IIT-JEE 2001Question 1
Let $-1\le p\le1$. Show that the equation $4x^{3}-3x-p=0$ has a unique root in the interval $[1/2,\,1]$ and identify it.
Differentiation and its Application
Answer$x=\cos\left(\dfrac13\cos^{-1}p\right)$
Solution
Concept. Show $f(x)=4x^3-3x-p$ is strictly increasing on $\left[\frac12,1\right]$ with a sign change; then find the root using $\cos3\theta$. Formulas used. (1) $\cos3\theta=4\cos^3\theta-3\cos\theta$Step 1 - Monotonic. $f'(x)=12x^2-3\ge0$ on $\left[\frac12,1\right]$, zero only at $\frac12$. So $f$ is strictly increasing. Step 2 - Sign change. $f\left(\frac12\right)=-1-p\le0$ and $f(1)=1-p\ge0$. So there is exactly one root in $\left[\frac12,1\right]$. Step 3 - Identify. Put $x=\cos\theta$. By (1) the equation is $\cos3\theta=p$. Take $\theta=\frac13\cos^{-1}p\in\left[0,\frac\pi3\right]$; then $x=\cos\theta\in\left[\frac12,1\right]$, so it is the unique root. Answer: $x=\cos\left(\dfrac13\cos^{-1}p\right)$