Question

the value of xfor the minimum value of√3cosx +sin x​

Answer 1

Answer:

$➜\sf\red{Answer:-}$

y=3 sin x + 4 cos x

dy/dx= 0 → x for Maximum & Minimum values of y

3cosx-4sinx =0

(-3/5)cosx+(4/5)sinx=0

if sinF = -3/5 and cosF = 4/5 then:

sinF cosx+ cosF sinx =0

sin(F+x)=0 → F+x= πK where k= ….,3,2,1,0,-1,-2,-3, ….

so x= -F +πK where k= ….,3,2,1,0,-1,-2,-3, ….

so sinx = sin(-F +πK)= sin(-F)cos(πK)+cos(-F)sin(πK)

sinx = sin(-F)cos(πK) +0 = -sin(F)cos(πK) = -(-3/5)*(-1)^(K+1)

sinx = (3/5)*(-1)^(K+1)

And cosx = cos(-F +πK)= cos(-F)cos(πK)-sin(-F)sin(πK)

cosx = cos(F)(-1)^(K+1)-0 = (4/5)(-1)^(K+1)

Result:

sinx = (3/5)*(-1)^(K+1) → Maximum = 3/5 , Minimum= -3/5

cosx = (4/5)(-1)^(K+1) → Maximum = 4/5 , Minimum= -4/5

so

for y=3 sin x + 4 cos x

Maximum Y= 3 (Maximum sinx) + 4 (Maximum cosx)

Maximum Y= 3 (3/5) + 4 (4/5) =25/5 =5

Minimum Y= 3 (Minimum sinx) + 4 (Minimum cosx)

Minimum Y= 3 (-3/5) + 4 (-4/5) =-5

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