mathematics-class-11-unit-05-chapter-04-problem-solving-quadratic-equations-l-4

<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Quadratic equation lecture 4</B></Success></p>
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<p>$f(1 / 2)$ & $f(1)$ have Opp. signs</p>
<p>$
\exists x_0 \in(1 / 2,1) \text { s.t } f\left(x_0\right)=0 \text {. }
$</p>
<p>Uniqueness.
$ f^{\prime}(x)=12 x^2-3$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Quadratic equation lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Solution</B></Primary></p>
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<p>$1 \geqslant \log _5 \frac{a x^2+4 x+a}{x^2+1} $</p>
<p>$\log _5 5 \geqslant \log \frac{a x^2+4 x+a}{x^2+1}$</p>
<p>$\Rightarrow 5 \geqslant \frac{a x^2+4 x+a}{x^2+1}$</p>
<p>Note that $a x^2+4 x+a>0$ & $x^2+1>0$</p>
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<p>======</p>
<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Quadratic equation lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Solution</B></Primary></p>
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<p>Consider $a x^2+4 x+a$</p>
<p>$D=16-4 a^2<0 \text { and } a>0$</p>
<p>$a^2>4, a>0 $</p>
<p>$\therefore a>2$</p>
<p>$a>2$ or $a<-2$</p>
<p>$\Rightarrow a x^2+4 x+a \leq 5\left(x^2+1\right)$</p>
<p>$(5-a) x^2-4 x+(5-a) \geqslant 0 \quad \forall x \in \mathbb{R}$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Quadratic equation lecture 4</B></Success></p>
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<p>This is the $\Leftrightarrow 16-4(5-a)^2 \leq 0$ & $5-a>0$</p>
<p>$(5-a)^2 \geqslant 4, a<5 $</p>
<p>$\therefore 5-a \geqslant 2 \text { or } 5-a \leqslant-2 ; a<5 \qquad \biggl[5-a \geqslant 2 \text { or } 5-a \leqslant-2\biggl]$</p>
<p>$a \leq 3 \text { or } a \geqslant 7 ; a<5$</p>
<p>$\Rightarrow a \leq 3$. The range of a is $(2,3)$.</p>
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