Q) A bird flies with a speed of v=| t-2 | m/s along a straight line , where t is in seconds. Find the distance traveled by the bird during the first 4 seconds of motion.
A) v = | t-2 | m/s
The critical point is t = 2 (At which the modulus becomes 0)
so from t=0 to t=2 , velocity is (t-2) m/s
and from t=2 to t= 4, velocity is (2 - t) m/s
Total distance covered = mod of ( integral (0 to 2) [ t-2] + integral (2 to 4) [ 2 - t] )
= [t^2/2 - 2t] 0 to 2 + [ 2t - t^2/2] 2 to 4
= | -2 - 2 |
= |-4 |
= 4 m
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Tuesday, September 16, 2008
Q) If f(x) is sum of all digits of x then find f(101) + f(102) + f(103) .... f(200) .
A)
f(101) + f(102) + f(103) .... f(200)
= 99(1) for hundreds places + 10(1+2+3+4+5+6+7+8+9) for tens places + 10(1+2+3+4+5+6+7+8+9) for units places + 2 (for 200)
= 99 + 20(45) + 2
= 101 + 900
= 1001
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A)
f(101) + f(102) + f(103) .... f(200)
= 99(1) for hundreds places + 10(1+2+3+4+5+6+7+8+9) for tens places + 10(1+2+3+4+5+6+7+8+9) for units places + 2 (for 200)
= 99 + 20(45) + 2
= 101 + 900
= 1001
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Q) Find the area contained within the graph of the equation: 
A)
x^2 - 6x + y^2 + 12y = 0
Form perfect squares:
x^2 - 6x + 9 - 9 + y^2 + 12y + 36 - 36 = 0
(x + 3)^2 + (y + 6)^2 = 45
(x + 3)^2 + (y + 6)^2 = [sqrt(45)]^2
This is the equation of a circle, with radius sqrt(45)
so area = pi.r^2
= 45.pi
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A)
x^2 - 6x + y^2 + 12y = 0
Form perfect squares:
x^2 - 6x + 9 - 9 + y^2 + 12y + 36 - 36 = 0
(x + 3)^2 + (y + 6)^2 = 45
(x + 3)^2 + (y + 6)^2 = [sqrt(45)]^2
This is the equation of a circle, with radius sqrt(45)
so area = pi.r^2
= 45.pi
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Q) Consider a sequence { 
} such that 
.
How many distinct pairs chosen from this sequence have g.c.d.= 6 ?
A) a1 = 2
a2 = 3
a3 = 7
a4 = 43
Now since
a(n+1) = a.n^2 - an + 1
and after a1, the next terms are odd...
so terms will be of the type
odd^2 - odd + 1
odd^2 is always odd... and odd^2 + 1 will become even.
then even - odd = odd always....
So there won't be any distinct pair with GCD = 6... 2 will be the only even number in the sequence.
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} such that .How many distinct pairs chosen from this sequence have g.c.d.= 6 ?
A) a1 = 2
a2 = 3
a3 = 7
a4 = 43
Now since
a(n+1) = a.n^2 - an + 1
and after a1, the next terms are odd...
so terms will be of the type
odd^2 - odd + 1
odd^2 is always odd... and odd^2 + 1 will become even.
then even - odd = odd always....
So there won't be any distinct pair with GCD = 6... 2 will be the only even number in the sequence.
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Monday, August 11, 2008
Q) A wire of length l is to be cut into two pieces, one of which is bent to form a circle and the other to form a square. How should the wire be cut so as to minimize the sum of the areas enclosed by the two pieces.
