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 (2x + 3) /  (x2 + 4x +3) dx
=  (2x + 4) /  (x2 + 4x +3) dx - dx / (x2 + 4x +3)
= d (x2 + 4x + 3 ) / (x2 + 4x +3) - dx / [ (x+2) 2 - 1 ] { since, (2x+4) dx = d (x2 + 4x + 3) }
= 2 (x2 + 4x + 3) - log [ (x+2) + [ (x+2) 2 - 1 ] ] + C
Cheers !!!
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The parametric equations for the parabola, y = x2 are x = t and y = t2
Therefore, let the coordinates of A be (t1, t12 ) and the coordinates of B be (t2,t22 )
Bold letters represent vectors.
Let, i and j be the unit vectors along OX and OY respectively.
Therefore,
OA = t1 i + t12 j
OB = t2 i + t22 j
Now, OA . i = 1  t1 = 1 ................... eqn. (1)
OB . i = - 2 t2 = - 2 ................ eqn (2)
So, OA = i + j
OB = - 2 i + 4 j
Hence, 2OA - 3OB = 8 i - 10 j
So, l 2OA - 3OB l = (82 + 102) = 164 units.
Ans : 164 units
Cheers !!!
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Let the time after which the helicopter switches off the engine after the start be t sec.
The distance traversed by the helicopter in the vertically upward direction
= speed of sound x time taken by sound to reach the ground once the helicopter switches off the engine
= 320 x (30 - t) m
Since the initial velocity of helicopter = 0 and the acceleration of the helicopter is 3m/s2, therefore,
1/2 . 3 t2 = 320 x (30 - t)
 (3t - 80)(t + 240) = 0
t = 80 / 3
Therefore, the velocity of the helicopter at the moment engine is switched off
= 3 x 80 / 3 m /s = 80 m /s
Ans: 80 m/s
Cheers !!!
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No: 1
Let the formula of the hydrated salt be Na2SO4 . n H2O
So, if 1 mole of the hydrated salt is heated the loss in weight is due to the n moles of water i.e 18 n gms.
Wt. of 1 mole of the hydrated salt = (23 x 2 + 32 + 16 x 4 + n x18) gms = (142+ 18n) gms
Therefore, 18n / (142+18n) x 100 = 55
i.e 18 n = 173.56 i.e n = 9.6 (approx)  10
Ans : n = 10
No: 2
1 atom weighs 6.64 x 10 - 23 gm
Therefore, 6.023 x 10 23 atoms weigh 6.64 x 6.023 gm
Hence, 1 gm atom of the element weighs 6.64 x 6.023 gm
Therefore, 40 Kg of the element has 40 x 1000 / (6.64 x 6.023 ) gm atoms i.e 1000.18 gm atoms
Ans: 1000.18 gm atoms
Cheers !!!!!
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Take, log x = t ; Therefore, x = et ; dx / x = dt i.e dx = et dt
(1 / log x) dx = e t / t dt
Now, e t can be expanded as the sum of an infinite series, given as below:
e t = 1 + t + t2 / 2 ! + t3 / 3 ! + t4 / 4 ! + ................................ 
Therefore, (1+t+t2/2 ! + t3/3 ! + .......) / t . dt
= dt / t + dt + t dt / 2 ! + t2 dt / 3 ! + t3 dt / 4 ! + ....................
= ln l t l + t + t2/ 2. 2 ! + t 3 / 3 . 3 ! + t4 / 4 . 4 ! + ........................
= ln l lnx l + ln x + (ln x)2/2.2 ! + (ln x)3 / 3 . 3 ! + ..............
= ln l ln x l + [r = 1 ] [r = ] ( ln x ) r / [ r . r ! ]
Cheers !!!!!
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Take 2x = y
Therefore,
22x + 2x . 22 - 4. 23 = 0 reduces to
y2 + 4y - 32 = 0
y2 + 8y - 4y - 32 = 0
(y+8)(y-4) = 0
y = 2x > 0
So, y = 8 is inadmissible.
Therefore y = 4 i.e 2x = 4 x = 2
Ans: x = 2
Cheers !!!
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4n  =  i.e 2n = / 2
Therefore, (2n - 1) + = / 2
Hence, tan (2n - 1) = cot i.e tan (2n -1) . tan = 1
Similarly, tan (2n -2) = cot 2 i.e tan (2n-2) . tan 2 = 1 and so on.....................
Now, n = / 4 ; tan n = 1
tantan2tan3tan4...........tan[2n-2] tan[2n-1]
= (tan . tan (2n-1) ) . (tan 2 . tan (2n - 2) ) . (tan3 . tan (2n -3)) .............. tan n
= 1x1x1............................................1 = 1
Ans: 1
Cheers !!!!!
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For, my convenience, I'll be writing the angles in degrees only.
