Minimum value of `cos2theta+costheta` for all real value of `theta` is

Options

A. `-2`
B. `-(9)/(8)`
C. 0
D. `-(9)/(16)`

Answer

B

Explanation

Let `A=costheta+cos2theta`
`therefore` On differentiating w .t .t to `theta` , we get
`(dA)/(d theta)=-sintheta-2sin2theta`
Put `(dA)/(d theta)=0` for maxima or minima .
`sintheta+2sin2theta=0rArrsintheta+4sinthetacostheta`
`rArrsintheta(1+4costheta)=0`
`rArrsintheta=0,or4costheta+1=0`
`rArrcostheta=1orcostheta=-(1)/(4)`
Now , `(d^(2)A)/(d theta^(2))=-costheta-4cos2theta`
`=-costheta-4(2cos^(2)theta-1)`
For `costheta=1`
`(d^(2)A)/(d theta^(2))=-costheta-4(2cos^(2)theta-1)`
`=-4(2(1)-1)=-1-4=-5lt0`
So , A is maximum at `costheta=1`
`rArr((d^(2)A)/(d theta^(2)))_(costheta=(-1)/(4))=(1)/(4)-4(2.(1)/(16)-1)gt0`
[ Since `cos2 theta=2cos^(2)theta-1]`
`therefore` A is minimum at `theta=cos^(-1)((-1)/(4))`.
Now minimum value of `costheta+cos2 theta`
or of `costheta +2cos^(2)theta-1`
`=((-1)/(4))+2((1)/(16))-1`
`=(-1)/(4)+(1)/(8)-1=(-2+1-8)/(8)=(-9)/(8)`