A)
Let the lengths be x and L-x
Area of square = (x/4)2
Area of circle : Given that circumference = L-x = 2.pi.r
r = (L-x) / 2.pi
Area of circle = pi.r^2 = (L-x)^2 / 4.pi
Total Area = x^2 / 16 + (L-x)^2 / 4.pi
Diff and equating to zero:
x/8 = (L-x)/2.pi
pi.x = 4L - 4x
x = 4L/(pi+4)
and thus L-x = L - [4L/(pi+4)] = L.pi / (pi+4)
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A)
Let the lengths be x and L-x
Area of square = (x/4)2
Area of circle : Given that circumference = L-x = 2.pi.r
r = (L-x) / 2.pi
Area of circle = pi.r^2 = (L-x)^2 / 4.pi
Total Area = x^2 / 16 + (L-x)^2 / 4.pi
Diff and equating to zero:
x/8 = (L-x)/2.pi
pi.x = 4L - 4x
x = 4L/(pi+4)
and thus L-x = L - [4L/(pi+4)] = L.pi / (pi+4)
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Q) If f(x) = ax^2 +bx+c AND g(x) = -ax^2+bx+c where ac is not equal to 0, then what can be said about the roots of the equation f(x) .g(x) = 0 ?
A)
f(x).g(x) = -a^2.x^4 + (bx + c)^2
f(x).g(x) = a^2.x^4 - b^2.x^2 - 2bc.x - c^2 = 0
Using Descartes' Rule of signs.... there is one positive root, and a max of 3 negative roots (can be 1 negative root and rest 2 imaginary). Remember that imaginary roots occur in pairs always.
So the least number of real roots is 1 positive and 1 negative... so at least 2 real roots.
So the answer is at least 2 real roots.
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A)
f(x).g(x) = -a^2.x^4 + (bx + c)^2
f(x).g(x) = a^2.x^4 - b^2.x^2 - 2bc.x - c^2 = 0
Using Descartes' Rule of signs.... there is one positive root, and a max of 3 negative roots (can be 1 negative root and rest 2 imaginary). Remember that imaginary roots occur in pairs always.
So the least number of real roots is 1 positive and 1 negative... so at least 2 real roots.
So the answer is at least 2 real roots.
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Q) Integrate [(x^2+1)/(x^2-5x+6)].dx
A)
x^2+1 / x^2-5x+6
= 1 + [ (5x-5) / x^2 -5x + 6 ]
= 1 + [ 5 (x-1+2 -2) / (x-2)(x-3)]
= 1 + 5[ (1 /x-3) + 1/(x-2)(x-3) ]
Notice that 1 in last term can be written as (x-2) - (x-3)
= 1 + 5[(1/x+3) + ( (x-2) - (x-3)) / (x-2)(x-3) ]
= 1 + (10/x-3) - (5/x-2)
integrating we get
x + 10.ln(x-3) - 5.ln(x-2) + c
= x + 5.ln ( (x-3)^2 / (x-2) ) + C
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A)
x^2+1 / x^2-5x+6
= 1 + [ (5x-5) / x^2 -5x + 6 ]
= 1 + [ 5 (x-1+2 -2) / (x-2)(x-3)]
= 1 + 5[ (1 /x-3) + 1/(x-2)(x-3) ]
Notice that 1 in last term can be written as (x-2) - (x-3)
= 1 + 5[(1/x+3) + ( (x-2) - (x-3)) / (x-2)(x-3) ]
= 1 + (10/x-3) - (5/x-2)
integrating we get
x + 10.ln(x-3) - 5.ln(x-2) + c
= x + 5.ln ( (x-3)^2 / (x-2) ) + C
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Q) An unbiased die is rolled until a number greater than 4 appears on the die. Find the probability that an even number of tosses are needed.
A)
prob. that a number greater than 4 appears = 2/6 = 1/3
prob. that a number less than or equal to 4 appears = 2/3
prob. that an even no. of tosses are needed =
(2/3)*1/3 + (2/3)^3*1/3 + (2/3)^5 * 1/3 ......
= 2/9 (1 + (2/3)^2 + (2/3)^4 + (2/3)^6 + ....)
= 2/9 [ 1 / ( 5/9) ]
= (2/9)*(9/5)
= 2/5
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A)
prob. that a number greater than 4 appears = 2/6 = 1/3
prob. that a number less than or equal to 4 appears = 2/3
prob. that an even no. of tosses are needed =
(2/3)*1/3 + (2/3)^3*1/3 + (2/3)^5 * 1/3 ......
= 2/9 (1 + (2/3)^2 + (2/3)^4 + (2/3)^6 + ....)
= 2/9 [ 1 / ( 5/9) ]
= (2/9)*(9/5)
= 2/5
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Q) If a set contains n elements and A and B are the subsets of that set, then find the probability that A intersection B is = n .