Let, P = sin18cos36
Therefore, 2Pcos18 = 2cos18 . sin18 cos36 = sin36cos36 [ since, 2sin18cos18 = sin36 ]
i.e 2x2P cos18 = 2 sin36cos36 = sin72 = cos 18
i.e 4P cos 18 = cos 18
i.e 4P = 1 i.e P = 1/4
Therefore, sin18 cos36 = 1/4
Now, 2 (cos36 - sin18) / sin18cos36
= 2 (sin54 - sin18) / (1/4)
= 8 . 2 sin18cos36 [ since, sin54 - sin18 = 2cos(54/2+18/2) sin(54/2 - 18/2) = 2cos36 sin18 ]
= 8x2x1/4 = 4
Ans: 4
Cheers !!!
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xlog(x+1) dx
= ln(x+1) x dx - x2/2 . 1/(x+1) dx [ integrating by parts ]
= x2 / [2.ln(x+1) ] - 1/2 [ (x2+x - (x+1) + 1] / (x+1) dx
= x2 / [2.ln(x+1) ] - 1/2 x dx + 1/2 dx - 1/(x+1) dx
=x2 / [2.ln(x+1) ] - x2 / 4 + x/2 - ln (x+1) + C
Cheers !!
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Check out the following link.
http://www.mitpune.com/out_look/outlook.htm
The above has all details regarding placements, faculties, salary, etc.
Prefer BIT Mesra to Thapar simply because of the 'Birla' Tag.
Cheers !!!
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The following arrangements will appear before COCHIN
CC - - - -
CH - - - -
CI - - - -
CN - - - -
In each case, the last four letters can be filled up in 4 ! ways ( as the last 4 letters in each of the case are distinct )
So, there are 4 ! x 4 words i.e 24 x 4 or 96 words before COCHIN.
Now, with CO as the two starting letters COCHIN is the first word as CHIN are already arranged alphabetically according to the dictionary.
So, in total 96 words will appear in dictionary before COCHIN
Ans : 96 words will appear in dictionary before COCHIN
Cheers !!!
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Q:1
w is a cube root of unity.
(1+w2) m = (1+w) m
[ (1+w2) / (1+w) ] m = 1
[ (1+w2)(1+w) / (1+w)2 ] m = 1
[ (1+w+w2 + w3 ) / (1+w)2 ] m = 1
1 / (1+w) 2m = 1 [ since, (1+w+w2) = 0 and w 3 = 1 ]
(1+w) 2m = 1
(-w2) 2m = 1 [ since, 1 + w = - w2 ]
(w 3)m . w m = 1
w m = 1
Therefore, the least positive integral value of m is 3
Ans: m = 3
Q:2
When three complex numbers z1, z2, z3 lie in A.P. i.e 2z2 = z1 + z3 ,
then z1, z2, z3 lie in a straight line with z2 being the midpoint of the two points z1 & z3.
Cheers !!!!!
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The gravitational force on the rocket will be zero when the gravitational force on the rocket due to the sun and that due to the earh are equal in magnitude.
Let mass of the earth be me , mass of the sun be ms , mass of the rocket be mr
Radius of the orbit = R
Let the distance of the rocket from the sun when it does not experience any gravitational force be r.
By the problem,
Gmrms / r2 = Gmrme / (R-r)2
i.e (R/r - 1)2 = m e / m s
i.e R/r - 1 = (m e / m s)
i.e r = R ms / (me + ms)
Put the values of R, me , ms to calculate r
The following diagram is attached for better explanation.
Cheers !!
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For the preparation of - hydroxy carboxylic acids, HCN addition comes into play. Here the nucleophile is CN - . As HCN is unstable it is prepared in the reaction stage from the mixture of NaCN & H2SO4. Now acid hydrolysis of CN converts it to carboxylic group (-COOH)
The following will illustrate best :
CH3 - CHO + NaCN + H2SO4  CH3 - CH (OH) - CN
(ethanal)
CH3 - CH (OH) - CN + H2O + H+ CH3 - CH (OH) - COOH
( 2-hydroxy propanoic acid / lactic acid )
Cheers !!
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x / (x+5) - 11x / (11x-8) + 7 / (6 - 4x) = 0
x(11x-8)(6 - 4x) - 11x (x+5)(6-4x) + 7(x+5)(11x-8) = 0
47x2 - 7x - 40 = 0 [ On simplification ]
47x2 - 47x + 40x - 40 = 0
(47x + 40)(x - 1) = 0
Therefore, x = 1, - 40 / 47
Cheers !!!
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From the conditions of the given problem,
p = q + 25 ......... (1)
10 + 500 / p = 500 / q
i.e 10 = 500 x (p-q) / pq
i.e pq = 500 x 25 / 10 = 50 x 25 .................. (2)
Therefore, q (q+25) = 50x25
i.e q2 + 25 q - 50x25 = 0
i.e q2 + 50q - 25q - 50x25 = 0
i.e (q+50)(q - 25) = 0
Therefore, q = 25, so, p = 50
Hence, p = 2q
Correct option is (b)
Cheers !!!
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This is a special class of problems. I have given the solution in details. Try to follow each of the steps.
