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Let -1lt=p<1. Show that the equation 4x^...

Let `-1lt=p<1.` Show that the equation `4x^3-3x-p=0` has a unique root in the interval `[1/2,1]` and identify it. (2001, 4M)

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The correct Answer is:
A, C
Let `f(x)=4x^(3)-3x-p" "...(i)`
`Now, f((1)/(2))=4((1)/(2))^(3)-3((1)/(2))-p=4/8-3/2-p=-(1+p)`
`f(1)=4(1)^(3)-3(1)-p=1-p`
`impliesf((1)/(2)),f (1)=-(1+p)(1-p)`
`=(p+1)(p-1)=p^(2)-1`
Which is `le0, AAp in[-1,1].`
`thereforef(x)` has atleast one root in `[(1)/(2),1].`
Now, `f'(x)=12x^(2)-3=3(2x-1)(2x-1)`
`=3/4(x-(1)/(2))(x+(1)/(2))gt0 "in"[(1)/(2),1]`
`implies f(x)` is an increasing function in [1/2,1]
Therefore, f(x) has exactly one root in [1/2,1] for any `p in [-1,1].`
Now, let `x=cos theta`
`therefore x in[(1)/(2),1]impliesthetain[0,(pi)/(3)]`
From Eq. (i),
`4 cos^(3)theta-3costheta=pimpliescos 3 theta=p`
`implies 3 theta=cos^(-1)p`
`impliestheta=1/3cos^(-1)p`
`impliescostheta=cos((1)/(3)cos^(-1)p)`
`impliesx=cos ((1)/(3)cos^(-1)p)`
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