Theory Of Equations Ques 83
- 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)
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Answer:
Correct Answer: 83.$x= \cos [\frac {1}{3} \cos ^{-1} p]$
Solution:
- Let $f(x)=4 x^{3}-3 x-p ……(i)$
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} $
BETA
