Question

Matching type questions:

$\begin{array}{cccc}& Column-I& & Column-II\\ \left(P\right)& No.ofpointsofintersectionofcurves|y|=ln|x|and(x-1{)}^2+y^2-4=0& \left(1\right)& 1\\ \left(Q\right)& S_n={\sum }_{r=1}^n(n>6)then{S}_n–7\left[\frac{S_n}{7}\right]is\left(where\right[x\left]denotesintegerlessthanorequaltox\right)& \left(2\right)& 2\\ \left(R\right)& Equationoftangentat{z}_{0}tothecircle|z|=risRe\left(\frac{z}{z_0}\right)=\lambda ,thenlis& \left(3\right)& 3\\ \left(S\right)& Normaldrawnatanypointofanellipse\frac{x^2}{25}+\frac{y^2}{9},istangenttothe& \left(4\right)& 5\\ & circle{x}^2+y^2=r^{2}thenmaximumvalueofris& & \\ & & \left(5\right)& 5\end{array}$

• A. P - 3, Q-1, R-2, S-4
• B. P - 3, Q-4, R-1, S-2
• C. P - 3, Q-4, R-2, S-1
• D. P - 3, Q-4, R-2, S-4

Answer 1

The correct option is B

P - 3, Q-4, R-1, S-2

From figure points of intersection is 3

(B) $S_n=1+2+6+24+120+720+7\lambda =873+7\lambda =7k+5\left[\frac{S_n}{7}\right]=K\Rightarrow S_n-7\left[\frac{S_n}{7}\right]=5$
(C) $z\overline{z_0}+z_0\overline{z}=2r^{2}or2rE\left(\frac{\overline{z_0}z}{r^2}\right)=2or,Re\left(\frac{z}{z_0}\right)=1\therefore \lambda =1$
(D) Equation of normal $\frac{5x}{cos\theta }-\frac{3y}{sin\theta }-16=0Fortangenttocircler=\frac{16}{\sqrt{\frac{25}{co{s}^{\theta }}+\frac{9}{si{n}^2\theta }}}\sqrt{\frac{25}{co{s}^{\theta }}+\frac{9}{si{n}^2\theta }}\ge 8r^{max}=2$