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Ex 6.3, 1 (iv) - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2023 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
Important
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv) You are here
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
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)
Transcript
Ex 6.3,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iv) f(π₯) = π₯3 + 1f(π₯)=π₯^3+1 Finding fβ(x) fβ(π₯)=π(π₯^3 + 1)/ππ₯ =3π₯^2 Putting fβ(π)=π 3π₯^2=0 π₯^2=0 π₯=0 Therefore by first derivate test, the point π₯=0 is neither a point of local maxima nor a point of local Minima Hence π=π is point of inflexion Hence, there is no minimum or maximum value Ex 6.3, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iv) f(π₯) = π₯3 + 1 π₯=0 Finding fββ(x) fβ(x) = 3x2 fββ(x) = 6x Finding fββ(x) at x = 0 fββ(0) = 6 Γ 0 = 0 Since fββ(x) = 0 at x = 0 β΄ The point π₯=0 is neither a point of local maxima nor a point of local Minima Hence π=π is point of inflexion Hence, there is no minimum or maximum value
