Theory Of Equations Ques 60
- If $5,5 r, 5 r^{2}$ are the lengths of the sides of a triangle, then $r$ cannot be equal to
(a) $\frac{5}{4}$
(b) $\frac{7}{4}$
(c) $\frac{3}{2}$
(d) $\frac{3}{4}$
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Answer:
Correct Answer: 60.(b)
Solution:
- Let $a=5, b=5 r$ and $c=5 r^{2}$
We know that, in a triangle sum of 2 sides is always greater than the third side.
$\therefore a+b>c ; b+c>a$ and $c+a>b$
Now, $\quad a+b>c$
$\Rightarrow 5+5 r>5 r^{2} \Rightarrow 5 r^{2}-5 r-5<0$
$\Rightarrow r^{2}-r-1<0$
$\Rightarrow \quad [r-(\frac{1-\sqrt{5}}{2})] \quad [r-(\frac{1+\sqrt{5}}{2})]<0$
$\left[\because\right.$ roots of $a x^{2}+b x+c=0$ are given by
$ \begin{array}{r} x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \text { and } r^{2}-r-1=0 \\ \left.\Rightarrow r=\frac{1 \pm \sqrt{1+4}}{2}=\frac{1 \pm \sqrt{5}}{2}\right] \end{array} $
$ \begin{aligned} \Rightarrow \quad r & \in (\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2})……(i) \\ & \frac{1}{\frac{1-\sqrt{5}}{2}}-\frac{1+\sqrt{5}}{2} \end{aligned} $
Similarly, $\quad b+c>a$
$ \Rightarrow \quad 5 r+5 r^{2}>5 $
$ \Rightarrow \quad r^{2}+r-1>0 $
$ [r-(\frac{-1-\sqrt{5}}{2})] \quad [r-(\frac{-1+\sqrt{5}}{2})]>0 $
$ [\because r^{2}+r-1=0 \Rightarrow r=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2}] $
$ \Rightarrow \quad r \in (-\infty, \frac{-1-\sqrt{5}}{2}) \cup (\frac{-1+\sqrt{5}}{2}, \infty) ……(ii) $
$ \frac{-1-\sqrt{5}}{2} \frac{-1+\sqrt{5}}{2}$
$ \text { and } \quad c+a>b $
$ \Rightarrow \quad 5 r^{2}+5>5 r $
$ \Rightarrow \quad r^{2}-r+1>0 $
$ \Rightarrow r^{2}-2 \cdot \frac{1}{2} r+(\frac{1}{2})^{2}+1-(\frac{1}{2})^{2}>0 $
$ \Rightarrow \quad (r-\frac{1}{2})^{2}+\frac{3}{4}>0 $
$\Rightarrow \quad r \in R ……(iii)$
From Eqs. (i), (ii) and (iii), we get
$ \begin{aligned} & r \in (\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}) \\ & \underset{-\infty}{\longleftarrow} \end{aligned} $
and $\frac{7}{4}$ is the only value that does not satisfy.
BETA
