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[n ] [ infinity] [r = 1] [r = n ] r2 / (n3 +1)
= [n ] [ infinity] [ n (n+1) (2n + 1)] / [ 6. (n3 + 1) ]
= [n ][ infinity] [ n (n+1) (2n + 1) ] / [ 6 . (n+1) . (n2 - n + 1) ]
= [n ][ infinity] [ n (2n + 1) ] / [ 6 . (n2 - n + 1) ]
= [n ][ infinity] [ 2 + 1 / n ] / [ 6 . ( 1 - 1 / n + 1 / n2 ) ]
= 2 / 6 = 1 / 3
Now, as said by Neeraj, we see that each term individually [n ][ infinity] r / (n3 + 1) tends to zero. But when such small positive quantities in infinite numbers are added the result tends to 1/3 as obtained. Remember that the infinite sum of infinitesimally small positive quantities will tend to a finite value.
Hope that removes all your doubt.
Cheers !
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Let the resultant be R which makes an angle of 45 deg with each of the vectors P and Q
Angle between the vectors P and Q = (45+45) degrees = 90 degrees
Therefore,
tan 45 = (P sin 90) / ( Q + P cos 90 )
i.e P = Q
Now, R2 = P2 + Q2 + 2PQ cos 90 = P2 + Q2 = 2 P2
Therefore, R =  2 P
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Avogadro's hypothesis states that
' Under the same conditions of temperature and pressure equal volumes of all gases will contain equal number of molecules. '
So Firstly, you should mention that all measurements are done under the same temperature and pressure.
Secondly, it was Dalton who proposed the law that,
' Under the same conditions of temperature and pressure equal volumes of all gases will contain equal number of atoms. '
The above one was rectified when Avogadro replaced the concept of atoms with that of the molecules.
Finally, let's take an example to clarify your doubt.
If V volume of a monoatomic gas contains n molecules, then under same temperature and pressure V volumes of a diatomic as well as a triatomic gas will conatin n number of molecules.
But the no. of atoms present in V vol. of monoatomic gas = n,
the no. of atoms present in V vol. of diatomic gas = 2n,
the no. of atoms present in V vol. of triatomic gas = 3n
[ under the same temperature and pressure ]
Hope, this clarifies ur doubt.
Keep Rockin !!
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1 / cos 70 - 1 / 3 . sin 70
= [  3. sin 70 - cos 70 ] / ( 3 . sin 70 . cos 70 )
= 4 . [ 3 / 2 sin 70 - 1/2 . cos 70 ] / ( 3 . 2 sin70. cos70 )
= 4 [ sin70.cos30 - cos70.sin30 ] / 3 . sin140
= 4 sin (70 - 30) / 3 . sin 40
= 4 / 3
Ans: 4 / 3
Cheers !!!!!!
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The reaction involved is
CO2 + C = 2 CO
Let x L of CO2 reacted to give 2x L of CO.
Remaining vol. of CO2 = (0.5 - x) L
Total vol. of the gases =( 0.5 - x + 2x ) L = (0.5 + x) L
By the problem,
0.5 + x = 0.7 i.e x = 0.2
Therefore,
vol. of CO2 remaining = 0.5 - 0.2 L = 0.3 L
Vol. of CO = 2 (0.2) L = 0.4 L
Required ratio of the vol. of CO2 : Vol. of CO = 3 : 4
Cheers !!!
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The graph of the function e - x is attached with the answer. As, the question wants the area of e - x with X axis and Y axis, it can mean two areas --- the area in the first quadrant and the area in the second quadrant.
Area in the first quadrant = [a ][ + infinity ] [ 0 ] [a] e - x dx = 1 sq. units.
Area in the second quadrant = [b ][ - infinity ] [ b]0[ ] e - x dx = 
Remark : Well, obviously, you can conclude from the graph that the area in the 1st quadrant will have finite value as the curve tends to zero as x tends to , and hence tends to a closed bounded figure. On the contrary, the area in the second quadrant has infinite value as the curve tends to as x tends to - and hence has an open unbounded figure.
Hope that satisfies and resolves all query.
Cheers !!!!!
