[2011][2010][2010]- Suppose log logab ba c .The smallest
possible integer value of c for all a b, 1 is
[2011]
(a) 4 (b) 3 (c) 2 (d)1- The number of roots of the equation
(^) cos sin 1^7 ^4 that lie in the interval
{0,2 } is [2010]
(a) 2 (b) 3 (c) 4 (d) 8
- Suppose m, n are positive integers such that
6 2 3 2 332m m n w n. The value of the
expression m mn n^2 ^2 is [2010]
(a) 7 (b) 13 (c) 19 (d) 21Sequence and Series- c 2. a 3. b 4. c 5. c
- a
Vectors- c 2. c 3. b 4. c 5. a
- c 7. a 8. d 9. d
Statistics- b 2. a 3. d
Miscellaneous- a 2. b 3. c 4. b 5. d
- c 7. c 8. c 9. d 10. c
- a 12. c
Sequence and SeriesVectorsStatisticsMiscellaneousSequence and Series
1.Sol: Method 1:
The problem is much simple than it looks
For simplicity, let f x( ) denote, for possible
real numbers x, the function
sin x 1 sin ( )x.
It is fairly easy to that asx tends to infinity,,f (x) tends to 0 by using mean value theorem.
(Not monotonically, it oscillates; but it tends to
0).
This implies that for each value of t 0 , there
is some value of x, say x 1 , such that if x x 1 ,
then f x t( ).
Therefore for each value of t 0 , there are
only finitely many positive integers n such that
f n t( ).
Therefore, in the language of the problem, for
each value of t 0 , the complement of Atis a
finite set.
Method 2:have infinite elements.
Method 3: Using mean value theorem, there
exist a number N in the intervalsin( 1) sin( ) 2cos^1 sin^1
2 2n n n n n n
1
2 sin 2
1n
n
1
n 1 n
Note that for^0 ,sin
2x x x
for all n 0 ,We have sin n 1 sin n is negligibilysmall.
The given expression is comparable to it, so
we get 0 for all n 0. So there exists
infinite natural numbers, which satisfies the
given condition. Therefore for all T,An N n 1 ,we haveSequence and Series
1.Sol: Method 1:
Method 2:Method 3:sin( 1) sin( )
'( ) 0
1n n
f N
n n
The given expression is comparable to it, so
we get 0 for all n 0. So there exits
infinite natural numbers, which satisfies the
given condition. Therefore for all ,A
have infinite elements.