Question

$\begin{array}{cc}ColumnI& ColumnII\\ \text{a. The value of}{\log }_2{\log }_2{\log }_{4}256+{\log }_{\sqrt{2}}4\text{is}& \text{p.}1\\ \text{b. If}{\log }_3(5x-2)-2{\log }_3\sqrt{3x+1}=1-{\log }_{3}4,\text{then}x=& \text{q.}6\\ \text{c. Product of roots of the equation}{7}^{{\log }_7(x^2-4x+5)}=(x-1)\text{is}& \text{r.}3\\ \text{d. Number of integers satisfying}{\log }_2\sqrt{x}-2{\left({\log }_{\frac{1}{4}}x\right)}^2+1>0\text{are}& \text{s.}5\end{array}$
Then which of the following is correct ?

• A. $a\to s,b\to r,c\to p,d\to q$
• B. $a\to s,b\to p,c\to q,d\to r$
• C. $a\to s,b\to p,c\to q,d\to r$
• D. $a\to p,b\to r,c\to p,d\to q$

Answer 1

The correct option is C $a\to s,b\to p,c\to q,d\to r$

a.

${\log }_2{\log }_2{\log }_{4}256+{\log }_{\sqrt{2}}4={\log }_2{\log }_2{\log }_4{4}^4+2{\log }_{2}4={\log }_2{\log }_{2}4+2\times 2={\log }_{2}2+4=5$

b.

${\log }_3(5x-2)-2{\log }_3\sqrt{3x+1}\Rightarrow {\log }_3(5x-2)=1-{\log }_{3}4\text{Now,}5x-2>0\Rightarrow x>\frac{2}{5}3x+1>0\Rightarrow x>\frac{-1}{3}\Rightarrow {\log }_3(5x-2)-2{\log }_3\sqrt{3x+1}=1-{\log }_{3}4\Rightarrow {\log }_3[\frac{5x-2}{3x+1}]+{\log }_{3}4=1\Rightarrow {\log }_3[\frac{4(5x-2)}{3x+1}]=1\Rightarrow 4(\frac{5x-2}{3x+1})=3\Rightarrow 20x-8=9x+3\Rightarrow 11x=11\Rightarrow x=1$

c.

$7^{{\log }_7(x^2-4x+5)}=x-1\cdots \left(1\right)\text{Now,}{x}^2-4x+5>0\Rightarrow x^2-4x+4+1>0\Rightarrow (x-2{)}^2+1>0\therefore x\in R\text{Also,}x-1>0\Rightarrow x>1\text{From (1),}{x}^2-4x+5=x-1\Rightarrow x^2-5x+6=0\Rightarrow (x-2\left)\right(x-3)=0\Rightarrow x=2\text{or}x=3$

d.

Given expression is defined if $x>0{\log }_2\sqrt{x}-2({\log }_{1/4}x{)}^2+1>0\Rightarrow {\log }_2{x}^{\frac{1}{\Rightarrow }}2-2({\log }_{2^{-2}}x{)}^2+1>0\Rightarrow \frac{1}{2}{\log }_{2}x-2\times \frac{1}{4}({\log }_{2}x{)}^2+1>0\Rightarrow \frac{1}{2}{\log }_{2}x-\frac{1}{2}({\log }_{2}x{)}^2+1>0\text{Let}{\log }_{2}x=t\Rightarrow \frac{y}{2}-\frac{y^2}{2}+1>0\Rightarrow y^2-y-2<0\Rightarrow (y-2)(y+1)<0\Rightarrow -1<y<2\Rightarrow -1<{\log }_{2}x<2\Rightarrow \frac{1}{2}<x<2\therefore x\in (\frac{1}{2},4)$
Integer values of $x$ are $1,2,3$