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Chapter 5: Problem 84

\(f(0)<0\) \(\Rightarrow \mathrm{a}^{2}+\mathrm{a}-2<0\) \(\Rightarrow a \in(-2,1)\) Integral values of a are \(-1,0\).

Short Answer

Expert verified
Answer: The possible integral values for a that satisfy the given condition are \(a = -1\) and \(a = 0\).

Step by step solution

01

Identify critical points

To find the critical points, we first need to set the quadratic function equal to zero: \(a^2 + a - 2 = 0\). Now, we will find the roots of this equation. There are several ways to do this, and we will use the quadratic formula: \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=1\), and \(c=-2\).
02

Calculate the roots

Plugging the values a, b, and c into the quadratic formula, we get: \(a = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)}\) \(a = \frac{-1 \pm \sqrt{1 + 8}}{2}\) \(a = \frac{-1 \pm \sqrt{9}}{2}\) \(a = \frac{-1 \pm 3}{2}\) Now, we have two possible values for a: \(a_1 = \frac{-1 + 3}{2} = \frac{2}{2} = 1\) \(a_2 = \frac{-1 - 3}{2} = \frac{-4}{2} = -2\) So the critical points are a = -2 and a = 1.
03

Identify where the function is negative

Now, we know that the quadratic function is positive between the critical points and negative outside of them. So the function will be negative if \(a<-2\) or \(a>1\). However, we need the condition \(f(0)<0\). Since the function is negative outside the critical points, \(f(0)<0\) will only be satisfied when \(a\) is between the critical points: \(a \in (-2, 1)\).
04

Determine integral values of a

Now, we need to find the integral values of a in the interval \((-2, 1)\). Based on the interval, the integral values in this range are -1 and 0. So, the possible integral values of a that satisfy \(f(0)<0\) are \(a = -1\) and \(a = 0\).

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Most popular questions from this chapter

\(\mathrm{x}=\mathrm{t}^{2}+\mathrm{t}+1\) \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=2 \mathrm{t}+1\) \(\mathrm{y}=\mathrm{t}^{2}-\mathrm{t}+1\) \(\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{t}-1\) \(\frac{d y}{\mathrm{~d} \mathrm{x}}=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\) $\Rightarrow \mathrm{y}-\left(\mathrm{t}^{2}-\mathrm{t}+1\right)=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\left(\mathrm{x}-\left(\mathrm{t}^{2}+\mathrm{t}+1\right)\right)$ $\Rightarrow\left(\mathrm{t}-\mathrm{t}^{2}\right)=\left(\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\right)\left(-\left(\mathrm{t}^{2}+\mathrm{t}\right)\right)$ $\Rightarrow 2 \mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-\mathrm{t}^{2}=\mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-2 \mathrm{t}^{2}$ \(\Rightarrow 2 \mathrm{t}^{2}=0\) \(\Rightarrow \mathrm{t}=0\)

\(f^{\prime}(x)= \begin{cases}-2 x, & x<0 \\ 2 x, & x \geq 0\end{cases}\) eqn of tangent $\rightarrow \quad y-y_{1}=2 x_{1}\left(x-x_{1}\right), x \geq 0$ \(y-y_{2}=-2 x_{2}\left(x-x_{2}\right) x<0\)

\(|\sin 2 \mathrm{x}|=|\mathrm{x}|-\mathrm{a}\) \(|\mathbf{x}|-\mathrm{a}>1 \quad\) for no solution \(|x|>1+a\)

\(y=\frac{x^{2}}{4}-3 x+10\) For \(y=2-\frac{x^{2}}{4}\) eqn of tangent \(y-2=m x+m^{2}\) Solving (II) \& (I) \(m x+2+m^{2}=\frac{x^{2}}{4}-3 x+10\) \(\Rightarrow \frac{x^{2}}{4}-(m+3) x+\left(8-m^{2}\right)=0\) \(\mathrm{D}=0\) \((m+3)^{2}-\left(8-m^{2}\right)=0\) \(2 m^{2}+6 m+1=0\) \(\mathrm{m}_{1}+\mathrm{m}_{2}=-3\)

Let \(\mathrm{P}\) be \(\left(\mathrm{t}^{2}, 2 \mathrm{t}\right)\) \(y^{2}=4 x\) \(2 y \frac{d y}{d x}=4\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{t}}\) If \(Q\) is a pt where normal meets the parabola again wit \(\mathrm{t}_{1}\) as parameter \(\mathrm{t}_{1}=-\mathrm{t}-\frac{2}{\mathrm{t}}\) Coordinate of $Q \rightarrow\left(\left(t+\frac{2}{t}\right)^{2},-2 t-\frac{4}{t}\right)$ Distance \(=\sqrt{\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right)^{4}+4\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right)^{2}}\) Distance $=\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right) \sqrt{\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right)^{2}+4}$ Min Distance \(=2 \sqrt{2} \sqrt{12}=4 \sqrt{6}\)
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