Question

$\begin{array}{cccc}& List-I& & List-II\\ P& P\left(t\right)=(\frac{1}{t^2+1},\frac{t}{t^2+1}).IfP\left(\alpha \right),B\left(\beta \right).C\left(\gamma \right)& 1.& 0\\ & \text{are the vertices of an equilateral triangle}& & \\ & (\alpha ,\beta ,gamma>0)\text{and its centroid is}(a,b)& & \\ & \text{then 2a+b=}\dots \dots & & \\ Q.& \text{If a complex number z satisfying}|z-2+i|\le 1,& 2.& 1^2\\ & \text{then the maximum distance of origin from}& & \\ & 4+i(2-z)is\dots \dots & & \\ R.& \text{Consider the curve}xy=25!\text{such that}(\alpha ,\beta )\text{is a}& 3.& 2^2\\ & \text{point on the curve},\alpha ,\beta ϵN.\text{Then the number of}& & \\ & \text{distinct ordered pairs}(\alpha ,\beta )\text{such that HCF}& & \\ & (\alpha ,\beta )=1is{2}^{k}thenk=\dots \dots & & \\ S.& If{2}^{(1+cos\pi x)lo{g}_2^5}+2^{x^2-1}+2^{2(1-|x|)}=3,& 4.& 3^2\\ & \text{then sum of the roots is}\dots \dots & & \end{array}$

• A. $\begin{array}{ccccc}& P& Q& R& S\\ \left(a\right)& 4& 3& 2& 1\end{array}$
• B. $\begin{array}{ccccc}& P& Q& R& S\\ \left(a\right)& 2& 3& 1& 4\end{array}$
• C. $\begin{array}{ccccc}& P& Q& R& S\\ \left(a\right)& 2& 3& 4& 1\end{array}$
• D. $\begin{array}{ccccc}& P& Q& R& S\\ \left(a\right)& 3& 2& 4& 1\end{array}$

Answer 1

The correct option is C

$\begin{array}{ccccc}& P& Q& R& S\\ \left(a\right)& 2& 3& 4& 1\end{array}$

P:Point lies on $x^2+y^2=x$
$\therefore$ centre is $(\frac{1}{2},0)$
$a=\frac{1}{2},b=0Q:|4+2i-iz|=|z-2+i+3i|\le 1+3=4R:xy=25!=2^{22}.3^{10}.5^6{7}^3{11}^2{13}^1{17}^1{19}^1{23}^1\therefore 2^9$
$S:1+cos\pi x=0,x^2-1=0$ $|x|=1\Rightarrow x=\pm 1sum=0$