Fax narayananpt- aits iit-jee

NPT-1 (AITS – IIT-JEE 2007)

(a) one–one onto	(b) one–one into
(c) many-one onto
	
(a) [0, ∞)	(b) [2, 1]
	(d) none of these

NPT-1 (AITS – IIT-JEE 2007)

					x	≠
4. f(x) =				0	, where {x} is the fractional part function, [x] is the step up
4. f(x) =					x	=
	−1	x	<		x	=
	0	x	=	0	, then f(x)
		x	>	0	, then f(x)

(a) f(0) = ln2 ⇒ f is continuous at x = 0 (c) f(0) = e2⇒ f is continuous at x = 0

(b) f(0) = 2 ⇒ f is continuous at x = 0
(d) f has an irremovable discontinuity at x = 0

7. The values of x for which the function f(x) = sin–1 sin |x| is increasing can be given by


8. Let	a	2	+	b	2	=	1
8. Let	x	2		y	2		1
(a) a4 + b4 (c) a4b4

Space for rough work

9. Let a (a < 0, a ∉ I) be a fixed constant and t be a parameter then the set of values of t for the function

					x	<	1	(b) [[a], [–a])
								(d) [[a – 1], [–a + 1])
10. Let f(x) =		2(2e	−	e ), x				, then at x = 1,
10. Let f(x) =					x	≥	1	, then at x = 1,

(a) x (|lnx| – (x – 1) + c (b) x |lnx| – x + c

(c) x (|lnx| + (x – 1) + c (d) x + x |lnx| + c

Given the function f(x) such that 2f(x) + xf 1								1		
Given the function f(x) such that 2f(x) + xf 1								4		
	+	xcos	π					4		
2			x

NARAYANA

(b) f(x) = tan (tan-1 x) & g(x) = cot (cot1 x)

(c)

x)dx

2ln

2

(b) differentiable at x = 1
(d) differentiable at x = 3

17. If f is an odd continuous function in [–1, 1] and differentiable in (–1, 1) then (a) f′(a) = f(1) for some a∈(–1, 0)
(b) f′(b) = f(1) for some b∈ (0, 1)
(c) n(f(α))n–1 f′(α) = (f(1))n for some α∈ (–1, 0), n∈N
(d) n(f(β))n–1 f′(β) = (f(1))n for some β∈ (0, 1), n∈N

NPT-1 (AITS – IIT-JEE 2007)

NPT-1 (AITS – IIT-JEE 2007)

NPT-1 (AITS – IIT-JEE 2007)

(c)	The value of a for which the area of the	(R)
	triangle included between the axes and

constant is

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these parameters for which the results given are correct. Match the integrals in List–I with the values of parameters A, B in List–II so that the given result is correct.

(a)		Column I					(P)	Column II				1
(a)		e	2x		−	x 2e dx	(P)		−		=−	1
	∫		e	2x								4