A)
For the given condition, A should have all n elements and B too should have all n elements.
Total number of subsets = 2^n
Prob. of selecting subset having all n elements = 1/2^n
so Prob. of selecting 2 subsets (sequentially) both having n elements = (1/2^n)*(1/2^n)
= 1 / 2^2n
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A)
For the given condition, A should have all n elements and B too should have all n elements.
Total number of subsets = 2^n
Prob. of selecting subset having all n elements = 1/2^n
so Prob. of selecting 2 subsets (sequentially) both having n elements = (1/2^n)*(1/2^n)
= 1 / 2^2n
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Q) What is the square root of the complex number 5+12i ?
A)
z = 5 + 12 i
z = 13 (5/13 + 12/13 i)
z = 13 (cosQ + i.sinQ)
cosQ = 5/13 ; sinQ = 12/13
Using DeMoivre's theorem:
z^(1/2) = +- sqrt (13) * (cos(Q/2)+ i.sin(Q/2) )
z^(1/2) = +- sqrt (13) * [ rt ( (cosQ + 1)/2) + i.rt ( (1 - cosQ)/2) ]
z^(1/2) = +- sqrt (13) * [ rt (9/13) + i.rt ( 4/13) ]
z^(1/2) = +- sqrt (13) * [ 3/rt(13) + i.2/rt(13) ]
z^(1/2) = +- ( 3 + 2i)
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A)
z = 5 + 12 i
z = 13 (5/13 + 12/13 i)
z = 13 (cosQ + i.sinQ)
cosQ = 5/13 ; sinQ = 12/13
Using DeMoivre's theorem:
z^(1/2) = +- sqrt (13) * (cos(Q/2)+ i.sin(Q/2) )
z^(1/2) = +- sqrt (13) * [ rt ( (cosQ + 1)/2) + i.rt ( (1 - cosQ)/2) ]
z^(1/2) = +- sqrt (13) * [ rt (9/13) + i.rt ( 4/13) ]
z^(1/2) = +- sqrt (13) * [ 3/rt(13) + i.2/rt(13) ]
z^(1/2) = +- ( 3 + 2i)
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Q) Find the sum of the series 1/2 + 1/6 + 1/12 + 1/20 .....n terms where n tends to infinity.
A)
The series is:
1/1.2 + 1/2.3 + 1/3.4 ........
= summation from 1 to inf. of [1/(n)(n+1)]
Now, 1/(n)(n+1) = n+1-n / (n)(n+1)
= (1/n) - 1/(n+1)
summation will give:
1 + 1/2 + 1/3 + 1/4+ 1/n....... - 1/2 - 1/3 - 1/4 - 1/5....-1/n - 1/(n+1)
= 1 - 1/(n+1)
put n tends to infinity and you get
1 - 0
= 1
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A)
The series is:
1/1.2 + 1/2.3 + 1/3.4 ........
= summation from 1 to inf. of [1/(n)(n+1)]
Now, 1/(n)(n+1) = n+1-n / (n)(n+1)
= (1/n) - 1/(n+1)
summation will give:
1 + 1/2 + 1/3 + 1/4+ 1/n....... - 1/2 - 1/3 - 1/4 - 1/5....-1/n - 1/(n+1)
= 1 - 1/(n+1)
put n tends to infinity and you get
1 - 0
= 1
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Q) The sum of the reciprocals of two real numbers is −1, and the sum of their cubes is 4. What are the numbers?
A)
1/x + 1/y = -1
x+y / xy = -1
x+y = -xy
[and (x+y)^2 = x^2.y^2 ]
Now x^3 + y^3 = 4
(x+y)(x^2 - xy + y^2) = 4
(-xy)(x^2.y^2 - 2xy - xy)
(xy)(3xy - x^2.y^2) = 4
(xy)(xy)(3-xy) = 4
Thus, xy = -1
x = -1/y
1/x + 1/y = -1
-y + 1/y = -1
y = (1 +- rt5) / 2
and x= -1/y
so you get the corresponding values of x.
