Application of Derivatives 2 Question 28

####28. If $-1 \leq p \leq 1$, then show that the equation $4 x^{3}-3 x-p=0$ has a unique root in the interval $[1 / 2,1]$ and identify it.

$(2001,5 M)$

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Answer:

Correct Answer: 28. ($ - \frac {2}{a} , \frac {a}{3} $)

Solution:

  1. Given, $-1 \leq p \leq 1$

Let $\quad f(x)=4 x^{3}-3 x-p=0$

Now, $\quad f(1 / 2)=\frac{1}{2}-\frac{3}{2}-p=-1-p \leq 0 \quad[\because p \geq-1]$

$ \text { Also, } \quad f(1)=4-3-p=1-p \geq 0 \quad[\because p \leq 1] $

$\therefore f(x)$ has atleast one real root between $[1 / 2,1]$.

Also, $\quad f^{\prime}(x)=12 x^{2}-3>0$ on $[1 / 2,1]$

$\Rightarrow \quad f^{\prime}(x)$ increasing on $[1 / 2,1]$

$\Rightarrow \quad f$ has only one real root between $[1 / 2,1]$.

To find a root, we observe $f(x)$ contains $4 x^{3}-3 x$, which is multiple angle formula for $\cos 3 \theta$.

$\therefore$ Put $\quad x=\cos \theta$

$\Rightarrow 4 \cos ^{3} \theta-3 \cos \theta-p=0$

$\Rightarrow \quad p=\cos 3 \theta \Rightarrow \theta=(1 / 3) \cos ^{-1}(p)$

$\therefore$ Root is $\cos \frac{1}{3} \cos ^{-1}(p)$.



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