



Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
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
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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 29, 2023 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 If cos π₯=cosβ‘π¦ Then general solution of π₯=2ππΒ±π¦ 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