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A)
1/x + 1/y = -1
x+y / xy = -1
x+y = -xy
[and (x+y)^2 = x^2.y^2 ]
Now x^3 + y^3 = 4
(x+y)(x^2 - xy + y^2) = 4
(-xy)(x^2.y^2 - 2xy - xy)
(xy)(3xy - x^2.y^2) = 4
(xy)(xy)(3-xy) = 4
Thus, xy = -1
x = -1/y
1/x + 1/y = -1
-y + 1/y = -1
y = (1 +- rt5) / 2
and x= -1/y
so you get the corresponding values of x.
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Q) The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to flow in the direction of motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same condition of projection , find the horizontal range of the projectile.
A)
The time taken by it to fall down to the ground will still remain the same, just that it will cover more distance due to additional horizontal acceleration:
Along horizontal axis, initial velocity = u.cos@
acceleration along same direction = (+)g/2
Time taken by projectile (to reach ground) = 2usin@/g
s = ut + 1/2.at^2
s = ucos@.2usin@/g + 1/2*g/2*4u^2sin^2@ / g^2
s = u^2.sin2@/g + u^2.sin^2@/g
s = (u^2 / g) [sin2@ + sin^2 @]
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A)
The time taken by it to fall down to the ground will still remain the same, just that it will cover more distance due to additional horizontal acceleration:
Along horizontal axis, initial velocity = u.cos@
acceleration along same direction = (+)g/2
Time taken by projectile (to reach ground) = 2usin@/g
s = ut + 1/2.at^2
s = ucos@.2usin@/g + 1/2*g/2*4u^2sin^2@ / g^2
s = u^2.sin2@/g + u^2.sin^2@/g
s = (u^2 / g) [sin2@ + sin^2 @]
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Q) A body is thrown vertically upwards from the top X of tower . It reaches the ground in A seconds . If it is thrown vertically downwards from X with the same speed, it reaches the ground in B seconds. If it is allowed to fall freely from X, the time taken by it to reach the ground is?
A)
Let's consider the top of X as origin
S = ut + 1/2 at^2
-X = uA - gA^2 / 2 -----------------------(1)
(Here we have taken -X because this is the final displacement after going up a certain distance coming back then going to the bottom of the building.)
and
-X = -uB - gB^2/2
X = uB + gB^2/2 --------------------------(2)
third case:(free fall, initial velocity is zero)
X = gT^2/2 ------------------------(3)
Multiply (1) and (2) with B and A respectively, then solving, we get:
X(A+B) = (g/2)(AB^2 + A^2.B)
X = ABg/2
Substitute this in (3):
ABg/2 = gT^2/2
T^2 = AB
T = root (AB)
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A)
Let's consider the top of X as origin
S = ut + 1/2 at^2
-X = uA - gA^2 / 2 -----------------------(1)
(Here we have taken -X because this is the final displacement after going up a certain distance coming back then going to the bottom of the building.)
and
-X = -uB - gB^2/2
X = uB + gB^2/2 --------------------------(2)
third case:(free fall, initial velocity is zero)
X = gT^2/2 ------------------------(3)
Multiply (1) and (2) with B and A respectively, then solving, we get:
X(A+B) = (g/2)(AB^2 + A^2.B)
X = ABg/2
Substitute this in (3):
ABg/2 = gT^2/2
T^2 = AB
T = root (AB)
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Q) 7 white balls & 3 black balls are randomly placed in a row. Find the probability that no two black balls are placed adjacently.
A)
First fix the seven white balls, then you have eight places for the black balls to choose from... and 8C3 is the number of combinations you can have ( how the places are chosen)
Total number of arrangements of the 10 balls is 10C3
So prob = 8C3 / 10C3 = 7/15
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A)
First fix the seven white balls, then you have eight places for the black balls to choose from... and 8C3 is the number of combinations you can have ( how the places are chosen)
Total number of arrangements of the 10 balls is 10C3
So prob = 8C3 / 10C3 = 7/15
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Q) Two stones are projected so as to reach the same distance from the point of projection on a horizontal plane. The maximum height reached by one exceeds the other by an amount equal to half the sum of the heights attained by them. The angles of projection for stones are?
