Question

$\begin{array}{ccccc}& \text{List - 1}& & \text{List - 2}& \\ \text{(I)}& \text{If}f\left(x\right)=e^{\left[x\right]}\text{and}g\left(x\right)=\frac{x^2-4x+3}{x^2-2x+3},& \text{(P)}& 0& \\ & \text{then number of integer(s) in the range of}(f\circ g)\left(x\right)\text{is}& & & \\ & \text{(where [.] represents the greatest integer function)}& & & \\ \text{(II)}& \underset{0}{\overset{4}{\int }}\frac{z^{18}-1}{\sum _{n=0}^{17}{z}^n}dz& \left(Q\right)& 1& \\ \text{(III)}& \text{In a}\Delta XYZ,y^2\sin \left(2Z\right)+z^2\sin \left(2Y\right)=2yz,& \text{(R)}& 2& \\ & \text{where}y=15,z=8\text{. Then the length of inradius is}& & & \\ \text{(IV)}& \text{The number of integers in the range of the function}& \text{(S)}& 3& \\ & f\left(x\right)=\sqrt{{\sin }^{-1}x-{\cos }^{-1}x}+\sqrt{{\tan }^{-1}x-{\cot }^{-1}x}\text{is}& & & \\ & & \text{(T)}& 4& \\ & & \text{(U)}& 5& \end{array}$

Which of the following has CORRECT pair of combination?

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

Answer 1

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

$\left(I\right)$
$(f\circ g)\left(x\right)=e^{\left[\frac{x^2-4x+3}{x^2-2x+3}\right]}$
Let $y=\frac{x^2-4x+3}{x^2-2x+3}$
$\Rightarrow (y-1)x^2+(4-2y)x+3y-3=0$
It is a quadratic equation and its discriminant should be $\ge 0$
$(4-2y{)}^2-4(y-1\left)\right(3y-3)\ge 0$
$\Rightarrow y\in [\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}]$
Now, for $y\in [\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}]$

Range of $(f\circ g)\left(x\right)$ $=\{e^{-1},e^0,e^1\}$
Only integer in the range of $(f\circ g)\left(x\right)$ is $e^0=1$
$\left(I\right)\to \left(Q\right)$

$\left(II\right)$
$\underset{0}{\overset{4}{\int }}\frac{z^{18}-1}{\sum _{n=0}^{17}{z}^n}dz=\underset{0}{\overset{4}{\int }}(z-1)dz$
$\Rightarrow \underset{0}{\overset{4}{\int }}\frac{z^{18}-1}{\sum _{n=0}^{17}{z}^n}dz=\frac{4^2}{2}-4=4$
$\left(II\right)\to \left(T\right)$

$\left(III\right)$
$y^2\sin \left(2Z\right)+z^2\sin \left(2Y\right)=2yz$
$\Rightarrow y^2\left(2\sin Z\cos Z\right)+z^2\left(2\sin Y\cos Y\right)=2yz$
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 $r=\frac{\Delta }{s}=\frac{15\times 8}{15+8+17}=\frac{120}{40}=3$
$\left(III\right)\to \left(S\right)$

$\left(IV\right)$
${\sin }^{-1}x-{\cos }^{-1}x\ge 0$
$\Rightarrow 2{\sin }^{-1}x\ge \frac{\pi }{2}$
$\Rightarrow x\in [\frac{1}{\sqrt{2}},1]$

${\tan }^{-1}x-{\cot }^{-1}x\ge 0$
$\Rightarrow 2{\tan }^{-1}x\ge \frac{\pi }{2}$
$\Rightarrow x\in [1,\infty ]$

Therefore, domain of $f\left(x\right)$ is $[\frac{1}{\sqrt{2}},1]\cap [1,\infty ]=\left\{1\right\}$

At $x=1$,
$\sqrt{{\sin }^{-1}1-{\cos }^{-1}1}=\sqrt{\frac{\pi }{2}}$
and $\sqrt{{\tan }^{-1}1-{\cot }^{-1}1}=0$
$\therefore f\left(1\right)=\sqrt{\frac{\pi }{2}}$
$\left(IV\right)\to \left(P\right)$