Question

If the roots of the equation $x^2-8kx+16(k^2-k+1)=0$ are real, distinct and have values at least $4$, then which of the following is (are) true?

• A. $m,m^2-1,m^3-4$ are in A.P., where $m$ is the minimum value of $k.$
• B. G.M. of $p$ and $p^2$ is $8$, where $p$ is the minimum value of $k^2.$
• C. The reciprocals of the integral values of $k$ are in H.P.
• D. The sum of the first ten integral values of $k$ is $55.$

Answer 1

Answer: (A), (B), (C)

The reciprocals of the integral values of $k$ are in H.P.

Given, $x^2-8kx+16(k^2-k+1)=0$
Roots are real and distinct $\Rightarrow D>0$
$64k^2-64(k^2-k+1)>0\Rightarrow k>1\cdots \left(1\right)$

Both the roots are at least $4$
$\Rightarrow f\left(4\right)\ge 0$
$\Rightarrow 16-32k+16(k^2-k+1)\ge 0$ and $\frac{-b}{2a}>4$
$\Rightarrow k^2-3k+2\ge 0$ and $4k>4$
$\Rightarrow k\in (-\infty ,1]\cup [2,\infty )$ and $k>1\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),k\ge 2$

Now, $m=2$
$\therefore m,m^2-1,m^3-4=2,3,4$ are in A.P.

$p=4$
G.M. of $p$ and $p^2$ is $\sqrt{p^3}=\sqrt{64}=8$

Reciprocal of integrals values of $k$ are $\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots$
Clearly, they are in H.P.

Sum of first $10$ integral values of $k$ is
$2+3+4+\cdots +11=\frac{10}{2}\times (2+11)=65$