SATHEE: Theory of Equations 5 Question 18

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)

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Answer:

Correct Answer: 18. $x = \cos [\frac{1}{3} \cos^{- 1} p]$

Solution:

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

Now, $f \frac{1}{2} = 4 {\frac{1}{2}}^{3} - 3 \frac{1}{2} - p = \frac{4}{8} - \frac{3}{2} - p$

$= - (1 + p)$

$f (1) = 4 (1)^{3} - 3 (1) - p = 1 - p$

$\Rightarrow f \frac{1}{2} \cdot f (1) = - (1 + p) (1 - p)$

$= (p + 1) (p - 1) = p^{2} - 1$

Which is $\leq 0, \forall p \in [- 1, 1]$ .

$∴ f (x)$ has atleast one root in $\frac{1}{2}, 1$ .

Now, $f^{'} (x) = 12 x^{2} - 3 = 3 (2 x - 1) (2 x + 1)$

$= \frac{3}{4} x - \frac{1}{2} x + \frac{1}{2} > 0 in \frac{1}{2}, 1$

$\Rightarrow f (x)$ 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 $x = \cos θ$

$∴ x \in \frac{1}{2}, 1 \Rightarrow θ \in 0, \frac{π}{3}$

From Eq. (i),

$\begin{aligned} \Rightarrow & 4 \cos^{3} θ - 3 \cos θ & = p \Rightarrow \cos 3 θ = p \\ \Rightarrow & 3 θ & = \cos^{- 1} p \\ \Rightarrow & θ & = \frac{1}{3} \cos^{- 1} p \\ \Rightarrow & \cos θ & = \cos \frac{1}{3} \cos^{- 1} p \\ \Rightarrow & x & = \cos \frac{1}{3} \cos^{- 1} p \end{aligned}$

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.

Answer:

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Theory of Equations 5 Question 18

18. Let −1≤p<1. Show that the equation 4x3−3x−p=0 has a unique root in the interval [1/2,1] and identify it.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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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.