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If, f ( x ) = l a (x) b (x) c (x) l
l d (x) e (x) f (x) l
l g (x) h (x) i (x) l
then f ' (x)
= l a' (x) b ' (x) c ' (x) l + l a (x) b (x) c (x) l + l a (x) b (x) c (x) l
l d (x) e (x) f (x) l l d ' (x) e ' (x) f ' (x) l l d (x) e (x) f (x) l
l g (x) h (x) i (x) l l g (x) h (x) i (x) l l g' (x) h' (x) i '(x) l
= 23 . l a (x) b (x) c (x) l + l a ' (x) b ' (x) c ' (x) l
l d (x) e (x) f (x) l + l d ' (x) e ' (x) f ' (x) l
l g (x) h (x) i (x) l + l g ' (x) h ' (x) i ' (x) l
So, here,
f ' (x) = 8 l cosx cos3x 3 cos2x l - l sinx 3sin3x 6 sin2x l
l sinx sin 2x 2 sin3x l l cos x 2cos2x 6cos3x
| cosx-sinx cosx+sinx cosx.sinx | | - cosx-sinx cosx-sinx cos2x. |
Put x =  / 3 to have the value of f ' (/3)
Cheers !!!!!
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A box function, say [ x ], where [x] denotes the greatest integer less than or equal to x, is not differentiable at the integral points or in other words, [ x ] is not differentiable when x is an integer.
Therefore, f (x) = [ n + p sinx ] will not be differentiable when n + p sinx is an integer.
0 < x <  , i.e 0 < sin x  1
Since, n is an integer, n + p sinx will be an integer only when p sinx is an integer.
Since, p is a prime = 19, p sinx will be an integer when,
sin x = 1/19, 2/19, 3/19,...........................18 / 19, 19/19
There are 18 values of x lying between 0 and  / 2, for which
sin x = 1/19, 2/19, 3/19 ......................18/ 19
At / 2 , sin x = 19 / 19 = 1
Again there are 18 values of x lying between / 2 and , for which
sin x = 1/19, 2/19, 3/19 ......................18/ 19
So, there are in total (18 + 1 + 18) = 37 values of x for given value of n and p = 19, for which f (x) is not differentiable.
Cheers !!!!!
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 sinX + cos 2X - 2 =0
 sin X + 1 - 2 sin2 X - 2 = 0
Let y = sin X
Therefore,
y + 1 - 2y2 - 2 = 0
2y2 - y + ( 2 - 1 ) = 0
y = sin X can have real solution, when the discriminant is  0
i.e 2 - 4. 2. (2 - 1) 0
i.e 2 - 2. 8. + 64 - 56 0
i.e ( - 8) 2 56
i.e - 8 56 or - 8 - 56
i.e 8 + 56 or 8 - 56
So, none of the options are correct but the closest option is (1)
Cheers !!!!!
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[ r = 1 ] [ r = 89 ] log e tan r = [ r = 1 ] [ r = 89 ] tan r [ since, ln e tan r = tan r ]
= tan1.tan2.tan3.tan4..............tan45. ..................tan86.tan87.tan88.tan89
= (tan1.cot1).(tan2.cot2) ........................... tan45
= 1.1.1........................1 = 1
If the question was, [ r = 1 ][r = 89 ] log e tan r = [ r = 1 ] [r = 89 ] tan r
= tan 1 + tan2 + ....................... tan 89
Cheers !!!!!
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The problem given is wrong.
We, prove an identity (an identity is satisfied by all values of the given variable)
We, solve an equation (an equation is satisfied by some specific values of the given variable)
The given problem is an equation as it is satisfied for some specific given values of x and not for all values of x.
Hence the problem is wrong.
Cheers !!!
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Bold letters represent the vector sign.
Let two vectors P and Q make an angle  with each other.
Let the resultant of the two vectors be R
R = P + Q
A vector has magnitude and direction.
Magnitude of R is l R l = l P l 2 + l Q l 2 + 2 l P l . l Q l cos
where, magnitude of P is l P l & magnitude of Q is l Q l
Now, the direction of R can be given by
tan  = P sin / ( Q + P cos )
where is the angle that R makes with Q on the side nearer to P
Tip: The resultant will lean towards the vector which has greater magnitude or in other words, < - when l Q l > l P l
i.e < / 2 when l Q l > l P l
Vice versa, > / 2 when l Q l < l P l
Cheers !!!!!!
