Question

$\begin{array}{cccc}& \text{Column -I}& & \text{Column -II}\\ \left(a\right)& \text{Probability that unit digits in}& \left(p\right)& 2\\ & \text{square of an even integer is 4 is}& & \\ \left(b\right)& \text{No. of integers in the domain of}f\left(x\right)=\sqrt{4-x^2}& \left(q\right)& 1\\ & +\csc (|x|-2)+\frac{1}{x}\text{is}& & \\ \left(c\right)& \text{Five times the minimum distance of}4x^2+y^2+4x-4y+5& \left(r\right)& \frac{2}{5}\\ & \text{=0 from the line}4x+3y=3\text{is}& & \\ \left(d\right)& \text{The remainder when}1!+2!+3!+....100!& \left(s\right)& \frac{2}{3}\\ & \text{is divided by 12 is k, then}\frac{2k}{45}\text{equals to}& & \end{array}$

• A. $\left(a\right)\to \left(p\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(r\right)$
• B. $\left(a\right)\to \left(r\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(q\right);\left(d\right)\to \left(r\right)$
• C. $\left(a\right)\to \left(p\right);\left(b\right)\to \left(q\right);\left(c\right)\to \left(r\right);\left(d\right)\to \left(s\right)$
• D. $\left(a\right)\to \left(s\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(r\right);\left(d\right)\to \left(q\right)$

Answer 1

The correct option is B $\left(a\right)\to \left(r\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(q\right);\left(d\right)\to \left(r\right)$

(a) Even integers ends in 0, 2, 4, 6, 8. Square of an even integer ends in 4 only when the integer ends either in 2 or 8 probability $=\frac{2}{5}$

(b) $x^2-4\le 0\Rightarrow xϵ[-2,2]\Rightarrow x=\pm 2,\pm 1,0$

$\left[x\right]-2\ne 0\Rightarrow x\ne \pm 2$

$\frac{1}{x}$ is undefined at $x=0$

so possible integers are -1, 1

(c) $4x^2+y^2+4x-4y+5=0$

$(2x+1{)}^2+(y-2{)}^2=0$

gives a point $(-\frac{1}{2},2)$ Min, distance of this point from the line is $\frac{1}{5}$.

(d) $1!+2!+3!+4!+\cdots 100!$

$1!+2!+3!+12k$, where k is integer $4!5!6!\cdots$ are all divisible by 12 so the remainder when 1 + 2 + 6 + 12k is divided by 12 is 9

So $\frac{2k}{45}=\frac{18}{45}\Rightarrow \frac{2}{5}$