/*! 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 56 Similar to ques 40 . \(f(x)=a x^... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 56

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

Short Answer

Expert verified
Question: Find the expression for the function h(x), which is the integral of the function g(x) satisfying the given conditions. Solution: Based on the given conditions and the integral of g(x), the function h(x) is expressed as \(h(x) = a^2 x^3 + ab x^2 + \left(\frac{b^2}{3} + c^2\right)x + C\).

Step by step solution

01

When a function is said to be increasing, its derivative is strictly positive for the entire domain of the function being considered. In this problem, we have \(f'(x) = 3ax^2 + 2bx + c > 0\). It's given that \(4b^2 - 12ac < 0\), which implies \(b^2 < 3ac\). #Step 2: Write the function g(x)#

The function g(x) is related to the function f(x) according to the condition: \(g(x) = a(3ax^2 + 2bx + c) + b(6ax + 2b) + c^2\). This function can be simplified and rewritten as \(g(x) = 3a^2x^2 + (6ab+b^2)x + (2ab^2+c^2)\). #Step 3: Apply the conditions given#
02

According to the given problem statement, we have \(b^20\), and \(3b^2\left(\sqrt{3} ax + \frac{4b}{\sqrt{3}}\right)^2 > 0\). Thus, we have ensured that g(x) is indeed positive. #Step 4: Find h(x) as the integral of g(x)#

Now that we have g(x), we can find h(x) since we know that \(h'(x) = g(x)\). So, \(h(x) = \int g(x) dx = \int \left(3a^2x^2 + (6ab+b^2)x + (2ab^2+c^2)\right) dx\). To calculate the integral, we simply apply the power rule to each term and update the exponents for the x terms: \(h(x) = a^2 x^3 + ab x^2 + \left(\frac{b^2}{3} + c^2\right)x + C\), where C is the constant of integration. In conclusion, the function \(h(x) = a^2 x^3 + ab x^2 + \left(\frac{b^2}{3} + c^2\right)x + C\) is the integral of the function \(g(x)\) that satisfies the given conditions.

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Most popular questions from this chapter

A) \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\frac{4}{\mathrm{x}^{2}}+7\) \(f^{\prime}(x)=3 x^{2}-\frac{8}{x^{3}}=\frac{3 x^{5}-8}{x^{3}}\) \(\mathrm{f}(\mathrm{x})\) is \(\uparrow\) in \((-\infty, \infty)\) B) \(g(x)=\sqrt{t}+\sqrt{1+t}-4\) \(g^{\prime}(x)=\frac{1}{2 \sqrt{t}}+\frac{1}{2 \sqrt{1+t}}>0\) \(\mathrm{g}(\mathrm{x})\) is increasing in \((0, \infty)\) C) \(\mathrm{r}(\theta)=\theta+\sin ^{2}\left(\frac{\theta}{3}\right)-8\) \(r^{\prime}(\theta)=1+\frac{1}{3} \sin \left(\frac{2 \theta}{3}\right)>0\) \(\mathrm{r}(\theta)\) is an increasing function D) \(r(\theta)=\tan \theta-\cot \theta-\theta\) \(r^{\prime}(\theta)=\sec ^{2} \theta+\operatorname{cosec}^{2} \theta-1\) \(=\tan ^{2} \theta+\operatorname{cosec}^{2} \theta\) \(\mathrm{r}(\theta)\) is an increasing function Hence, $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ is correct

\(f(x)=x \sqrt{4 a x-x^{2}}\) \(f^{\prime}(x)=\sqrt{4 a x-x^{2}}+\frac{x(2 a-x)}{\sqrt{4 a x-x^{2}}}\) \(=\frac{6 a x-2 x^{2}}{\sqrt{4 a x-x^{2}}}=\frac{2 x(3 a-x)}{\sqrt{x(4 a-x)}}\)

\(f^{\prime \prime}(x)>0 \Rightarrow f^{\prime}(x)\) is increasing \(g(x)=f\left(\tan ^{2} x-2 \tan x+4\right)\) $g^{\prime}(x)=f^{\prime}\left(\tan ^{2} x-2 \tan x+4\right)\left(2 \tan x \sec ^{2} x-2 \sec ^{2} x\right)$ \(=f^{\prime}\left((\tan x-1)^{2}+3\right) 2 \sec ^{2} x(\tan x-1)\) As \(\left.f^{\prime}(\tan x-1)^{2}+3\right)\) is \(+v e\) \(\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})>0\) if $\mathrm{x} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ Hence, D is correct

\(f^{\prime}(x)=g(x)(x-a)^{2}\) \(f(x)\) is increasing if \(g(a)>0\) \& \(\mathrm{f}(\mathrm{x})\) is decreasing if \(\mathrm{g}(\mathrm{a})<0\) Hence, A, D is correct

$\mathrm{f}(\mathrm{x})=\sin ^{-1}\left(\frac{2 \phi(\mathrm{x})}{1+\phi^{2}(\mathrm{x})}\right)=2 \tan ^{-1} \phi(\mathrm{x})\( if \)\mid \phi(\mathrm{x})<1$ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{1+\phi^{2}(\mathrm{x})} \times \phi^{\prime}(\mathrm{x})<0$ So, \(\mathrm{f}(\mathrm{x})\) is decreasing when \(|\phi(\mathrm{x})|<1\) Hence, B is correct
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