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Since we are talking about a car's registration number, the beginning digits can be zero (s) [ for e.g registration number 0123, 0001 or 0015, etc ]
Each place of the four digit number can be filled up in 10 ways i.e 0 to 9
Total such four digit numbers that are possible are = 10 4
(a) When same digit occurs twice
Two places can be selected from four-digit no. in 4C2 ways and the repetitive digit can be sel. in 10 ways & the remaining two places can be filled up in 9 x 8 = 72 ways since the repeated digit can't be selected again.
Total such four digit numbers that are possible whose same digit occurs twice
= 72 x 4C2 x 10
Total probability = 72 x 4C2 / 10 3
(b) When same digit occurs thrice
3 places can be selected from four-digit no. in 4C3 ways and the repetitive digit can be sel. in 10 ways & the remaining 1 place can be filled up in 9 ways.
Total such four digit numbers that are possible whose same digit occurs twice
= 9 x 4C3
Total probability = 9 x 4C3 / 10 3
(c) When the same digit occupies all the four places.
Only one such number is possible and the repetitive digit can be sel. in 10 ways
Total probability = 10 / 10 4 = 1 / 10 3
Cheers !!!!!
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4x - 3x - 1/2 = 3x + 1/2 - 22x - 1
4x + 22x - 1 = 3x [ 3 + 1 / 3 ]
2 2x ( 1 + 1/2 ) = 3 x . 4 / 3
2 2x . 3 / 2 = 3 x . 4 / 3
2 2x - 3 = 3 x - 3 / 2
4 x - 3 / 2 = 3 x - 3 / 2
(4 / 3) x - 3 / 2 = 1
x - 3/2 = 0 x = 3 / 2
Ans: x = 3 / 2
Cheers !!!!!
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Without any doubt, you should opt for ECE at once.
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Consider a room. The rooms has a length, breadth and height. Now, consider that from one of the corners of the room three mutually perpendicular axes originate, one along the length of the room, one along the breadth of the room and one along the height of the room.
Let, the axes along the length of the room be X - axes, that along the breadth be Y - axes, and that along the height be Z - axis.
Now, imagine yourself to be standing at a particular point in the floor of the room.
( i.e in the X-Y plane )
Case when only 1 coordinate changes
Now, if you move only along the length of the room, you'll find that your x-coordinate is changing only whereas your y and z coordinates remain unaltered.
Now, if you move only along the breadth of the room, you'll find that your y-coordinate is changing only whereas your x and z coordinates remain unaltered.
Now, if you jump only along the height of the room, you'll find that your z-coordinate is changing only whereas your x and y coordinates remain unaltered.
Case when all 2 Coordinates Change
Now, if you move diagonally on the floor, you'll find that your x and y coordinates are changing whereas the z- coordinate remains constant.
Case when all 3 Coordinates Change
Now, if you jump diagonally, you ' ll find that your x,y and z co ordinates , all change simultaneously.
So, now, you can see that " "the position of an object i space is specified by 3 coordinates, x,y, and z."The position of the object changes due to change in 1 or 2 or all the 3 coordinates." "
Hope this solves your doubt. But I have also attached a diagram for your convenience.
Cheers !!
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tan (  / 3) =  3  [ tan 2 / 5 - tan / 15 ] / [ 1 + tan 2 / 5 . tan / 15 ] = 3 tan 2 / 5 - tan / 15 = 3 + 3 [ tan 2 / 5 . tan /15 ] tan 2 / 5 - tan / 15 - 3 [ tan 2 / 5 . tan /15 ] = 3 Hence, tan 2 / 5 - tan / 15 - 3 [ tan 2 / 5 . tan /15 ] = 3 Cheers !!!!!
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cos  + cos  = 0 cos 2 + cos 2 + 2cos cos = 0 ............ eqn (1) Again, sin + sin = 0 sin 2 + sin 2 + 2sin sin = 0 ............ eqn (2) Now, eqn (1) - eqn (2) ( cos 2 - sin 2 ) + ( cos 2 - sin 2 ) + 2 (cos cos - sin sin ) = 0 cos 2 + cos 2 + 2cos ( + ) = 0 cos 2 + cos 2 = - 2cos ( + ) Hence, cos 2 + cos 2 = - 2cos ( + ) Cheers !!!!!
