Question

The sum of an infinitely decreasing geometric progression whose first term is a and common ratio $r$, is equal to least value of the quadratic trinomial $P(x)=3x^2-x+\frac{25}{12}$, in [0,2]. Also the first term of the geometric progression is equal to the square of its common ratio.
If the minimum value of $P(x)$ lies between roots of equation $x^2+(k+1)x+a+2\sqrt{3}=0$, then the maximum integral value of $k$ is

• A. - 6
• B. - 1
• C. - 7
• D. - 10

Answer 1

Answer: (A)

$\therefore r=(\sqrt{3}-1)[\text{ As }r\in (0,1)]$
$\Rightarrow a=r^2=4-2\sqrt{3}$
[As $r\in (0,1)]$
$\Rightarrow a=r^2=4-2\sqrt{3}$
Let $f(x)=x^2+(k+1)x+a+2\sqrt{3}$
Now, $f(x)=0\Rightarrow a^2+(k+1)x+4=0$.
Since 2 lies between the roots, hence $f(2)<0$
$\Rightarrow 4+2k+2+4<0\Rightarrow 2k+10<0\Rightarrow k<-5$
So, $k_{\max .}=-6$.