/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 \(f(x)=x^{3}+4 x^{2}+\lambda x+1... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 34

\(f(x)=x^{3}+4 x^{2}+\lambda x+1\) \(f^{\prime}(x)=3 x^{2}+8 x+\lambda<0\) roots are \(-2 \& \frac{-2}{3}\) Product \(=+\frac{4}{3}\) \(\Rightarrow \lambda=+4\) Hence, A is correct

Short Answer

Expert verified
Based on the given cubic function \(f(x) = x^3 + 4x^2 + \lambda x + 1\), which has roots -2 and \(-\frac{2}{3}\), and its derivative \(f'(x) = 3x^2 + 8x + \lambda < 0\), find the value of \(\lambda\). Answer: \(\lambda = 4\).

Step by step solution

01

Relationship between roots and coefficients of a polynomial

Using Vieta's formulas, we know that for a cubic function \(f(x) = ax^3 + bx^2 + cx + d\) with roots \(\alpha, \beta, \gamma\), the sum of the roots is equal to \(-(\frac{b}{a})\), and the product of pairs of roots is equal to \((\frac{c}{a})\)
02

Use the given roots to find a polynomial function

We are given that two of the roots of \(f(x)\) are \(-2\) and \(\frac{-2}{3}\). Let the third root be \(\gamma\). Then: Sum of roots: \(\alpha + \beta + \gamma = -(4)\) Product of roots: \(\alpha \times \beta = +(\frac{4}{3})\) Calculating the sum and product gives us: \((-2) + (-\frac{2}{3}) + \gamma = -4\) \((-2) \times (-\frac{2}{3}) = \frac{4}{3}\) This gives us \(\gamma = -\frac{2}{3}\) and \(\frac{4}{3}\) as the product of the roots.
03

Derivative of the polynomial function

We are given \(f'(x) = 3x^2 + 8x + \lambda\). Since the roots of \(f(x)\) are known, we can write the function as follows: \(f(x) = (x - (-2))(x - (-\frac{2}{3}))(x - \gamma)\) By expanding and comparing coefficients, we obtain the following equation: \(\lambda = 4\gamma - 12\)
04

Use the inequality condition to find the value of \(\lambda\)

To satisfy the condition \(f'(x) < 0\), we must ensure that \(3x^2 + 8x + \lambda < 0\). Since the polynomial is negative, all its critical points must lie to the left of the maximum turning point. As the product of roots is positive (\(\frac{4}{3}\)), it means \(x^2 + 4x + \lambda > 0\). Therefore, we must have: \(\lambda = 4\gamma - 12 = 4\) This gives us the value of \(\lambda = 4\). Hence, option A is correct.

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Most popular questions from this chapter

$g^{\prime}(\theta)=\left(f\left(\sin ^{2} \theta\right)-f\left(\cos ^{2} \theta\right)\right) \sin 2 \theta$ In $\left(-\frac{\pi}{4}, \theta\right), \sin 2 \theta<0 \& \sin ^{2} \theta<\cos ^{2} \theta$ \& \(f\left(\sin ^{2} \theta\right)0\) Hence, D is correct

A) \(f(x)=\frac{x}{1+x^{2}}\) $f^{\prime}(x)=\frac{\left(1+x^{2}\right)-2 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{1-x^{2}}{\left(1+x^{2}\right)^{2}}<0$ \(\Rightarrow x \in(-\infty,-1) \cup(1, \infty)\) B) \(f(x)=\tan ^{-1} x-x\) \(f^{\prime}(x)=-\frac{x^{2}}{1+x^{2}}<0\) \(\Rightarrow x \in R\) C) \(f(x)=x-e^{x}+\tan \frac{2 \pi}{7}\) \(f^{\prime}(x)=1-e^{x} \geq 0\) \(\quad x \in(-\infty, 10]\) D) \(f(x)=x^{3}-\ln \left(1+x^{3}\right)\) \(f^{\prime}(x)=3 x^{2}\left(1-\frac{1}{1+x^{3}}\right)\) \(=\frac{3 x^{5}}{1+x^{3}}\)

\(f^{\prime}(x)=\frac{-2}{x^{2}}\left(\frac{\sqrt{x^{2}+4}+2}{\sqrt{x^{2}+4}}+2\right)<0\) Hence, A is correct

\(g^{\prime}(x)=2\left(6 x^{2}-6 x\right)\) $f^{\prime}\left(2 x^{3}-3 x^{2}\right)+\left(12 x-12 x^{2}\right)+f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)$ $g^{\prime}(x)=\left(12 x^{2}-12 x\right)\left[f^{\prime}\left(2 x^{3}-3 x^{2}\right)-f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)\right]$ $=12 x(x-1)\left[f^{\prime}\left(2 x^{3}-3 x^{2}\right)-f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)\right]$ For increasing, If \(x \in(1, \infty)\) \(2 x^{3}-3 x^{2}>6 x^{2}-4 x^{3}-3\) \(6 x^{3}-9 x^{2}+3>0\) \(2 x^{3}-3 x^{2}+1>0\) \(\mathrm{x} \in(1, \infty)\) If \(\mathrm{x} \in(0,1)\) \(6 x^{2}-4 x^{3}-3<2 x^{3}-3 x^{2}\) \(2 x^{3}-3 x^{2}+1>0\) no soln If \(x<0\) \(2 x^{3}-3 x^{2}+1>0\) \(x \in\left(-\frac{1}{2}, 0\right)\) Hence, B is correct

Similar to ques 40 . \(f(x)=a x^{3}+b x^{2}+c x+d\) \(f^{\prime}(x)=3 a x^{2}+2 b x+c>0\) \(\Rightarrow 4 b^{2}-12 a c<0 \& a>0\) \(\Rightarrow \mathrm{b}^{2}-3 \mathrm{ac}<0\) \(b^{2}<3 a c \& a>0,3 b^{2}\left(\sqrt{3} a x+\frac{4 b}{\sqrt{3}}\right)^{2}>0\) Hence, \(\mathrm{B}\) is correct
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