Question

pls answer following questions​

Answer 1

Answer:

8 is the value

Step-by-step explanation:

Answer

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f(x)=cos3x−15cosx+8 where xϵ[3π,23π]

f′(x)=−3sin3x+15sinx=0

⇒sin3x=5sinx

⇒3sinx−4sin3x=5sinx

⇒−4sin3x=2sinx

⇒2sinx+4sin3x=0

⇒2sinx[1+2sin2x]=0

Now, sinx=0 and sin2x=2−1 Not possible

x=π is the only choice because

xϵ[2π,23π]

f′′(x)=−9cos3x+15cosx

f′′(π)=−6<0, therefore x=π is the point of maxima.

f(2π)=cos