mathematics-class-11-unit-05-chapter-02-quadratic-equations-l-2

<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Quadratic equation lecture-2</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Question 7</B></Primary></p>
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<li>Let $a \in \mathbb{R}$ and the equation $-3(x-[x])^2+2(x-[x])+a^2=0$ has no integral solutions. Here $[x]$ denotes the greatest integer $\leq x$. Then all the possible values of a lie in :</li>
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<p>$(1) (-2,1)$         <br>
$(2) (-\infty,-2) \cup(2, \infty)$         <br>
$(3) (-1,0) \cup(0,1)$         <br>
$(4) (1,2)$</p>
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<p>$  a \neq 0 $</p>
<p>$ [\text { if } a=0$</p>
<p>$ x \in \mathbb{Z}$</p>
<p>$ x-[x]=\{x\}=0]$</p>
<p>$  x-[x]=\{x\}$</p>
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<li>For $x \geq 0$, let $g(x)=\cos \left(x^2\right)$ and $f(x)=\sqrt{x}$.
Let $\alpha<\beta$ be the solutions of the equation $18 x^2-9 \pi x+\pi^2=0$.
Then the area (in sq. units) bounded by the curve $y=(g \circ f)(x)$ and the lines $x=\alpha, x=\beta$ and $y=0$ is :</li>
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<p>$(1) \frac{\sqrt{3}+1}{2}$</p>
<p>$(2) \frac{\sqrt{3}-\sqrt{2}}{2}$</p>
<p>$(3) \frac{\sqrt{2}-1}{2}$</p>
<p>$(4) \frac{\sqrt{3}-1}{2}$</p>
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<p>$\frac{9 \pi \pm \sqrt{81 \pi^2-72 \pi^2}}{36}=\frac{9 \pi \pm 3 \pi}{36} $</p>
<p>$\alpha=\frac{9 \pi-3 \pi}{36}=\frac{\pi}{6}, \beta=\frac{9 \pi+3 \pi}{36}=\frac{\pi}{3}$</p>
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<li>For $0 \leq \alpha \leq 1$, if the line $x=\alpha$ divides the area of the region</li>
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<p>${R}=\{{(x, y)} \in \mathbb{R}^2: {x^3 \leq y \leq x} \text { and } 0 \leq x \leq 1 \}$</p>
<p>into two equal parts, then</p>
<p>(1) $ 0<\alpha \leq \frac{1}{2}$</p>
<p>(2) $ \frac{1}{2}<\alpha< 1$</p>
<p>(3) $ 2 \alpha^4-4 \alpha^2+1=0$</p>
<p>(4) $ \alpha^4+4 \alpha^2-1=0$</p>
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<p>Area of A</p>
<p>$ =\int_\alpha^\alpha x d x-\int_0^\alpha x^3 d x $</p>
<p>$ =\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^\alpha$</p>
<p>$ =\frac{\alpha^2}{2}-\frac{\alpha^4}{4}$</p>
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<li>If, for a positive integer $n$, the quadratic equation</li>
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<p>$ {x(x+1)+(x+1)(x+2)+\cdots+(x+n-1)(x+n)=10 n}$</p>
<p>has two consecutive integral solutions, then $n$ is equal to</p>
<p>(1) $9$     <br>
(2) $10$     <br>
(3) $11$     <br>
(4) $12$</p>
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<p>coefficient of $x^2$ is $n$</p>
<p>$ x^2+x \quad x^2+x+2 x+2=x^2+3 x+2$</p>
<p>$=x^2+(n-1) x+n x+n(n-1)$</p>
<p>$ =  x^2+(2 n-1) x+n(n-1)$</p>
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<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Question 11</B></Primary></p>
