A)
=
=
By Applying partial fractions -
=
=
=
=
BY using the statement that f(0) = 0 ........We get C = 0
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A compilation of various questions in Mathematics and Physics solved by me on GoIIT
log_{7} 10 = 1 + log_{7} (10/7) = 1 + log_{7} 1.428
log_{11} 13 = 1 + log_{11} (13/11) = 1 + log_{11 } 1.18
Let A = log_{7} 1.428
and B = log_{11} 1.18
So, 7^{A} = 1.428
and 11^{B} = 1.18
Clearly A and B will both be fractional, and B will be smaller than A; (Since, in case of A, 7 is being reduced to 1.428 but with B, a larger number 11 is being reduced to a correspondingly less value than for A. )
Thus, 1 + log_{7} 1.428 > 1 + log_{11} 1.18
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Q) Find the integral of :
A)
=
=
This is of the form : | 2ab / (a^2 + b^2) | where a is 2x and b is 3. And we know: (a - b)^2 > = 0a^2 + b^2 > = 2ab so | 2ab / (a^2 + b^2) | < = 1 so y < = 1 and y > = 0 ....(since it is absolute value of something) So the two boundary lines are y = 1 and y = 0 . |
Q) A fly wheel rotates about its central axis . Due to friction at the axis, it experiences angular retardation proportional to its angular velocity. If its angular velocity falls to half its initial value while its makes n revolutions , how many more revolutions will it make before coming to rest?
A) - a = k.w .........(k is a constant of proportionality)
(dw/dQ)(dQ/dt) = k.(dw/dt)
dw = - k.dQ
w = -kQ + c
Initially Q = 0, and w = W0
W0 = C
w = -kQ + W0
After n revolutions,
W0 / 2 = -k(2.n.pi) + W0
W0 / 2 = 2.k.n.pi
k = W0 / 4.n.pi
For final stop, w = 0
w = -kQ + W0
0 = -(W0 / 4.n.pi). Q + W0
Q = 4.n.pi
means 2n revolutions.
And given that n revolutions have already occurred. So number of revolutions that will further occur before the flywheel stops = 2n - n = n revolutions.
Q) The figure below shows the acceleration-time graph of a particle moving along a straight line. After what time does the particle acquire its initial velocity?
Let D = distance between A and B Let velocity of the buses = v Relative velocity of buses with respect to the cyclist (going from A to B) v-20 km/hr for buses going from A to B v+20 km/hr for Buses going from B to A Now use Time = distance/velocity Use the 2 equations to get d and v D/(v-20) = 18/60 ---------------------1 D/(v+20) = 6/60 -----------------------2 =>v=40 km/hr and D=6 km T = 6/40 hr = 9 mins Hence the speed of the buses is 40km/hr & the period of bus service is 9 mins . (A new bus is sent after every 9 mins) |
The 3 groups are A, B and C. The prob. we have to find is of getting BGG where B is boy and G is girl. Case 1 - boy is selected from A, and the girls from B and C. prob. of BGG = 1/4*1/2*1/4 = 1/32 Case 2 - boy is selected from B, and girls from A and C. prob. of BGG = 1/2*3/4*1/4 = 3/32 Case 3 - boy is selected from C, and girls from A and B. prob. of BGG = 3/4*3/4*1/2 = 9/32 Total prob. = Sum of prob.= 1/32 + 3/32 + 9/32 = 13/32 |
So for 2003 terms the summation is = (2003)(2004)(4007) / 6 = (2003)(4007)(334)
Now,
(1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) =
= [ 2004(2003)(2003 + 1) / 2 ] - [ (2003)(4007)(334) ]
= [ (1002)(2003)(2004) ] - [ (2003)(4007)(334) ]
= 3 [(334)(2003)(2004) ] - [ (2003)(4007)(334)]
= (2003)(334)( 3*2004 - 4007)
= (2003)(334)(2005)
So the answer is 2005.
1 + (1+2) + (1+2+4) + ...... till n terms = 1 + [1*(2^2 -1)/(2-1)] + [1*(2^3 - 1)/(2 - 1)] + ....(1 + 2 + 4 + ...2^(n-1) ) ---->( Concept of sum of infinite Geometric Progression ) = 1 + [1*(2^2 -1)/(2-1)] + [1*(2^3 - 1)/(2 - 1)] + ...[1*(2^n - 1) / (2 -1) ] = 1 + [2^2 - 1 + 2^3 - 1 + 2^4 - 1 .... + 2^n - 1] = 1 + [ 2^2.(2^(n-1) - 1) / (2-1) - (n-1) ] = 1 + [ 4. 2^(n-1) - 4 - n + 1] = 2^(n+1) - 2 - n |