A)
Since Range = ucos@ . 2usin@/g = u^2.sin2@/g
therefore for angle of projection @ and 90-@, range is same.
so the angles are X, and 90 - X
Height = u^2.sin^2@ / 2g
H - h = 1/2 * (H +h)
H - h = H/2 +h/2
H/2 = 3 h / 2
H = 3h
H/h = 3 = (u^2.sin^2 X / 2g ) (2g / u^2. sin^2 Y)
3 = sin^2 X/sin^2 Y
3 = sin^2 X / sin^2 (90 - X) = tan^2 X
tan X = root (3)
X = 60 degrees
and Y = 30 degrees
So the angles of projection are 30 and 60 degrees with the horizontal.
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A)
Since Range = ucos@ . 2usin@/g = u^2.sin2@/g
therefore for angle of projection @ and 90-@, range is same.
so the angles are X, and 90 - X
Height = u^2.sin^2@ / 2g
H - h = 1/2 * (H +h)
H - h = H/2 +h/2
H/2 = 3 h / 2
H = 3h
H/h = 3 = (u^2.sin^2 X / 2g ) (2g / u^2. sin^2 Y)
3 = sin^2 X/sin^2 Y
3 = sin^2 X / sin^2 (90 - X) = tan^2 X
tan X = root (3)
X = 60 degrees
and Y = 30 degrees
So the angles of projection are 30 and 60 degrees with the horizontal.
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Q) A curve passes through (1, 1) such that triangle formed by the coordinate axes and tangent at any point of the curve is in the first quadrant and has its area equal to 2. Then the curve can be?
A)
Let the tangent equation be:
x/a + y/b = 1 ...........(where a and b are x and y intercepts respectively)
and ab/2 = 2 so ab = 4 ............(area of triangle)
bx + ay = ab
bx + ay = 4
one solution of this is x=1, y=1 as per question.
so
b + a = 4
Also, ab = 4
On solving, we get a = b = 2
So it can be a rectangular hyperbola, OR a straight line whose x intercept and y intercept both are 2 (ie. the line is x + y = 2).
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A)
Let the tangent equation be:
x/a + y/b = 1 ...........(where a and b are x and y intercepts respectively)
and ab/2 = 2 so ab = 4 ............(area of triangle)
bx + ay = ab
bx + ay = 4
one solution of this is x=1, y=1 as per question.
so
b + a = 4
Also, ab = 4
On solving, we get a = b = 2
So it can be a rectangular hyperbola, OR a straight line whose x intercept and y intercept both are 2 (ie. the line is x + y = 2).
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Q) If cos(x-y), cos(x),cos(x+y) are in H.P.,then cos(x).sec(y/2) is equal to?
A)
Solving the whole thing using AM(arithmetic mean) concept we finally get
cos^2x.cosy = cos^2y + cos^2x - 1
cos^2 x = (1-cos^2y) / (1-cosy)
cos^2 x = 1 + cos y
cos^2 x = 2cos^2(y/2)
cos^2 x. sec^2 (y/2) = 2
cosx.sec(y/2) = +- rt (2)
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A)
Solving the whole thing using AM(arithmetic mean) concept we finally get
cos^2x.cosy = cos^2y + cos^2x - 1
cos^2 x = (1-cos^2y) / (1-cosy)
cos^2 x = 1 + cos y
cos^2 x = 2cos^2(y/2)
cos^2 x. sec^2 (y/2) = 2
cosx.sec(y/2) = +- rt (2)
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Q) A projectile is thrown horizontally with a velocity 30m/s . After what time will it be moving at an angle of 30 degree with the horizontal?
A)
Since horizontal velocity remains constant, the body acquires some downward vertical velocity due to gravity which causes resultant velocity to be inclined 30 degrees with horizontal.
Tan30=v/30
v=10.sqrt(3) m/s
now this v=at
v=gt
t= 10.sqrt(3) /10 = sqrt(3) secs
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A)
Since horizontal velocity remains constant, the body acquires some downward vertical velocity due to gravity which causes resultant velocity to be inclined 30 degrees with horizontal.
Tan30=v/30
v=10.sqrt(3) m/s
now this v=at
v=gt
t= 10.sqrt(3) /10 = sqrt(3) secs
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Q) What is the remainder when 538 is divided by 11?
