Question

$\begin{array}{cccc}& \text{List- I}& & \text{List-II}\\ \text{(I)}& \text{If}{\log }_{|\cos x|}(1+\sin x)=2,x\in [-2\pi ,2\pi ],& \text{(P)}& -2\\ & \text{then the number of value(s) of}x\text{is/are}& & \\ \text{(II)}& \text{If max}||x-4|-|x-3||=a\text{and}& \text{(Q)}& -1\\ & \text{min}|x-4|+|x-3|=b,\text{then}a+b=& & \\ \text{(III)}& \text{If}f(x^3-6x^2+11x-3)=x-1,\text{then}& \text{(R)}& 0\\ & f\left(3\right)=& & \\ \text{(IV)}& \text{If}x\cos \theta -y\sin \theta +z=1+\cos \varphi & \text{(S)}& 1\\ & x\sin \theta +y\cos \theta +z=1-\sin \varphi & & \\ & x\cos (\theta +\varphi )-y\sin (\theta +\varphi )=z,& & \\ & \text{then}{x}^2+y^2+z^2=& & \\ & & \text{(T)}& 2\\ & & \text{(U)}& 3\end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

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

$\left(I\right)$
${\log }_{|\cos x|}(1+\sin x)=2$
So,
$1+\sin x\ne 0\cdots \left(1\right)$
$|\cos x|\ne 1\cdots \left(2\right)$
We can write,
$1+\sin x=|\cos x{|}^2$
$\Rightarrow 1-{\cos }^{2}x+\sin x=0$
$\Rightarrow \sin x(\sin x+1)=0$
$\Rightarrow \sin x=0$
$\sin x=0\Rightarrow \cos x=\pm 1\Rightarrow |\cos x|=1$
Which is not possible,
Hence the equation will not have any solution
$\left(I\right)\to \left(R\right)$

$\left(II\right)$
$x\ge 4\Rightarrow ||x-4|-|x-3||=|x-4-x+3|=1x\le 3\Rightarrow ||x-4|-|x-3||=|-x+4+x-3|=1$
$3<x<4$
$||x-4|-|x-3||=|-x+4-x+3|=|-2x+7|$
So the maximum will be
$a=|4-3|=1$

$x\ge 4|x-4|+|x-3|=x-4+x-3=2x-7x\le 3|x-4|+|x-3|=-x+4-x+3=-2x+7$
$3<x<4|x-4|+|x-3|==-x+4+x-3=1$
So the minimum will be,
$b=1$
Hence $\Rightarrow a+b=2$
$\left(II\right)\to \left(T\right)$

$\left(III\right)$
$f(x^3-6x^2+11x-3)=x-1$
$x^3-6x^2+11x-3=3\Rightarrow (x-1)(x^2-5x+6)=0\Rightarrow (x-1)(x-2)(x-3)=0$
$\Rightarrow x=1,2,3$
So,
$x=1\Rightarrow f\left(3\right)=1-1=0$
$x=2\Rightarrow f\left(3\right)=2-1=1$
$x=3\Rightarrow f\left(3\right)=3-1=2$
$\left(III\right)\to \left(R\right),\left(S\right),\left(T\right)$

$\left(IV\right)$
Putting $\theta =0$ and $\varphi =0$
$x+z=2y+z=1x=z$
$x=z=1,y=0$
$\therefore x^2+y^2+z^2=1+1=2$
$\left(IV\right)\to \left(T\right)$