Question

If three successive term of a G.P with common ratio $\left( {{\rm{r}} > 1} \right)$ from the sides of a triangle ABC and [r] denotes greatest integer function, then modulus of [-r] + [r] is.

Answer 1

Hint:
Suppose the sides of the triangle be a, ar and ${\rm{a}}{{\rm{r}}^2}$ (being the largest side as r>1). Now, using ${\rm{a}} + {\rm{ar}} > {\rm{a}}{{\rm{r}}^2}$(sum of two side is greater than third side) find the range of r and then [r] and [-r].

Complete step by step solution:
Let us suppose the sides of the triangle be a, ar and ${\rm{a}}{{\rm{r}}^2}$ (being the largest side as r > 1).
Now
We know that, sum of two sides is greater than the third side
So,
      ${\rm{a}} + {\rm{ar}} > {\rm{a}}{{\rm{r}}^2}$
     ${\rm{a}} + {\rm{ar}} - {\rm{a}}{{\rm{r}}^2} > 0$

Multiply both sides by ($ \times 1$) and the inequality sign will be changed.
then , ${\rm{a}}{{\rm{r}}^2} - {\rm{ar}} - {\rm{a}} < 0$
Take a, will be common
 $ \Rightarrow {\rm{a}}\left( {{{\rm{r}}^2} - {\rm{r}} - 1} \right) < 0$
 $ \Rightarrow {{\rm{r}}^2} - {\rm{r}} - 1 < 0$
We know that quadratic equation.
 ${\rm{a}}{{\rm{x}}^2} - {\rm{bx}} + {\rm{c}} < 0$
then, $\dfrac{{ - {\rm{b}} - \sqrt {\rm{D}} }}{{2{\rm{a}}}} < {\rm{x}} < \dfrac{{ - {\rm{b}} + \sqrt {\rm{D}} }}{{2{\rm{a}}}}$
Where $\rm{D}={b}^{2}-4ac$
So,
 $ \Rightarrow \dfrac{{1 - \sqrt {1 - 4\left( 1 \right)\left( { - 1} \right)} }}{2} < {\rm{r}} < \dfrac{{ - 1 + \sqrt {1 - 4\left( 1 \right)\left( { - 1} \right)} }}{2}$
$ \Rightarrow \dfrac{{1 - \sqrt 5 }}{2} < {\rm{r}} < \dfrac{{1 + \sqrt 5 }}{2}$
 $ \Rightarrow \left[ {\rm{r}} \right] = 1$
 Also,
 $ - \dfrac{{1 + \sqrt 5 }}{2} < - {\rm{r}} < - 1$
 $ \Rightarrow \left[ { - {\rm{r}}} \right] = - 2$

Hence
$\therefore \left[ {\rm{r}} \right] + \left[ { - {\rm{r}}} \right] = 1 - 2 = - 1$

Note:
Greatest integer function is that function rounds of the real number down to the integer less than the number. When negative is multiplied, the inequality sign changes. It is also known as the floor function. it’s one of the mostly used piecewise continuous functions. Whenever a problem needs to be made more complex just add floor function in it.