0 \Rightarrow x=4$ Now, $f^{\prime \prime}(x)=2>0$ Therefore, $f(4)=4^{2}-8(4)+17=1$

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

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