Question

If all values of $x$ which satisfies the inequality ${\log }_{\frac{1}{3}}\left(x^2+2px+p^2+1\right)\ge 0$ also satisfy the inequality $k{x}^2+kx-k^2\le 0$ for all real values of $k$, then all possible values of $p$ lies in the interval

• A. $[-1,1]$
• B. $[0,1]$
• C. $[0,2]$
• D. $[-2,0]$

Answer 1

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

We have
${\log }_{\frac{1}{3}}\left(x^2+2px+p^2+1\right)\ge 0$
$\Rightarrow x^2+2px+p^2+1\le 1\Rightarrow x^2+2px+p^2\le 0\Rightarrow (x+p{)}^2\le 0,\Rightarrow x=-p$
Now, $x=-p$ satisfy the inequality, $k{x}^2+kx-k^2\le 0$
$\Rightarrow k{p}^2-kp-k^2\le 0\forall k\in R\Rightarrow -k^2+\left(p^2-p\right)k\le 0\forall k\in R$
$\Rightarrow k^2-\left(p^2-p\right)k\ge 0\forall k\in R$
then discriminant $\le 0\Rightarrow {\left(p^2-p\right)}^2\le 0\Rightarrow p^2-p=0\Rightarrow p=0,1$.
Now verifvaltematives,