Exam

JEE Advanced

Year

2010

Paper

Paper 1

Subject

Mathematics

Chapter

Properties of Triangles

JEE Advanced 2010 · Paper 1 · Mathematics · Properties of Triangles

Let A B C be a triangle such that A C B=(π)/(6) and let a, b and c...

Let $A B C$ be a triangle such that $\angle A C B=\frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s)$of x$ for which $a=x^2+x+1, b=x^2-1$ and $c=2 x+1$ is (are)

Step-by-step solution

Using, $\cos C=\frac{a^2+b^2-c^2}{2 a b}$
$\begin{gathered} \Rightarrow\frac{\sqrt{3}}{2}=\frac{\left(x^2+x+1\right)^2+\left(x^2-1\right)^2-(2 x+1)^2}{2\left(x^2+x+1\right) \left( x^2-1\right)}\\ \Rightarrow(x+2)(x+1)(x-1) x+\left(x^2-1\right)^2 \\ =\sqrt{3}\left(x^2+x+1\right)\left(x^2-1\right) \\ \Rightarrow x^2+2 x+\left(x^2-1\right)=\sqrt{3}\left(x^2+x+1\right) \\ \Rightarrow(2-\sqrt{3})x^2+(2-\sqrt{3})x-(\sqrt{3}+1)=0 \end{gathered} ⇒ x=-(2+\sqrt{3}) and x=1+\sqrt{3} But, x=-(2+\sqrt{3}) \Rightarrow$ c$is negative.$\therefore \quad x=1+\sqrt{3}$is the only solution. Hence, (b) is the correct option.$