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6651.

If bara=2bari+3barj-bark,barb=-bari+2barj-4bark, barc=bari+barj+bark then abs((baraxxbarb)· (barbxxbarc))=

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104
106
108
112

Answer :C
6652.

Assume that P ( A ) = P ( B ). Show that A = B

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ANSWER :`A= B`
6653.

cot((pi)/4-2cot^(-1)3)=

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`3`
5
7
9

Answer :C
6654.

The vertices of Delta PQR " are " P (2, 1) , Q (-2, 3) " and " R (4, 5). Find equation of the median through the vertex R.

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ANSWER :`3X - 4Y + 8 = 0 `
6655.

A line AB of length '2l' moves with the end 'A' always on x - axis and the end 'B' on the line y = 6x find the equation of locus of middle point of AB

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`9X^(2)+10Y^(2)-6xy-9l^(2)=0`
`9x^(2)-10y^(2)+6xy+9l^(2)=0`
`9x^(2)+10y^(2)+6xy+9l^(2)=0`
`9x^(2)-10y^(2)-4xy+6l^(2)=0`

ANSWER :A
6656.

Evaluate the following limit : Lim_(x to pi/2) (1+cos 2x)/(pi-2x)^(2)

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ANSWER :`1/2 `
6657.

Solve :tan 3 theta = 1 .

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ANSWER :`THETA = 1/2 NPI, N in I`
6658.

If a le 3 cos x+ 5 sin ( x - pi//6) le b for all x , then ( a, b) is

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`(-SQRT(19), sqrt(19))`
(-17, 17)
`(-sqrt(21), sqrt(21))`
NONE of these

Answer :A
6659.

If sin(sin x + cos x) = cos (cos x - sin x) then the largest possible value of sin x is

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`(1)/(sqrt(2))`
1
`sqrt(16 - PI^(2))/(4)`
`(pi)/(4)`

Answer :C
6660.

A straight line through the origin O meets the parallel lines 4x+2y=9 and 2x+y+6=0 at points P and Q respectively. Then the point O divides the segment PQ in the ratio

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`1 : 2`
`3 : 4`
`2 : 1`
`4 : 3`

ANSWER :B
6661.

If the coefficients of 4^(th) and 13^(th) terms in the expansion of (a+b)^(n) are equal then find n .

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ANSWER :15
6662.

Find sum_(n=1)^(oo)(1)/(n^(2)+5n+6)

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ANSWER :`=(1)/(3)`
6663.

A tree is broken by wind, its upper part touches the ground at a point 10 metres from the foot of the tree and makes an angle of45^(@)with the ground . The entire length of the tree is

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15 metres
20 metres
` 10 ( 1+ SQRT2) ` metres
` 10 ( 1+ (sqrt3)/(2))` metres

ANSWER :C
6664.

Find the mean deviation about the mean for the following data:12,3,18,17,4,9,17,19,20,15,8,17,2,3,16,11,3,1,0,5 .

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ANSWER :6.2
6665.

Find r if""^(5)P_(r)=""^(6)P_(r-1)

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ANSWER :4
6666.

There are 8 white and 7 black balls in a bag . Three-three balls are drawn twice from the bag and balls drawn first time are not being replaced when the balls are drawn second time. Find the probability that first three balls drawn are white and second thre balls drawn are black.

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SOLUTION :N/a
6667.

If sin x sin h y = cos theta and cos x cos h y = sin theta then cos h^(2) y + cos^(2)x =

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`-1`
0
1
2

Answer :D
6668.

Differentiate the following function w.r.t. x: x^(2) cos 5x

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ANSWER :`-5 X^(2) SIN 5X + 2 x cos 5x`
6669.

Ifalpha + beta + gamma = 2 theta , " then " cos theta+ cos ( theta - alpha ) + cos (theta - beta ) + cos ( theta - gamma ) =

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`4SIN ""(alpha )/(2).cos""(beta)/(2).SIN""(gamma)/(2)`
`4 cos ""(alpha)/(2).cos""(beta)/(2).cos""(gamma)/(2)`
`4 sin ""alpha /2.sin""beta/2.sin""gamma2`
`3 sin alpha.sinbeta.singamma`

ANSWER :B
6670.

Write down the negation of following compound statements. |x| is equal to eighter x or -x.

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ANSWER :not EQUAL to -X
6671.

