Question

$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ \text{(I)}& \text{If A and B are two sets with elements 4 and 3},& \text{(P)}& 1\\ & \text{respectively. Then the number of onto function}& & \\ & \text{from A to B is}& & \\ \text{(II)}& \text{The number of points in the interval}[-\sqrt{13},\sqrt{13}]& \left(Q\right)& 3\\ & \text{at which}f\left(x\right)=\sin \left(x^2\right)+\cos \left(x^2\right)\text{attains its}& & \\ & \text{maximum value, is}& & \\ \text{(III)}& \text{In a}\Delta XYZ,y^2\sin \left(2Z\right)+z^2\sin \left(2Y\right)=2yz,& \text{(R)}& 4\\ & \text{where}y=15,z=8\text{. Then the length of inradius is}& & \\ \text{(IV)}& {\log }_{x+1}(4x^3+9x^2+6x+1)+{\log }_{4x+1}(x^2+2x+1)& \text{(S)}& 8\\ & =5,\text{then the number of solution is}& & \\ & & \text{(T)}& 24\\ & & \text{(U)}& 36\end{array}$

Which of the following has CORRECT pair of combination?

• A. $\left(I\right)\to \left(T\right)$
• B. $\left(II\right)\to \left(S\right)$
• C. $\left(III\right)\to \left(R\right)$
• D. $\left(IV\right)\to \left(P\right)$

Answer 1

The correct option is D $\left(IV\right)\to \left(P\right)$

$\text{(I)}$
The number of onto functions from $A$ to $B$ is :
$=\sum _{k=0}^3(-1{)}^k{}^n{C}_k(3-k{)}^4$
$=3^4-{}^3{C}_1\times (3-1{)}^4+{}^3{C}_2\times (3-2{)}^4$
$=81-48+3=36$

$\text{(II)}$
$f\left(x\right)=\sin \left(x^2\right)+\cos \left(x^2\right)$
$x^2\in [0,13]$
Maximum value of $f\left(x\right)=\sqrt{2}$ and its attans at
$\sin \left(x^2\right)=\cos \left(x^2\right)$
$\Rightarrow \tan \left(x^2\right)=1$
$\Rightarrow x^2=\frac{\pi }{4},\frac{9\pi }{4}<13\Rightarrow x=\pm \frac{\sqrt{\pi }}{2},\pm \frac{3\sqrt{\pi }}{2}$
Hence, there are $4$ points in $x\in [-\sqrt{13},\sqrt{13}]$

$\text{(III)}$
$y^2\sin \left(2Z\right)+z^2\sin \left(2Y\right)=2yz$
$\Rightarrow y^2\left(2\sin Z\cos Z\right)+z^2\left(2\sin Y\cos Y\right)=2yz$
Multiplying both sides by $R$ where $R$ is the circumradius.
$y^2\left(z\cos Z\right)+z^2\left(y\cos Y\right)=2Ryz$
$\Rightarrow y\cos Z+z\cos Y=2R$
$\Rightarrow x=2R$
$\Rightarrow \angle X={90}^{\circ }$

$x^2=y^2+z^2=225+64=289$
$\Rightarrow x=17$

Now, inradius is, $r=\frac{\Delta }{s}=\frac{15\times 8}{15+8+17}=\frac{120}{40}=3$

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

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

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

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

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