Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important You are here
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at May 26, 2023 by Teachoo
Since NCERT Books are changed, we are still changing the name of content in images and videos. It would take some time.
But, we assure you that the question is what you are searching for, and the content is the best -Teachoo Promise. If you have any feedback, please contact us.
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 6.3, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(π₯) = 4π₯ β 1/2 π₯2 , π₯ β [β2, 9/2] f(π₯) = 4π₯ β 1/2 π₯2 Finding fβ(π) fβ(π₯)=π(4π₯ β 1/2 π₯^2 )/ππ₯ = 4 β 1/2 Γ2π₯ = 4 β π₯ Putting fβ(π)=π 4 β π₯=0 π₯=4 β΄ π₯=4 is only critical point Since given interval π₯ β [β2 , 9/2] Hence , calculating f(π₯) at π₯ = β 2 , 4 , 9/2 Absolute Maximum value of f(x) is 8 at π = 4 & Absolute Minimum value of f(x) is β10 at π = β2
