Question 10 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by

If x is real, the minimum value of x 2 – 8x + 17 is

(A)–1                                (B) 0

(C) 1                                 (D) 2

 

This question is similar to Ex 6.5, 1 (i) - Chapter 6 Class 12 - Application of Derivatives

If x is real, the minimum value of x2 – 8x + 17 is - Teachoo Class 12 - NCERT Exemplar - MCQs

part 2 - Question 10 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Question 10 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Question 10 If x is real, the minimum value of x2 – 8x + 17 is –1 (B) 0 (C) 1 (D) 2 𝑓(𝑥) =𝑥^2−8𝑥+17 Finding 𝒇^′ (𝒙) 𝒇^′ (𝒙)=2𝑥−8+0 𝑓^′ (𝑥)=2𝑥−8 𝑓′(𝑥)=𝟐(𝒙−𝟒) Putting f’ (𝒙)=𝟎 2(𝑥−4)=0 (𝑥−4)=0 𝒙=𝟒 Thus, x = 4 is the minima Finding Minimum value 𝑓(𝑥) =𝑥^2−8𝑥+17 Putting 𝑥 =4 f(𝟒)=4^2−8(4)+17 =16−32+17 =33−32 =𝟏 Hence, minimum value of 𝑓(𝑥) is 1. So, the correct answer is (C)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo