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if f(x) = (cos(x) +5cos(3x)+cos(5x))/(cos(6x)+6cos(4x)+15cos(2x)+10), then find the value of f(0)+ f’(0) +f”(0).

if f(x) = (cos(x) +5cos(3x)+cos(5x))/(cos(6x)+6cos(4x)+15cos(2x)+10), then find the value of f(0)+ f’(0) +f”(0).

Grade:12th pass

2 Answers

Rinkoo Gupta
askIITians Faculty 81 Points
10 years ago
f(x)=(cosx+5cos3x+cos5x)/cos6x+6cos4x+15cos2x+10)
f(0)=(1+5+1)/(1+6+15+10)=7/32
f'(x)={(cos6x+6cos4x+15cos2x+10)(-sinx-15sin3x-5sin5x)-(cosx+5cos3x+cos5x)(-6sin6x-24sin4x-30sin2x)}/(cos6x+6cos4x+15cos2x+10)^2
=>f(0)={(32)(0)-(7)(0)}/(32)^2=0
f"(x)=[(cos6x+6cos4x+15cos2x+10)^2 {(cos6x+6cos4x+15cos2x+10)(-cosx-45cos3x-25cos5x)+(-sinx-15sin3x-5sin5x)(-6sin6x-24sin4x-30sin2x)+(cosx+5cos3x+cos5x)(36cos6x+96cos4x+60cos2x)}-{cos6x+6cos4x+15cos2x+10)(-sinx-15sin3x-5sin5x)+(cosx+5cos3x+cos5x)(6sin6x+24sin4x+30sin2x).2(cos6x+6cos4x+15cos2x+10)(-6sin6x-24sin4x-30sin2x)]/(c0s6x+6cos4x+15cos2x+10)^4

f"(0)=[(32)^2{(32)(-71)+(7)(152)}-{(32)(0)+0}]/(32)^4

=(32)^2{-2272+1064}/(32)^4
=-1208/1024
f(0)+f'(0)+f"(0)=7/32+0-1208/1024

=7/32-1208/1024

=(224-1208)/1024
=-984/1024


Thanks & Regards

Rinkoo Gupta
AskIITians Faculty

noogler
489 Points
8 years ago
ans. is  6240/3068=1560/767
 
approve if it is right
 

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