If d/(dx) (f(x))^(n)=n(f(x))^(n-1)(df(x))/(dx) then (d)/(dx) (sin^3x)=3sin^(2)x.cosx.

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ANSWER :TRUE STATEMENT
6672.

Given that, z=(1+2i)/(1-i)=((1+2i)(1+i))/((1-i)(1+i))

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ANSWER :`((-1)/(2),(-3)/(2))`
6673.

If f(x) = cos x , a = ( - pi )/(2) then the value of 'c' of Rolle's theorem in (a,b)

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POSITIVE
Negative
Neither positive Nor Negative
Irrational

Answer :C
6674.

If sin^(3) theta cos theta - cos ^(3) theta sin theta = 1 //4 " then " theta =

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`THETA = (n PI)/( 2) + (pi)/(8) n in Z`
`theta = ( n pi)/(2) - (pi)/(8) n in Z`
`theta = (n pi)/(2) + (pi)/(4) n in Z`
`theta = (n pi)/(2) - (pi)/(4) n in Z`

Answer :B
6675.

If the area enclosed by the curves f(x)=cos^(-1)(cosx) and g(x)=sin^(-1)(cosx) in x in [(9pi)/4,(15pi)/4] is api^(2),//b (where a and b are coprime), then the value of (a-b) is

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ANSWER :1
6676.

A line passing through the points (7,2) , (-3,2) then the image of the line in the x-axis is

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y=4
y=9
y=-1
y=-2

Answer :D
6677.

Find the sum of odd integers from 1 to 2001.

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ANSWER :1002001
6678.

Find the orthocentre of the triangle formed by the lines 2x-y=2,3x-2y=1,x-y=0

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ANSWER :`(11,-4)`
6679.

If |{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=5, then the value of |{:(b_(2)c_(3)-b_(3)c_(2),a_(3)c_(2)-a_(2)c_(3),a_(2)b_(3)-a_(3)b_(2)),(b_(3)c_(1)-b_(1)c_(3),a_(1)c_(3)-a_(3)c_(1),a_(3)b_(1)-a_(1)b_(3)),(b_(1)c_(2)-b_(2)c_(1),a_(2)c_(1)-a_(1)c_(2),a_(1)b_(2)-a_(2)b_(1)):}| is

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5
25
125
0

Answer :B
6680.

cos theta + cos^(2) theta = 1, cos theta + cos^(2) theta = 1 + d sin^(6) theta = 1 rArr (b+c)/(a+d)=1

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2
3
4
6

Answer :B
6681.

If D is the midpoint of the side BC of a triangle ABC then bar(AB)^(2)+bar(AC)^(2)=

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`bar(AD)^(2)+bar(BD)^(2)`
`2(bar(AD)^(2)+bar(BD)^(2))`
`3(bar(AD)^(2)+bar(BD)^(2))`
`4(bar(AD)^(2)+bar(BD)^(2))`

Answer :B
6682.

The value of alpha^(3)/2cosec^(2)(1/2tan^(-1)""alpha/beta)+beta^(3)/2sec^(2)(1/2tan^(-1)(beta/alpha)) is equal to

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`(ALPHA-BETA)(alpha^(2)+beta^(2))`
`(alpha-beta)(alpha^(2)+beta^(2))`
`(alpha+beta)(alpha^(2)+beta^(2))`
`(alpha-beta)(alpha^(2)-beta^(2))`

ANSWER :C
6683.

Find the area of the parallelogram whose sides are 3x+4y+5=0, 3x+4y-2=0,2x+3y+1=0, 2x+3y-7=0

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ANSWER :56 SQ. UNITS
6684.

The number of values of x for which |||x^(2) -x+4| -2| =3|=x^(2)+x- 12is …………………

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ANSWER :1
6685.

If cosA=(3)/(4) then 32sin((A)/(2))sin((5A)/(2)) = ……………

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`-11`
`-SQRT(11)`
`sqrt(11)`
11

Answer :D
6686.

If n (A) =10, n (B) =6 and n (C ) =5 for three disjoint sets A,B,C, then n (A uu B uuC)equals

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11
21
1
9

Answer :B
6687.

Compute the missing frequencies in the following distribution , given that is sum f_(i)= 100 and the median is 32{:("Marks",0-10,10-20,20-30,30-40,40-50,50-60),("Number of students "," "9," "?, " "26, " "30, " " ?, " "10):}

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ANSWER :`9 . 16`
6688.