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Sine function is positive when the angle lies in the 1st or 2nd quadrant. Now, sin = sin = sin  = sin  = k (where, k is a + ve quantity  1 ) Since, , , and are the least positive angles, then lies in the interval [ 0, / 2 ] lies in the interval [ / 2, ] and = - ................ equation (1) lies in the interval [ 2 , 5 / 2 ] & = 2 + ................ equation (2) lies in the interval [ 5 / 2 , 3 ] & = 2 + = 3 - --------- equation (3) Now, (sin /2 + cos /2) 2 = (sin 2 /2 + cos 2/2) + 2sin /2 . cos /2 = 1 + sin = 1 + k i.e sin /2 + cos /2 = (1+k) ................... equation (4) The given expression 4sin /2 +3sin /2 + 2sin /2 + sin /2 = 4sin /2 + 3sin ( /2 - /2) + 2sin ( + /2) + sin (3 /2 - /2) [ from eqn. (1), (2), (3) ] = 4sin /2 + 3cos /2 - 2sin /2 - cos /2 = 2 (sin /2 + cos /2) = 2 (1+k) [ from eqn. (4) ] Ans: 2 (1+k) Cheers !!!
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Consider two mutually perpendicular axes, X - axis and Y - axis.
Let the point A be at ( -k, 0 ) ; B be at (0,0) & C be at (k,0)
Point A is moving with a constant velocity of v along positive Y axis keeping its x-coordinate constant.
Therefore, at any instant of time t, the coordinates of A is given by (-k, vt)
where vt is the distance traversed in the positive y direction i.e its ordinate.
Point C moves with 0 initial velocity and constant acceleration 'a' along the negative Y - axis keeping its abscissa constant.
Therefore, at any instant of time t, the coordinates of C is given by (k, - 1/2 at2)
where - 1/2 a t2 is the displacement of C in the negative Y direction i.e its ordinate.
Let (x,y) be any point on the straight line AC at any instant of time t
Therefore,
(y - vt) / (x + k) = (vt +1/2 at2) / (-k - k)
(y - vt) / (x+k) = ( vt + 1/2 at2) / - 2k ............ equation (1)
Now, the point B moves in such a manner that it will always lie on the straight line AC and as it moves along the Y - axis only, its x - coordinate i.e its abscissa will always be 0
Putting x = 0 in equation (1)
(y - vt) / k = ( vt + 1/2 at2) / -2k
y - vt = - 1/2 vt - 1/4 at2
y = (v / 2) t + 1/2 . ( - a / 2 ) t 2 .................... equation (2)
Now, compare the equation (2) with the standard equation of displacement of a particle moving with an initial velocity u and constant acceleration f, namely,
S = ut + 1/2 ft2
Therefore, from comparison, we have,
u = v / 2 & f = - a /2
Therefore, the point B was moving with an initial velocity of v / 2 along the positive Y - axis and the constant acceleration was ( - a / 2 ). The negative sign in the acceleration shows that the acceleration was directed along the negative Y - axis i.e vertically downwards.
Now, y = vt / 2 - 1/4 at2 = v2 / 4a - (1/2 t - v / 2a )2
Since, (1/2 t - v / 2a)2 > = 0 , the maximum value of y = v2 / 4a attained when
1/2 t - v / 2a = 0 i.e t = v / a
Now, differentiate (2) w.r.t ' t' ,
dy / dt = v / 2 - 1/2 at , dy / dt < 0 for t > v / a ; dy / dt = 0 at t = v/a ; dy / dt > 0 for t < v / a
So, the velocity of B i.e dy / dt remains positive and hence it moves upward as long as t < v / a.
After the attainment of maximum height at t = v / a, the velocity of the point B i.e dy / dt reverses in sign becomes negative and hence the point B starts moving downwards.
So, finally we can conclude that
B moves with initial velocity V/2 directed upwards and a constant acceleration a/2 directed downwards. After reaching the height v^2/4a , the point will move downward
Cheers !!!!!
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