Ex 6.3, 3 (iv) - 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 You are here
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: (iv) f (π₯)=sinβ‘π₯ βcosβ‘π₯, 0<π₯<2 πf (π₯)=sinβ‘π₯ βcosβ‘π₯, 0<π₯<2 π Finding fβ(π) fβ(π₯)=cosβ‘π₯β(βsinβ‘π₯ ) fβ(π₯)=cosβ‘π₯+sinβ‘π₯ Putting fβ(π)=π cosβ‘π₯+sinβ‘π₯ = 0 cosβ‘π₯=βsinβ‘π₯ 1=(βsinβ‘π₯)/cosβ‘π₯ (βsinβ‘π₯)/cosβ‘π₯ =1 β tan π₯=1 tan π₯=β1 Since 0 < π₯ < 2Ο & tan π₯ is negative tan ΞΈ lies in either 2nd or 4th quadrant So, value of π₯ is π₯=3π/4 ππ 7π/4 Now finding fββ(π₯) fββ(π₯)=π(cosβ‘π₯ + sinβ‘π₯ )/ππ₯ fββ(π₯)=βsinβ‘π₯+cosβ‘π₯ Putting π = ππ /π fββ(3π/4)=βπ ππ(3π/4)+πππ (3π/4) =βπ ππ(πβπ/4)+πππ (πβ π/4) =βπ ππ(π/4)+(βπππ π/4) =(β1)/β2β1/β2 =(β2)/β2 =ββ2 < 0 Hence fββ(π₯)<0 when π₯ = 3π/4 Thus π₯ = 3π/4 is point of local maxima β΄ f(π₯) is maximum value at π₯ = 3π/4 The local maximum value is f(π₯)=sinβ‘π₯βcosβ‘π₯ f(3π/4)=π ππ(3π/4)βπππ (3π/4) =π ππ(πβπ/4)βπππ (πβπ/4) =π ππ(π/4)β(βπππ π/4) =π ππ π/4+πππ π/4 =1/β2+1/β2 =2/β2 =β2 Now, for π = ππ /π fββ(π₯)=βsinβ‘π₯+cosβ‘π₯ Putting π₯ = 7π/4 fββ(7π/4)=βsinβ‘(7π/4)+cosβ‘(7π/4) fββ(7π/4)=βπ ππ(2πβπ/4)+πππ (2πβπ/4) =β(βπ ππ(π/4))+πππ (π/4) =π ππ π/4+πππ π/4 =1/β2 + 1/β2 =2/β2 =β2 > 0 fββ(π₯)>0 when π₯ = 7π/4 Thus π₯ = 7π/4 is point of local minima f(π₯) has minimum value at π₯ = 7π/4 Local minimum value is f(π₯)=π πππ₯βπππ π₯ f(7π/4)=π ππ(7π/4)βπππ (7π/4) =π ππ(2πβπ/4)βπππ (2πβπ/4) =βπ ππ(π/4)βπππ (π/4) =(β1)/β2 β 1/β2 =(β2)/β2 =ββ2 Thus, f(π₯) is maximum at x = ππ /π and maximum value is βπ & f(π₯) is minimum at x = ππ /π and maximum value is ββπ