Ex 6.3,8 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
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
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8 You are here
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 April 16, 2024 by Teachoo
Ex 6.3, 8 At what points in the interval [0, 2Ο ], does the function sin 2π₯ attain its maximum value? Let f(π₯)=sinβ‘2π₯, π₯ β [0 , 2π] Finding fβ(π) fβ(π₯)=π(sinβ‘2π₯ )/ππ₯ fβ(π₯)=2 cosβ‘2π₯ Putting fβ(π)=π 2cos 2π₯=0 cos 2π₯=0 cos 2π₯=cosβ‘γπ/2γ 2π₯=2ππ+π/2 π₯=(2π+1) π/4 Putting n = 0 π₯=(0+1) π/4= π/4 Putting n = 1 π₯=(2+1) π/4= 3π/4 Putting n = 2 π₯=(2(2)+1) π/4= 5π/4 Putting n = 3 π₯=(2(3)+1) π/4= 7π/4 Putting n = 4 π₯=(2(4)+1) π/4= 9π/4 Putting n = 5 π₯=(2(5)+1) π/4 =11π/4>2π Since π₯ β [0 , 2π] So, Critical Point are π₯ = 0 , π/4 , 3π/4 , 5π/4 , 7π/4 & 2Ο Finding value of f(x) at critical points i.e. π₯ = 0 , π/4 , 3π/4 , 5π/4 , 7π/4 & 2Ο Finding value of f(x) at critical points i.e. π₯ = 0 , π/4 , 3π/4 , 5π/4 , 7π/4 & 2Ο