Question

$\begin{array}{cccc}& \text{List- I}& & \text{List-II}\\ \text{(I)}& {\log }_{x+1}(4x^3+9x^2+6x+1)+{\log }_{4x+1}(x^2+2x+1)& \text{(P)}& -1\\ & =5,\text{then the number of solution is}& & \\ \text{(II)}& \text{If}f\left(x\right)=\{\begin{array}{cc}{\left(\frac{3}{4}\right)}^{\frac{\tan 4x}{\tan 3x}},& 0<x<\frac{\pi }{2}\\ a+2,& x=\frac{\pi }{2}\\ (1+|\cot x|{)}^{ab|\tan x|},& \frac{\pi }{2}<x<\pi \end{array}& \text{(Q)}& 0\\ & \text{is continuous at}x=\frac{\pi }{2},\text{then}a+b=& & \\ \text{(III)}& \text{If}I=\underset{-1}{\overset{3}{\int }}\frac{1}{x^2}dx,\text{then}|3I|=& \text{(R)}& 1\\ \text{(IV)}& \text{If x-z plane divides the line joining of the points}& \text{(S)}& 2\\ & P(3,a,-4)\text{and}Q(2,-\frac{2}{3},5)\text{in the ratio}3:1,& & \\ & \text{then,}a=& & \\ & & \text{(T)}& 3\\ & & \text{(U)}& \text{Does not}\\ & & & \text{exist}\end{array}$

Which of the following is the only CORRECT combination?

• A. $\left(I\right)\to \left(U\right)$
• B. $\left(II\right)\to \left(P\right)$
• C. $\left(IV\right)\to \left(Q\right)$
• D. $\left(IV\right)\to \left(T\right)$

Answer 1

The correct option is B $\left(II\right)\to \left(P\right)$

$\left(I\right)$
${\log }_{x+1}(4x^3+9x^2+6x+1)+{\log }_{4x+1}(x^2+2x+1)=5$
$x+1>0,x\ne 0\text{and}4x+1>0,x\ne 0\cdots \left(1\right)$

$\text{Now,}{\log }_{x+1}\left[\right(x+1{)}^2(4x+1)]+{\log }_{4x+1}(x+1{)}^2=5$
$\Rightarrow 2{\log }_{x+1}(x+1)+{\log }_{x+1}(4x+1)+2{\log }_{x+1}(4x+1)=5$
$\Rightarrow {\log }_{x+1}(4x+1)+\frac{2}{{\log }_{x+1}(4x+1)}=3$

$\text{Put}{\log }_{x+1}(4x+1)=t$
$\Rightarrow t+\frac{2}{t}=3$
$\Rightarrow t=1,2$

$\text{When}t=1\Rightarrow {\log }_{x+1}(4x+1)=1$
$\Rightarrow 4x+1=3x+1\Rightarrow x=0$
Which is not possible [from eqn(1)]

$\text{When}t=2\Rightarrow {\log }_{x+1}(4x+1)=2$
$\Rightarrow 4x+1=(3x+1{)}^2\Rightarrow x=0,2\text{but}x\ne 0$
$\therefore x=2\text{is the only solution.}$

$\left(II\right)$
$f\left(x\right)=\{\begin{array}{cc}{\left(\frac{3}{4}\right)}^{\frac{\tan 4x}{\tan 3x}},& 0<x<\frac{\pi }{2}\\ a+2,& x=\frac{\pi }{2}\\ (1+|\cot x|{)}^{ab|\tan x|},& \frac{\pi }{2}<x<\pi \end{array}$

$f\left(\frac{\pi }{2}\right)=a+2$

$LHL=\underset{x\to \frac{{\pi }^{-}}{2}}{lim}f\left(x\right)=\underset{h\to 0^{+}}{lim}f(\frac{\pi }{2}-h)$
$=\underset{h\to 0^{+}}{lim}{\left(\frac{3}{4}\right)}^{\frac{\tan 4(\frac{\pi }{2}-h)}{\tan 3(\frac{\pi }{2}-h)}}$
$=\underset{h\to 0^{+}}{lim}{\left(\frac{3}{4}\right)}^{\frac{\tan (2\pi -4h)}{\tan (\frac{3\pi }{2}-3h)}}$
$=\underset{h\to 0^{+}}{lim}{\left(\frac{3}{4}\right)}^{\frac{\tan 4h}{\cot 3h}}$
$={\left(\frac{3}{4}\right)}^{\underset{h\to 0^{+}}{lim}\tan 4h\tan 3h}$
$={\left(\frac{3}{4}\right)}^0=1$

$RHL=\underset{h\to 0^{+}}{lim}f(\frac{\pi }{2}+h)$
$=\underset{h\to 0^{+}}{lim}(1+|\cot (\frac{\pi }{2}+h)|{)}^{ab|\tan (\frac{\pi }{2}+h)|}$
$=\underset{h\to 0^{+}}{lim}(1+\tan h{)}^{ab\cot h}$
$=e^{\underset{h\to 0^{+}}{lim}\left(\tan h\right)\left(ab\cot h\right)}\left[1^{\infty }\text{form}\right]$
$=e^{ab}$
$\Rightarrow a+2=1+e^{ab}\Rightarrow a=-1,b=0\Rightarrow a+b=-1$

$\left(III\right)$
$I=\underset{-1}{\overset{3}{\int }}\frac{1}{x^2}dx$
Integral diverges at $x=0$
Hence, integral does not exist.

$\left(IV\right)$
$P(3,a,-4),Q(2,-\frac{2}{3},5)$
$\text{Ratio}=3:1\Rightarrow (\frac{2\times 3+3}{3+1},\frac{-\frac{2}{3}\times 3+a}{3+1},\frac{5\times 3-4}{3+1})$
Since, the above points in $x-z$ plane, then its $y$-coordinate is zero.
$\Rightarrow \frac{-2+a}{4}=0\Rightarrow a=2$

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