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Ex 6.3,28 (MCQ) - Chapter 6 Class 12 Application of Derivatives
Last updated at August 2, 2026 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
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Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima
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Transcript
Ex 6.3, 28 For all real values of x, the minimum value of (1 ā š„ + š„2)/(1 + š„ + š„2) is (A) 0 (B) 1 (C) 3 (D) 1/3Let š(š„)=(1 ā š„ + š„2)/(1 + š„ + š„2) Finding šā²(š) š(š„)=(1 ā š„ + š„2)/(1 + š„ + š„2) š^ā²(š„) =((1 ā š„ + š„^2 )^ā² (1 + š„ + š„^2 ) ā (1 ā š„ + š„^2 ) (1 + š„ + š„^2 )^ā²)/(1 + š„ + š„^2 )^2 šā²(š„)=(ā1 ā š„ ā š„^2 + 2š„ + 2š„^2+ 2š„^3 ā (1 ā š„ + š„^2+ 2š„ ā 2š„^2 + 2š„^3 ))/(1 + š„ + š„^2 )^2 š(š„)=(ā1 + š„ + š„^2 + 2š„^3 ā (1 + š„ ā š„^2 + 2š„^3 ))/(1 + š„ + š„^2 )^2 šā²(š„)=(ā2 + 2š„^2)/(1 + š„ + š„^2 )^2 Putting š^ā² (š)=š (ā2 + 2š„^2)/(1 + š„ + š„^2 )^2 =0 2š„^2ā2=0 2š„^2=2 š„^2=1 š„=Ā±1 Hence, x = 1 or x = ā1 are the critical points Finding value of š(š) at critical points Hence, minimum value of f(x) is 1/3. So, (D) is the correct answer
