Question 23
If x is real, the minimum value of x2 β 8x + 17 is
β1 (B) 0
(C) 1 (D) 2
π(π₯) =π₯^2β8π₯+17
Finding π^β² (π)
π^β² (π)=2π₯β8+0
π^β² (π₯)=2π₯β8
πβ²(π₯)=π(πβπ)
Putting fβ (π)=π
2(π₯β4)=0
(π₯β4)=0
π=π
Thus, x = 4 is the minima
Finding Minimum value
π(π₯) =π₯^2β8π₯+17
Putting π₯ =4
f(π)=4^2β8(4)+17
=16β32+17
=33β32
=π
Hence, minimum value of π(π₯) is 1.
So, the correct answer is (C)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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