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Please solve RD Sharma class 12 Chapter Maxima and Minima exercise Multiple choice question, question 31 maths textbook solution.

Answers (1)

Answer: option(c)1

Hint: For local maxima or minima, we must have f'(x)=0.

Given: y=x^2-8x+17

Solution:

We have,

y=x^2-8x+17                                                          

\Rightarrow f'(y)=2x-8

For maxima and minima f'(x)=0

\Rightarrow 2x-8=0

\Rightarrow 2x=8

\Rightarrow x=4

f(4)=4^{2}-8(4)+17=16-32+17=33-32=1>0

Hence, minimum value f(x) of is 1 at x = 4

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