A)
Remainders of exponential powers of 5 when divided by 11 come as 5, 3, 4, 9, 1 repetitively.
( Rem. 5 = 5
Rem. 25 = 3
Rem. 125 = 4
Rem. 625 = 9
Rem. 3125 = 1
And again Rem. 15625 = 5)
repetition occurs after every 5 terms....so 38th power will come as a third term of the sequence (7x5 + 3)
so the remainder is 4.
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A)
Remainders of exponential powers of 5 when divided by 11 come as 5, 3, 4, 9, 1 repetitively.
( Rem. 5 = 5
Rem. 25 = 3
Rem. 125 = 4
Rem. 625 = 9
Rem. 3125 = 1
And again Rem. 15625 = 5)
repetition occurs after every 5 terms....so 38th power will come as a third term of the sequence (7x5 + 3)
so the remainder is 4.
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Q) A bird flies with a speed of v=|t-2|m/s along a straight line , where t is in seconds. What is the distance traveled by the bird during the first 4 seconds?
A)
v=|t-2|m/s
critical point is t = 2
so from t=0 to t=2 , velocity is t-2
and from t=2 to t= 4, velocity is 2 - t
Total distance covered = mod of ( integral (0 to 2) [ t-2] + integral (2 to 4) [ 2 - t] )
= [t^2/2 - 2t] 0 to 2 + [ 2t - t^2/2] 2 to 4
= | -2 - 2 |
= |-4 |
= 4 m ... (since area under graph cannot be negative)
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A)
v=|t-2|m/s
critical point is t = 2
so from t=0 to t=2 , velocity is t-2
and from t=2 to t= 4, velocity is 2 - t
Total distance covered = mod of ( integral (0 to 2) [ t-2] + integral (2 to 4) [ 2 - t] )
= [t^2/2 - 2t] 0 to 2 + [ 2t - t^2/2] 2 to 4
= | -2 - 2 |
= |-4 |
= 4 m ... (since area under graph cannot be negative)
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Q) Solve for x:
x^2 + [x^2/(x^2 + 1)] = 3
A)
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x^2 + [x^2/(x^2 + 1)] = 3
A)
| x^2 + [x^2/1+x^2 = 3 x^4 - x^2 = 3 x^2(x^2 - 1) = 3 Let x^2 = y .....thus x = root y y(y-1)=3 y^2 - y - 3 = 0 y = 1+- root(1 + 12) / 2 y = 1 +-rt(13) /2 now x = root y so x = root (1 + rt(13) / 2) other value of y is truncated since x cannot be negative due to root. x = root [ (1 + rt(13) ) / 2] |
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Q) If f(x) is sum of all digits of x then
f(101)+f(102)+f(103).................f(200) is ?
A) 99(1) for hundreds places + 10(1+2+3+4+5+6+7+8+9) for tens places + 10(1+2+3+4+5+6+7+8+9) for units places + 2 for 200
= 99 + 20(45) + 2
= 101 + 900
= 1001
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f(101)+f(102)+f(103).................f(200) is ?
A) 99(1) for hundreds places + 10(1+2+3+4+5+6+7+8+9) for tens places + 10(1+2+3+4+5+6+7+8+9) for units places + 2 for 200
= 99 + 20(45) + 2
= 101 + 900
= 1001
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Monday, May 19, 2008
Q) Find the sum to 2n terms of the series whose every even term is 'a' times the term before it and every odd term is 'c' times the term before it, the first term being unity.
A) Let T indicate a term of the progression.
T1 T2 T3.......Tn......T2n
T1 = 1
T2 = a
T3 = ca
T4 = c.a^2
T5 = c^2.a^2
Tk if k is even = a^(k/2). c^(k/2 - 1)
T2n = a^(2n/2).c^(2n/2 -1)
T2n = a^n. c^(n-1)
S 2n = 1 + a + ca + c.a^2 + c^2.a^2 + c^2.a^3 .....a^n. c^(n-1)
= 1 + [ a + ca^2 + c^2.a^3 ....+ a^n.c^(n-1) ] + [ ca + c^2.a^2 + c^3.a^3..... + a^(n-1). c^(n-1) ]
= 1 + [ a.(a^n.c^n - 1) / (ac - 1) ] + [ ac( a^(n-1).c^(n-1) - 1) / (ac - 1) ]
..