Consider the lines L _(1) : (x +1)/( 3) = (y+2)/(1) = (z+1)/(2), L_(2): (x-2)/(1) = (y+2)/(2) = (z-3)/(3) The shortest distance between L _(1) and L _(2) is

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0
`(17)/(sqrt3)`
`(41)/(5SQRT3)`
`(17)/(5sqrt3)`

Answer :D
6689.

If 25, x - 6 and x - 12 are in G.Pthen x = .......... or ......... .

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ANSWER :21
6690.

A box without lid having maximum volume is made out of square metel sheet of edge 60 cms by cutting equal square pieces from the four corners and turing up the projecting pieces tomake the sides of the box . The height of the box is

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60
10
15
12

Answer :B
6691.

Find equation of the line perpendicular to the linex - 7y + 5 = 0 and having x intercept 3.

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ANSWER :`y + 7X = 21`
6692.

Solve the equation sin^(4)x + cos^(4)x =7/2 sinx.cosx.

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Solution :`SIN^(4)x+COS^(4)x=7/2sinx.cosx rArr (sin^(2)x +cos^(2)x)^(2) =7/2sinx.cosx`
`rArr 1-1/2(SIN2X)^(2)=7/4(sin2x) rArr 2SIN^(2)2x+7sin2x-4=0`
`rArr (2sin2x-1)(sin2x+4) =0 rArr sin2x =1/2` or `sin2x = -4` (Which is not possible)`rArr 2x = npi+(-1)^(n)pi/6 , n in I`
i.e, `x=(npi)/(2)+(-1)^(n)pi/12, n in I` ANS.
6693.

The difference betweentworootsof theequation x^3 -13 x^2 + 15 x+189=0is2thentherootsof theequation are

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2
1
` SQRT(2)`
`2 sqrt(2)`

ANSWER :A
6694.

A and B are events such that P(A)= 0.42, P(B)= 0.48 and P(A and B)= 0.16, Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B).

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ANSWER :(i) 0.58, (II) 0.52, (III) 0.74
6695.

Perpendicular distance from origin to the plane barr.barm = bara.barm is

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`ABS(bara.barm)/absbarm`
`abs(bara.barm)/ABSBARA`
`(bara.barm)/absbara`
`(bara.barm)/absbarm`

ANSWER :A
6696.

Let f : (-1, 1) to R be a differentiable function with f(0) = -1 and f'(0) = 1 . Let g(x) = [f(2f(x)+ 2)]^2 . Then g'(0) is equal to

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`-2`
`4`
`-4`
`0`

ANSWER :B
6697.

If f(x) = sqrt(x + 2sqrt(2x - 4)) + sqrt(x - 2sqrt(2x - 4)), then f'(3) = f'(6) =

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0
`1`
`2`
`1/2`

ANSWER :D
6698.

A single letter is selected at random from the word 'Probability' . Find the probability that it is a vowel.

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ANSWER :`=(4)/(11)`.
6699.

Sum of the power of a and b in expansion (a+b)^n is n .

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ANSWER :TRUE STATEMENT
6700.

Define loss of kinetic energy in inelastic collision

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Solution :In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like SOUND, thermal, heat, light etc. LET `KE_i` be the TOTAL kinetic energy before collision and `KE_f` be the total kinetic energy after collision.
Total kinetic energy before collision,
`KE_i = 1/2 m_1u_1^2 + m_2 u_2^2 "" ...(1)`
Total kinetic energy after collision,
`KE_f = 1/2 (m_1 + m_2) v^2 "" ...(2)`
Then the loss of kinetic energy is
Loss of KE ,` Delta Q = KE_f - KE_i`
` =1/2 (m_1 + m_2) v^2 - 1/2 m_1 u_1^2 + 1/2 m_2 u_2^2 "" ...(3)`
SUBSTITUTING equation
` v = (m_1u_1 + m_2 u_2)/(m_1 + m_2)`
in equation (3), and on simplifying (expand v by using the algebra `(a + b)^2 =a ^2 + b^2 + 2ab` we get
Loss of KE , ` DeltaQ = 1/2 ((m_1 m_2)/(m_1 + m_2)) (u_1 - u_2)^2`