Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

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₯ βcosβ‘π₯, 0<π₯<2 πf (π₯)=sinβ‘π₯ βcosβ‘π₯, 0<π₯<2 π Finding fβ(π) fβ(π₯)=cosβ‘π₯β(βsinβ‘π₯ ) fβ(π₯)=cosβ‘π₯+sinβ‘π₯ Putting fβ(π)=π cosβ‘π₯+sinβ‘π₯ = 0 cosβ‘π₯=βsinβ‘π₯ 1=(βsinβ‘π₯)/cosβ‘π₯ (βsinβ‘π₯)/cosβ‘π₯ =1 β tan π₯=1 tan π₯=β1 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 ππ 7π/4 Now finding fββ(π₯) fββ(π₯)=π(cosβ‘π₯ + sinβ‘π₯ )/ππ₯ fββ(π₯)=βsinβ‘π₯+cosβ‘π₯ Putting π = ππ/π fββ(3π/4)=βπ ππ(3π/4)+πππ (3π/4) =βπ ππ(πβπ/4)+πππ (πβ π/4) =βπ ππ(π/4)+(βπππ  π/4) =(β1)/β2β1/β2 =(β2)/β2 =ββ2 < 0 Hence fββ(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₯βcosβ‘π₯ f(3π/4)=π ππ(3π/4)βπππ (3π/4) =π ππ(πβπ/4)βπππ (πβπ/4) =π ππ(π/4)β(βπππ  π/4) =π ππ π/4+πππ  π/4 =1/β2+1/β2 =2/β2 =β2 Now, for π = ππ/π fββ(π₯)=βsinβ‘π₯+cosβ‘π₯ Putting π₯ = 7π/4 fββ(7π/4)=βsinβ‘(7π/4)+cosβ‘(7π/4) fββ(7π/4)=βπ ππ(2πβπ/4)+πππ (2πβπ/4) =β(βπ ππ(π/4))+πππ (π/4) =π ππ π/4+πππ  π/4 =1/β2 + 1/β2 =2/β2 =β2 > 0 fββ(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 πππ₯βπππ π₯ f(7π/4)=π ππ(7π/4)βπππ (7π/4) =π ππ(2πβπ/4)βπππ (2πβπ/4) =βπ ππ(π/4)βπππ (π/4) =(β1)/β2 β 1/β2 =(β2)/β2 =ββ2 Thus, f(π₯) is maximum at x = ππ/π and maximum value is βπ & f(π₯) is minimum at x = ππ/π and maximum value is ββπ