Question

If $p^{th},q^{th}$ and $r^{th}$ terms of a G.P. be 27, 8, and 12 respectively, then the equation $p{x}^2+2qx-2r=0$ has

• A. only one root in (0, 1)
• B. no root in (0, 1)
• C. both roots are in (0, 1)
• D. roots are imaginary

Answer 1

Answer: (A)

According to problem
$27=A{R}^{p-1}\dots (i)$
$8=A{R}^{q-1}\dots (ii)$
$12=A{R}^{r-1}\dots (iii)$
$\frac{27}{8}=R^{p-q}\dots (iv)$
[Using (i) & (ii)]
$\frac{12}{8}=R^{r-q}\dots (v)$
[Using (ii) & (iii)]
Now from (iv) and (v), we get
${\left(\frac{3}{2}\right)}^3=R^{p-q}$ and ${\left(\frac{3}{2}\right)}^1=R^{r-q}$
${\left({\left(\frac{3}{2}\right)}^1\right)}^3={\left(R^{r-q}\right)}^3$
$\Rightarrow R^{p-q}=R^{3r-3q}$
$\Rightarrow \left(p-q\right)lo{g}_{R}R=\left(3r-3q\right)lo{g}_{R}R$
$\Rightarrow p+2q-3r=0\dots \left(vi\right)$
Now, Assume $f\left(x\right)=p{x}^2+2qx-2r$
$\therefore f\left(0\right)=-2r$
and $f\left(1\right)=p+2q-2r$
and $f\left(1\right)=p+2q-2r$

$\therefore f\left(1\right)=r$ (From (vi))
Now, $f\left(0\right)f\left(1\right)=-2r^2<0$
$\therefore f\left(0\right)$ and $f\left(1\right)$ are having opposite signs, which means there exist only one root between $\left(0,1\right)$