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<ul>
<li>Suppose the quadratic equation with real coefficients $p(x)=0$ has purely imaginary solutions (ie. the real part of the solutions are zero). Then the equation $p(p(x))=0$ has</li>
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<p>$i \alpha \quad \alpha \in \mathbb{R}$</p>
<p>(1) one purely imaginary solution</p>
<p>(2) all real solutions</p>
<p>(3) two real and two purely imaginary solutions</p>
<p>(4) neither real nor purely imaginary solutions</p>
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<p>$ a x^2+b x+c=0 \quad a>0$</p>
<p>$ \frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$</p>
<p>$-4 a c<0$</p>
<p>$ b^2-4 a c<0  $</p>
<p>$ b=0   $</p>
<p>$ \Rightarrow a c>0$</p>
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<p>$\rArr ~ \frac{-2 c^{\prime} \pm \sqrt{4 c^2-4\left(c^{\prime 2}+c^{\prime}\right)}}{2} $</p>
<p>$\rArr ~ \beta^2 \text { is a solution of } P(P(x))=0$</p>
<p>$\rArr ~ -c^{\prime} \pm i \sqrt{c^{\prime}}$</p>
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<p>$ \text { if } \beta=i \alpha \text { or } \beta=\alpha \quad \alpha \in \mathbb{R}$</p>
<p>$ \text { then } \beta^2=-\alpha^2 \text { or } \beta^2=\alpha^2$</p>
<p>$ \Rightarrow \beta^2 \rightarrow \text { is not real }$</p>
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<li>Let $a, b, c, d$ be four distinct numbers. Let $a, b$ be the solutions of $x^2-10 c x-11 d=0$ and $c, d$ be the solutions of $x^2-10 a x-11 b=0$. Then the value of $a+b+{c+d}$ is :</li>
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<p>(1) 0</p>
<p>(2) 11</p>
<p>(3) 121</p>
<p>(4) 1210</p>
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<p>$a, b$ solutions of $x^2-10 e x-11 d=0$,</p>
<p>a+b=10 c</p>
<p>$c, d$ solutions of $x^2-10 a x-11 b=0$, $c+d=10$ a</p>
<p>a+b+c+d=10(a+c)</p>
<p>$  a^2-10 c a-11 d=0 $</p>
<p>$ c^2-10 a c-11 b=0 $</p>
<p>$ a^2-c^2-11 d+11 b=0 $</p>
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<li>Let $S$ be the set of all nonzero real numbers $\alpha$ such that the quadratic equation $\alpha x^2-x+\alpha=0$ has two distinct real solutions $x_1$ and $x_2$ with $\left|x_1-x_2\right|<1$. Identify the possible subsets of $S$ from the ${\text { list given below }}$</li>
</ul>
<p>(1) $ \left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right)$</p>
<p>(2) $ \left(-\frac{1}{\sqrt{5}}, 0\right)$</p>
<p>(3) $ \left(0, \frac{1}{\sqrt{5}}\right)$</p>
<p>(4) $ \left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$</p>
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<p>$a x^2+b x+c=0$</p>
<p>$ b^2-4 a c>0$</p>
<p>$ 1-4 \alpha^2>0 $</p>
<p>$ \Rightarrow 4 \alpha^2< 1  $</p>
<p>$ \Rightarrow \alpha^2< 1 / 4 $</p>
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<p>$ \alpha \in\left(-\frac{1}{2}, \frac{1}{2}\right) \backslash\{0\}$</p>
<p>$ \left|x_ 1-x_2\right|< 1$</p>
<p>$ \Leftrightarrow \underbrace{\left(x_ 1-x_ 2\right)^2}_ {\left(x_ 1+x_ 2\right)^2-4 x_ 1 x_2 < 1}< 1$</p>
<p>$\alpha x^2-x+\alpha=0$</p>
