SATHEE: Application of Derivatives 2 Question 28

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:

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

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

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

$Also, f (1) = 4 - 3 - p = 1 - p \geq 0 [∵ p \leq 1]$

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

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

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

$\Rightarrow f$ 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 $\cos 3 θ$ .

$∴$ Put $x = \cos θ$

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

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

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

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