Theory of Equations 5 Question 18
18. Let $-1 \leq p<1$. Show that the equation $4 x^{3}-3 x-p=0$ has a unique root in the interval $[1 / 2,1]$ and identify it.
$(2001,4$ M)
Show Answer
Solution:
- Let $f(x)=4 x^{3}-3 x-p$
Now, $\quad f \frac{1}{2}=4 \frac{1}{2}^{3}-3 \frac{1}{2}-p=\frac{4}{8}-\frac{3}{2}-p$
$$ =-(1+p) $$
$$ f(1)=4(1)^{3}-3(1)-p=1-p $$
$\Rightarrow \quad f \frac{1}{2} \cdot f(1)=-(1+p)(1-p)$
$$ =(p+1)(p-1)=p^{2}-1 $$
Which is $\leq 0, \forall p \in[-1,1]$.
$\therefore f(x)$ has atleast one root in $\frac{1}{2}, 1$.
Now, $f^{\prime}(x)=12 x^{2}-3=3(2 x-1)(2 x+1)$
$$ =\frac{3}{4} \quad x-\frac{1}{2} \quad x+\frac{1}{2}>0 \text { in } \frac{1}{2}, 1 $$
$\Rightarrow f(x)$ 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 $x=\cos \theta$
$$ \therefore \quad x \in \frac{1}{2}, 1 \quad \Rightarrow \quad \theta \in 0, \frac{\pi}{3} $$
From Eq. (i),
$$ \begin{array}{rlrl} & \Rightarrow & 4 \cos ^{3} \theta-3 \cos \theta & =p \Rightarrow \quad \cos 3 \theta=p \\ \Rightarrow & 3 \theta & =\cos ^{-1} p \\ \Rightarrow & & \theta & =\frac{1}{3} \cos ^{-1} p \\ \Rightarrow & & \cos \theta & =\cos \frac{1}{3} \cos ^{-1} p \\ \Rightarrow & & x & =\cos \frac{1}{3} \cos ^{-1} p \end{array} $$
