Column -IColumn -II(a)Probability that unit digits in(p)2square of an even integer is 4 is(b)No. of integers in the domain of f(x)=√4−x2(q)1+csc(|x|−2)+1x is(c)Five times the minimum distance of 4x2+y2+4x−4y+5(r)25=0 from the line 4x+3y=3 is(d)The remainder when 1!+2!+3!+....100!(s)23is divided by 12 is k, then 2k45 equals to

Column -IColumn -II(a)Probability that unit digits in(p)2square of an

1.	Column -IColumn -II(a)Probability that unit digits in(p)2square of an even integer is 4 is(b)No. of integers in the domain of f(x)=√4−x2(q)1+csc(\|x\|−2)+1x is(c)Five times the minimum distance of 4x2+y2+4x−4y+5(r)25=0 from the line 4x+3y=3 is(d)The remainder when 1!+2!+3!+....100!(s)23is divided by 12 is k, then 2k45 equals to
Answer» $\begin{matrix} Column -I & Column -II (a) & Probability that unit digits in & (p) & 2 square of an even integer is 4 is (b) & No. of integers in the domain of f (x) = \sqrt{4 - x^{2}} & (q) & 1 + csc (\| x \| - 2) + \frac{1}{x} is (c) & Five times the minimum distance of 4 x^{2} + y^{2} + 4 x - 4 y + 5 & (r) & \frac{2}{5} =0 from the line 4 x + 3 y = 3 is (d) & The remainder when 1! + 2! + 3! + . . . .100! & (s) & \frac{2}{3} is divided by 12 is k, then \frac{2 k}{45} equals to \end{matrix}$