Let 2 / 7 = x
sin 2/7 + sin 4/7 + sin 8/7
= sinx + sin2x + sin4x ................(1)
Let P = (sinx + sin2x + sin4x)2 + (cosx + cos2x + cos4x)2
= 3 + 2 [ (sinxsin2x + cosxcos2x) + (sin2xsin4x + cos2xcos4x) + (sin4xsinx+cos4xcosx) ]
[ expanding, then arranging and grouping the terms ]
= 3 + 2 (cosx + cos2x + cos3x)
= 3 + 2( cosx + cos2x + cos4x)
[ since 7x = 2 , 3x = 2 - 4x, cos3x = cos(2 - 4x) = cos4x ]
Now, let us calculate the value of cosx + cos2x + cos4x
Let Q = cosx + cos 2x + cos4x
Therefore, 2Q sinx = 2cosxsinx + 2cos2xsinx + 2cos4xsinx
= sin2x + sin3x - sinx + sin 5x - sin 3x = - sinx
[ since , sin 5x = sin (7x - 2x) = sin(2 - 2x) = - sin2x ]
Therefore, 2Q sinx = - sinx i.e Q = - 1 / 2 ............ (2)
Finally,
P = (sinx + sin2x + sin4x)2 + Q2
= 3 + 2Q
(sinx + sin2x + sin4x)2 + 1 / 4 = 3 + 2 (-1/2) = 2 [since , Q = - 1/2 ]
(sinx + sin2x + sin4x)2 = 2 - 1/4 = 7 / 4
(sinx + sin 2x + sin4x) =  7 / 2
sin 2 / 7 + sin 4 / 7 + sin 8 / 7 = 7 / 2
Now, we have to reject the negative value i.e - 7 / 2
sin 2 / 7 > sin / 7 ( 2 / 7 = 51 deg and / 7 = 25 deg approx. )
i.e sin 2 / 7 - sin / 7 > 0
i.e sin 2 / 7 + sin 8 / 7 > 0 [ since, sin 8 / 7 = - sin / 7 ]
i.e sin 2 / 7 + sin 4 / 7 + sin 8 / 7 > 0 [ since, sin 4/7 > 0 as 0 <4/7 < ]
Therefore, negative value for (sin 2 / 7 + sin 4 / 7 + sin 8 / 7) is inadmissible.
Hence, sin 2 / 7 + sin 4 / 7 + sin 8 / 7 = 7 / 2
Cheers !!
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The first term of the A.P is 2 and the common difference is 3.
Sn denotes the sum of the first n terms of the A.P.
Sn = n / 2 [ 2. 2 + (n-1) . 3 ] = n (3n+1) / 2 = 3n2 / 2 + n / 2
Therefore,
S1 + S2 + ...........................+ S2n - 1
= [1 ][2n - 1 ] Sn = 3/2 [1 ] [2n - 1 ] n 2 + 1/2 [1 ][2n - 1 ] n
= 3 / 2 . (2n - 1) ( 2n - 1 + 1) ( 4n - 2 + 1) / 6 + 1/2 (2n - 1) ( 2n - 1 + 1) / 2
= n (2n - 1)(4n - 1) / 2 + n (2n - 1) / 2
= n (2n - 1) 4n / 2 = 2 n2 (2n - 1)
Cheers !!!
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The water drop is at rest as the net force acting on it is 0
Weight if water bubble
= Force on the charged water bubble
i.e mass of water bubble x g
= excess charge on water bubble x electric field
i.e Volume of water bubble x density of water x g
= excess charge on water bubble x electric field
i.e 4/3  (10 - 6) 3 x 1000 x 9.8
= excess charge in Coulombs x 5.1 x 104
[ radius of water bubble = 1 micron = 10 - 6 m
Vol. of water bubble = 4/3 (10 - 6) 3 m3 ]
Therefore, excess charge in Coulombs = 8 x 10 - 19 C
Now, 1 e = 1.6 x 10 - 19 C
So, excess electrons in water bubble = 8 x 10 - 19 / (1.6 x 10 - 19) = 5
Ans : 5
Cheers !!
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The diagrams attached with the answers will give a clear description.
Case I
N + qE sin = m cos (a+g) ......... eqn (1)
Net unbalanced force acting along the plane = (mg sin + masin + qEcos)
Therefore acceleration of the block along the plane, f
= (g sin + a sin + qE/m cos )
Calculate f putting = 30 deg
If the distance between the block and the lowest point of the plane be S,
then, S = 1/2 f t2 ( since initially the block was at rest )
i.e t = (2S/f)
Case II
N = m cos (a+g) + qE sin ......... eqn (2)
Net unbalanced force acting along the plane = (mg sin + masin - qEcos)
Therefore acceleration of the block along the plane, f
= (g sin + a sin - qE/m cos )
Calculate f putting = 30 deg
If the distance between the block and the lowest point of the plane be S,
then, S = 1/2 f t2 ( since initially the block was at rest )
i.e t = (2S/f)
Cheers !!
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