`{:(,"ColumnI",,"ColumnII"),((A),(K.E.)/(P.E.),(P),2),((B),P.E+2K.E.,(Q),-(1)/(2)),((C),(P.E.)/(T.E.),(R),1),((D),(K.E.)/(T.E.),(S),0):}`

Let a_(1),a_(2),a_(3), …, a_(10) be in G.P. with a_(i) gt 0 for i=1, 2, …, 10 and S be te set of pairs (r, k), r, k in N (the set of natural numbers) for which |(log_(e)a_(1)^(r)a_(2)^(k),log_(e)a_(2)^(r)a_(3)^(k),log_(e)a_(3)^(r)a_(4)^(k)),(log_(e)a_(4)^(r)a_(5)^(k),log_(e)a_(5)^(r)a_(6)^(k),log_(e)a_(6)^(r)a_(7)^(k)),(log_(e)a_(7)^(r)a_(8)^(k),log_(e)a_(8)^(r)a_(9)^(k),log_(e)a_(9)^(r)a_(10)^(k))| = 0. Then the number of elements in S is

If x(t) is a solution of ((1+t)dy)/(dx)-t y=1 and y(0)=-1 then y(1) is (a) ( b ) (c)-( d )1/( e )2( f ) (g) (h) (i) (b) ( j ) (k) e+( l )1/( m )2( n ) (o) (p) (q) (c) ( d ) (e) e-( f )1/( g )2( h ) (i) (j) (k) (d) ( l ) (m) (n)1/( o )2( p ) (q) (r) (s)

A sample space consists of 9 elementary outcomes E_(1),E_(2),…..,E_(9) whose probabilities are P(E_(1))=P(E_(2))=0.08,P(E_(3))=P(E_(4))=P(E_(5))=0.1 P(E_(6))=P(E_(1))=0.2,P(E_(8))=P(E_(9))=0.07 "Suppose"" "A={E_(1),E_(5),E_(8)},B={E_(2),E_(5),E_(8),E_(9)} (i) Calculate P(A), P(B) and P(AcapB) . (ii) Using the addition law of probability, calculate P (AcupB). (iii) List the composition of the event AcupB and calculate P(AcupB) by adding the probabilities of the elementary outcomes. Calculate P(overset(-)B) from P(B), also calculate P(overset(-)B) directly from the elementary outcomes of overset(-)B ,

In Fig.97, If A B is parallel to CD, then the value of /_B P E is (a) ( b ) (c) (d) (e) 106^(( f )0( g ))( h ) (i) (j) (b) ( k ) (l) (m) (n) 76^(( o )0( p ))( q ) (r) (s) (c) ( d ) (e) (f) (g) 74^(( h )0( i ))( j ) (k) (l) (d) ( m ) (n) (o) (p) 84^(( q )0( r ))( s ) (t) (u)

In Fig.88 P R is a straight line and /_P Q S :/_S Q R=7: 5. The measure of /_S Q R is (a) ( b ) (c) (d) (e) 60^(( f )0( g ))( h ) (i) (j) (b) ( k ) (l) 62 (m) (n) (o)1/( p )2( q ) (r)^(( s )0( t ))( u ) (v) (w) (c) ( d ) (e) 67 (f) (g) (h)1/( i )2( j ) (k)^(( l )0( m ))( n ) (o) (p) (d) ( q ) (r) (s) (t) 75^(( u )0( v ))( w ) (x) (y)

The value of int_0^1lim_(n->oo)sum_(k =0)^n(x^(k+2)2^k)/(k!)dx is: (a) e ^(2) -1 (b) 2 (c) (e ^(2)-1)/(2) (d) (e ^(2) -1)/(4)

For two events E_(1) and E_(2), P(E_(1))=1/2,P(E_(2))=1/3 and P(E_(1)nnE_(2))=1/10 . Find: (i) P(E_(1) "or" E_(2)) (ii) P(E_(1) "but not" E_(2)) (iii) P(E_(2)"but not" E_(1))

Solution of (dy)/(dx)+2x y=y is (a) ( b ) (c) y=c (d) e^( e ) (f) x-( g ) x^((( h )2( i ))( j ) (k))( l ) (m) (n) (b) ( o ) (p) y=c( q ) e^( r ) (s) (t) x^((( u )2( v ))( w )-x (x))( y ) (z) (aa) (c) ( d ) (e) y=c( f ) e^(( g ) x (h))( i ) (j) (k) (d) ( l ) (m) y=c (n) e^( o ) (p)-( q ) x^((( r )2( s ))( t ) (u))( v ) (w) (x)

Let E and F be two independent events. The probability that exactly one of them occurs is 11/25 and the probability if none of them occurring is 2/25 . If P(T) denotes the probability of occurrence of the event T , then (a) P(E)=4/5,P(F)=3/5 (b) P(E)=1/5,P(F)=2/5 (c) P(E)=2/5,P(F)=1/5 (d) P(E)=3/5,P(F)=4/5

The sum of series 1/2!+1/4!+16!+………. is (A) (e^2-1)/2 (B) (e^2-2)/e (C) (e^2-1)/(2e) (D) )(e-1)^2)/(2e)