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Q.
Column I Column II
A If $\log _{\frac{1}{x}}\left(\frac{2(x-2)}{(x+1)(x-5)}\right) \geq 1$, then $x$ can belongs to P $(0,1 / 3]$
B If $\log _{1 / 2}(4-x) \geq \log _{1 / 2} 2-\log _{1 / 2}(x-1)$, then $x$ can belongs to Q $(1,2]$
C If $\log _3 x-\log _3^2 x \leq \frac{3}{2} \log _{(1 / 2 \sqrt{2})} 4$, then $x$ can belongs to R $(3,4)$
D Let $\alpha$ and $\beta$ are the roots of the quadratic equation$\left(p^2-3 p+4\right) x^2-4(2 p-1) x+16=0$If $\alpha$ and $\beta$ satisfy the condition $\beta>1>\alpha$, then $p$ can lie in S $(3,8)$

Complex Numbers and Quadratic Equations

Solution:

(A) Let $1 / x>1$ i.e. $0< x< 1$
$\log _{\frac{1}{x}}\left(\frac{2(x-2)}{(x+1)(x-5)}\right) \geq 1 \Rightarrow \frac{2(x-2)}{(x+1)(x-5)} \geq \frac{1}{x} \Rightarrow \frac{2(x-2)}{(x+1)(x-5)}-\frac{1}{x} \geq 0 $
$\frac{2 x(x-2)-(x+1)(x-5)}{x(x+1)(x-5)} \geq 0 \Rightarrow \frac{x^2+5}{x(x+1)(x-5)} \geq 0 $
image
Hence no solution in this case
now let $0< \frac{1}{x}<1 ; x>1$
in this case $0< x< 5$
its intersection with $x >1$ and the domain of the equation gives $x \in(1,2) \Rightarrow $ (Q)
(B)
$\log _{1 / 2}(4-x) \geq \log _{1 / 2} 2-\log _{1 / 2}(x-1)$
$(4-x)(x-1) \leq 2 $
$5 x - x ^2-4 \leq 2 \Rightarrow x ^2-5 x +6 \geq 0$
$(x-3)(x-2) \geq 0 \Rightarrow x \geq 3 \text { or } x \leq 2$
$\log _{1 / 2}(4-x)(x-1) \geq \log _{1 / 2} 2$
but $x \in(1,4)$
Hence final answer is $(1,2] \cup[3,4) \Rightarrow(Q) \&$ (R)
(C) Given: $\log _3 x-\log _3^2 x \leq \frac{3}{2} \log _{(1 / 2 \sqrt{2})} 4$
$\log _3 x-\log _3^2 x \leq-\frac{3}{2} \frac{\log _2 4}{\log _2 2 \sqrt{2}} $
$\log _3 x-\log _3^2 x \leq-2 $
$\text { put } \log _3 x=y$
$y-y^2 \leq-2 $
$y^2-y-2 \geq 0$
$(y-2)(y+1) \geq 0$
image
$x \leq \frac{1}{3} \Rightarrow (0,1 / 3] \cup[9, \infty) \Rightarrow(P)$
(D) $ f(x)=\left(p^2-3 p+4\right) x^2-4(2 p-1) x+16=0$
1 lies between the roots hence $f (1)<0$
image
$p^2-3 p+4-8 p+4+16<0 $
$p^2-11 p+24<0 $
$(p-8)(p-3)<0$
$\Rightarrow p \in(3,8) \Rightarrow ( R ) \&( S )$