<p>$ x_ 1+x_2=\frac{1}{\alpha}$</p>
<p>$  x_ 1 x_2=1$</p>
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<p>$ \frac{1}{\alpha^2}-4 < 1 \Rightarrow ~ \frac{1}{\alpha^2} < 5 $</p>
<p>$ \Rightarrow ~ \alpha^2 > \frac{1}{5}$</p>
<p>$ \Rightarrow ~ \alpha > \frac{1}{\sqrt{5}} \quad \text { or } \quad \alpha < -\frac{1}{\sqrt{5}} $</p>
<p>$S=\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$</p>
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<li>For $p \neq 0$, let $\alpha, \beta$ be the solutions of $p x^2+q x+r=0$. If $p, q, r$ are in arithmetic progression and $\frac{1}{\alpha}+\frac{1}{\beta}=4$, then the value of $|\alpha-\beta|$ is :</li>
</ul>
<p>(1) $\frac{\sqrt{34}}{9}$</p>
<p>(2) $ \frac{2 \sqrt{ } 13}{9}$</p>
<p>(3) $\frac{\sqrt{61}}{9}$</p>
<p>(4) $ \frac{2 \sqrt{34}}{9}$</p>
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<p>$ \alpha+\beta=-\frac{q}{p} \ldots (i) $</p>
<p>$ \alpha \beta=\frac{r}{p}$</p>
<p>$ \frac{1}{\alpha}+\frac{1}{\beta}=4 $</p>
<p>$ \alpha+\beta=4 \alpha \beta$</p>
<p>$ =\frac{4 r}{p} \ldots (ii) $</p>
<p>from (i) and  (ii)</p>
<p>$-\frac{q}{p}=\frac{4 r}{p}$</p>
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<p>$\operatorname{as} ~ {p \neq 0} $</p>
<p>$[{q}=-{4 p}]$</p>
<p>$ q=\frac{p+r}{2} \quad $</p>
<p>$ \Rightarrow-4 r=\frac{b+r}{2} \quad $</p>
<p>$ \Rightarrow p=-9 r \quad $</p>
<p>$ \frac{r}{p}=-\frac{1}{9}$</p>
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<p>$ \alpha+\beta = -\frac{4}{9} $</p>
<p>$ {(\alpha-\beta)}^2 ={\alpha+\beta}^2 -4\alpha\beta $</p>
<p>$ =\frac{16}{81}-4\left(-\frac{1}{9}\right) ~ =\frac{16}{81}+\frac{4}{9} ~ =\frac{52}{81} $</p>
<p>$ \alpha-\beta ~ = \pm \frac{2 \sqrt{13}}{9}$</p>
<p>$ \therefore|\alpha-\beta| = \frac{2 \sqrt{13}}{9}$</p>
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<li>Let $a, b, c$ be real numbers and $a \neq 0$. If $\alpha$ is a solution of $a^2 x^2+b x+c=0, \beta$ is a solution of $a^2 x^2-b x-c=0$ and $0<\alpha<\beta$, then the equation $a^2 x^2+2 b x+2 c={0}$ has a solution $\gamma$ which always satisfies :</li>
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<p>(1) $\gamma=\frac{\alpha+\beta}{2}$</p>
<p>(2) $\gamma=\alpha+\frac{\beta}{2}$</p>
<p>(3) $\gamma=\alpha$</p>
<p>(4) $\alpha<\gamma<\beta$</p>
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<p>$ a^2 \alpha^2+b \alpha+c=0$</p>
<p>$ a^2 \beta^2-b \beta-c=0$</p>
<p>$ f(x)=a^2 x^2+2 b x+2 c$</p>
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<p>$f(\alpha)=a^2 \alpha^2+2 b \alpha+2 c$</p>
<p>$=\underbrace{a^2 \alpha^2+b \alpha+e}+(b \alpha+e)$</p>
<p>$=b \alpha+c  $</p>
<p>$=-a^2 \alpha^2<0  $</p>
<p>$\gamma \neq \alpha$</p>
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<p>$ f(\beta)={a^2 \beta^2}+2 b \beta+2 c $</p>
<p>$ = 3 b \beta + 3 c ~ =3 a^2 \beta^2 > 0 $</p>
<p>$ b \alpha+c=-a^2 \alpha^2$</p>
<p>$ a^2 \beta^2=b \beta+c$</p>
<p>$0 < \alpha < \beta$</p>
<p>$\exists \alpha<\gamma<\beta$</p>
<p>$ \text { so that } f(\gamma)=0 $</p>
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