Ex 6.3, 3 (iii) - 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) You are here
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)
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (iii) β(π₯)=sinβ‘π₯+cosβ‘π₯, 0<π₯<π/2 β(π₯)=sinβ‘π₯+cosβ‘π₯, 0<π₯<π/2 Finding πβ²(π) ββ²(π₯)=π(sinβ‘π₯ + cosβ‘π₯" " )/ππ₯ β^β² (π₯)=cosβ‘π₯βsinβ‘π₯ Putting πβ²(π)=π cosβ‘π₯βπ πππ₯=0 cosβ‘γπ₯=π ππ π₯γ 1 = sinβ‘π₯/cosβ‘π₯ 1 = tan π₯ tan π₯=1 β΄ π₯=45Β°= π/4 Finding hββ(π) hβ(π₯)=cosβ‘π₯βsinβ‘π₯ hββ(π₯)=βsinβ‘π₯βcosβ‘π₯ Putting π=π /π hββ(Ο/4)=βsin(Ο/4)βπππ (Ο/4) = β 1/β2β1/β2 = (β2)/β2 = β β2 Since hββ(π₯)<0 when π₯=Ο/4 β΄ π₯=Ο/4 is point of Local Maxima f has Maximum value at π=π /π f(π₯)=π πππ₯+πππ π₯ f(Ο/4)=π ππ(Ο/4)+πππ (Ο/4) = 1/β2+1/β2 = 2/β2 = βπ