Match the column : Column I Column IIa)∫−11[x+[x+[x]]]dx(p)3b)2∫25(|x|+[−x])dx(q)1/5c)∫−13sgn(x−[x])dx(r)4d)∫0π/4(tan6(x−[x])+tan4(x−[x]))dx(s)-3

∫−11[x+[x+[x]]]dx=3∫−11[x]dx = -3

Match the column : Column I Column IIa)∫−11[x+[x+[x]]]dx(p)3b)2∫25(|x|+[−x])dx(q)1/5c)∫−13sgn(x−[x])dx(r)4d)∫0π/4(tan6(x−[x])+tan4(x−[x]))dx(s)-3

∫−11[x+[x+[x]]]dx=3∫−11[x]dx = -3

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Match the column :
Column I Column II
a) $\int_{- 1}^{1} [x + [x + [x]]] d x$ (p) 3
b) $2 \int_{2}^{5} (| x | + [- x]) d x$ (q) 1/5
c) $\int_{- 1}^{3} sgn (x - [x]) d x$ (r) 4
d) $\int_{0}^{π / 4} (\tan^{6} (x - [x]) + \tan^{4} (x - [x])) d x$ (s) -3

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a→r, b→p, c→s, d→q

a→p, b→r, c→q, d→s

a→s, b→p, c→r, d→q

a→s, b→s, c→ r, d→q

answer is D.

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$\int_{- 1}^{1} [x + [x + [x]]] d x = 3 \int_{- 1}^{1} [x] d x = - 3$

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	Column I		Column II
a)	$\int_{- 1}^{1} [x + [x + [x]]] d x$	(p)	3
b)	$2 \int_{2}^{5} (\| x \| + [- x]) d x$	(q)	1/5
c)	$\int_{- 1}^{3} sgn (x - [x]) d x$	(r)	4
d)	$\int_{0}^{π / 4} (\tan^{6} (x - [x]) + \tan^{4} (x - [x])) d x$	(s)	-3

Match the column : Column I Column IIa)∫−11[x+[x+[x]]]dx(p)3b)2∫25(|x|+[−x])dx(q)1/5c)∫−13sgn(x−[x])dx(r)4d)∫0π/4(tan6(x−[x])+tan4(x−[x]))dx(s)-3

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Match the column :
Column I Column II
a) $\int_{- 1}^{1} [x + [x + [x]]] d x$ (p) 3
b) $2 \int_{2}^{5} (| x | + [- x]) d x$ (q) 1/5
c) $\int_{- 1}^{3} sgn (x - [x]) d x$ (r) 4
d) $\int_{0}^{π / 4} (\tan^{6} (x - [x]) + \tan^{4} (x - [x])) d x$ (s) -3