Column−I Column−II(P)The shortest distance between(1)6 origin and the curve is x2+y2+xy=60 is(Q)The value of ∫2011−2011dx1+x9+√1+x18−2011 is(2)0(R)On[0,2]the maximum value of(3)3 f(x)=max{x,x−1,3x}is (S)Letf:R→R be given by(4)√40 f(x)={|x−[x]|when[x] is odd |x−[x]−1when[x] is even Thenthevalueof∫4−2 f(x)dx−3 is

Column−I Column−II(P)The shortest distance between(1)6 origin and

1.	Column−I Column−II(P)The shortest distance between(1)6 origin and the curve is x2+y2+xy=60 is(Q)The value of ∫2011−2011dx1+x9+√1+x18−2011 is(2)0(R)On[0,2]the maximum value of(3)3 f(x)=max{x,x−1,3x}is (S)Letf:R→R be given by(4)√40 f(x)={\|x−[x]\|when[x] is odd \|x−[x]−1when[x] is even Thenthevalueof∫4−2 f(x)dx−3 is
Answer» $\begin{matrix} C o l u m n - I & C o l u m n - I I (P) & T h e s h o r t e s t d i s t a n c e b e t w e e n & (1) & 6 o r i g i n a n d t h e c u r v e i s x^{2} + y^{2} + x y = 60 i s (Q) & T h e v a l u e o f \int_{- 2011}^{2011} \frac{d x}{1 + x^{9} + \sqrt{1 + x^{18}}} - 2011 i s & (2) & 0 (R) & O n [0, 2] t h e m a x i m u m v a l u e o f & (3) & 3 f (x) = m a x {x^{,} x - 1, 3 x} i s (S) & L e t f : R \to R b e g i v e n b y & (4) & \sqrt{40} f (x) = {\begin{matrix} \| x - [x] \| & w h e n [x] i s o d d \| x - [x] - 1 & w h e n [x] i s e v e n \end{matrix} T h e n t h e v a l u e o f \int_{- 2}^{4} f (x) d x - 3 i s \end{matrix}$