IITJEE problem

Let $x = \cos{\theta}$ ; then we have:

$\begin{aligned} 4x^3-3x-p & = 0 \\ \Rightarrow 4\cos^3{\theta} - 3\cos{\theta} & = p \\ \cos{3\theta} & = p \\ 3\theta & = \cos^{-1}{p} \\ \Rightarrow x & = \cos{\theta} = \cos {\left( \frac{1}{3} \cos^{-1}{p} \right)} \end{aligned}$

As $\space p \in [-1,1]\quad \Rightarrow 3 \theta \in [0,\pi] \quad \Rightarrow \theta \in [0,\frac{\pi}{3}] \quad \Rightarrow x = \cos{\theta} \in [\frac{1}{2},1]$ , there is only one root lies in the interval $[\frac{1}{2},1]$ .

Therefore, the sum of roots in this interval is the single root $\boxed{\cos {\left( \dfrac{1}{3} \cos^{-1}{p} \right)} }$ .