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. (b)
Solution:
- 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 $\cos \frac{1}{3} \cos ^{-1}(p)$.
