You are here
Ex 6.3,28 (MCQ) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Chapter 6 Class 12 Application of Derivatives
Concept wise
- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
-
Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima
Transcript
Ex 6.3, 28 For all real values of x, the minimum value of (1 − 𝑥 + 𝑥2)/(1 + 𝑥 + 𝑥2) is (A) 0 (B) 1 (C) 3 (D) 1/3Let 𝑓(𝑥)=(1 − 𝑥 + 𝑥2)/(1 + 𝑥 + 𝑥2) Finding 𝒇′(𝒙) 𝑓(𝑥)=(1 − 𝑥 + 𝑥2)/(1 + 𝑥 + 𝑥2) 𝑓^′(𝑥) =((1 − 𝑥 + 𝑥^2 )^′ (1 + 𝑥 + 𝑥^2 ) − (1 − 𝑥 + 𝑥^2 ) (1 + 𝑥 + 𝑥^2 )^′)/(1 + 𝑥 + 𝑥^2 )^2 𝑓′(𝑥)=(−1 − 𝑥 − 𝑥^2 + 2𝑥 + 2𝑥^2+ 2𝑥^3 − (1 − 𝑥 + 𝑥^2+ 2𝑥 − 2𝑥^2 + 2𝑥^3 ))/(1 + 𝑥 + 𝑥^2 )^2 𝑓(𝑥)=(−1 + 𝑥 + 𝑥^2 + 2𝑥^3 − (1 + 𝑥 − 𝑥^2 + 2𝑥^3 ))/(1 + 𝑥 + 𝑥^2 )^2 𝑓′(𝑥)=(−2 + 2𝑥^2)/(1 + 𝑥 + 𝑥^2 )^2 Putting 𝒇^′ (𝒙)=𝟎 (−2 + 2𝑥^2)/(1 + 𝑥 + 𝑥^2 )^2 =0 2𝑥^2−2=0 2𝑥^2=2 𝑥^2=1 𝑥=±1 Hence, x = 1 or x = –1 are the critical points Finding value of 𝒇(𝒙) at critical points Hence, minimum value of f(x) is 1/3. So, (D) is the correct answer
