Question

If all values of $x$ which satisfies the inequality $\log _{\frac{1}{3}}\left(x^2+2 p x+p^2+1\right) \geq 0$ also satisfy the inequality $kx ^2+ kx - k ^2 \leq 0$ 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+2 px + p ^2+1\right) \geq 0 $
$\Rightarrow x ^2+2 px + p ^2+1 \leq 1 \Rightarrow x ^2+2 px + p ^2 \leq 0 \Rightarrow( x + p )^2 \leq 0, \Rightarrow x =- p$
Now, $x =- p$ satisfy the inequality, $kx ^2+ kx - k ^2 \leq 0$
$\Rightarrow kp ^2- kp - k ^2 \leq 0 \forall k \in R \Rightarrow- k ^2+\left( p ^2- p \right) k \leq 0 \forall k \in R $
$\Rightarrow k ^2-\left( p ^2- p \right) k \geq 0 \forall k \in R$
then discriminant $\leq 0 \Rightarrow\left( p ^2- p \right)^2 \leq 0 \Rightarrow p ^2- p =0 \Rightarrow p =0,1$.
Now verifvaltematives,