..
.. solving further..
---> S 2n = (a^n.c^n - 1)(a + 1) / (ac - 1)
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A) Let T indicate a term of the progression.
T1 T2 T3.......Tn......T2n
T1 = 1
T2 = a
T3 = ca
T4 = c.a^2
T5 = c^2.a^2
Tk if k is even = a^(k/2). c^(k/2 - 1)
T2n = a^(2n/2).c^(2n/2 -1)
T2n = a^n. c^(n-1)
S 2n = 1 + a + ca + c.a^2 + c^2.a^2 + c^2.a^3 .....a^n. c^(n-1)
= 1 + [ a + ca^2 + c^2.a^3 ....+ a^n.c^(n-1) ] + [ ca + c^2.a^2 + c^3.a^3..... + a^(n-1). c^(n-1) ]
= 1 + [ a.(a^n.c^n - 1) / (ac - 1) ] + [ ac( a^(n-1).c^(n-1) - 1) / (ac - 1) ]
..
..
.. solving further..
---> S 2n = (a^n.c^n - 1)(a + 1) / (ac - 1)
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Q) Let x, y and z be three positive numbers such that x + y + z = 12. Then the maximum value of x2y3z is ?
A) x+y+z = 12
x/2 + x/2 + y/3 + y/3 + y/3 + z = 12
Use AM > GM inequality:
(x + y + z) / 6 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2^6 > (x^2 / 4)(y^3 / 27)(z)
2^6 * 4*27 > x^2.y^3.z
So x2y3z < 6912
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A) x+y+z = 12
x/2 + x/2 + y/3 + y/3 + y/3 + z = 12
Use AM > GM inequality:
(x + y + z) / 6 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2^6 > (x^2 / 4)(y^3 / 27)(z)
2^6 * 4*27 > x^2.y^3.z
So x2y3z < 6912
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Q) If log a, log b ,log c are in Arithmetic Progression and also loga - log2b ,log2b - log3c , log3c - loga are in Arithmetic Progression then :
1)a, b, c are in AP
2)a, 2b, 3c are in AP
3)a, b, c are the sides of the triangle
4)none of the above
A)
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1)a, b, c are in AP
2)a, 2b, 3c are in AP
3)a, b, c are the sides of the triangle
4)none of the above
A)
| log a, log b ,log c are in AP loga + logc = 2logb logac = logb^2 ac = b^2 .....................so a,b,c are in GP. log a-log 2b ,log2b-log3c , log3c-loga are in AP loga - log2b + log3c - loga = 2 (log2b - log3c) log(3c/2b) = 2 (log (2b/3c)) 3c/2b = (2b/3c)^2 THUS, 3c/2b = 1 3c = 2b b = 3c/2 ac= b^2 = 9c^2/4 a = 9c/4 b = 3c/2 c = c since this is a pythagorean triplet (sum of any 2 is greater than 3rd) So they are the sides of a triangle. |
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Q) Find domain of 2x + 2y = 2 .
A) 2^x + 2^y = 2
2^x = 2 - 2^y
x log2 = log (2 - 2^y)
x = log(2 - 2^y) / log 2
x = log (2 - 2^y) to base 2
2 - 2^y is always <= 2...... as 2^y least value will tend to 0.
log (2 - 2^y) to base 2 will lie in (-infinity, 1)
So domain is (- infinity, 1)
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A) 2^x + 2^y = 2
2^x = 2 - 2^y
x log2 = log (2 - 2^y)
x = log(2 - 2^y) / log 2
x = log (2 - 2^y) to base 2
2 - 2^y is always <= 2...... as 2^y least value will tend to 0.
log (2 - 2^y) to base 2 will lie in (-infinity, 1)
So domain is (- infinity, 1)
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Q) A piece of hair shows the 14C activity falling by 75% of the initial activity. Given that 14C has half life of 5760 yrs, the age of the hair is ?
A) Activity is proportional to amount of nuclei.
If activity falls by 75% it means nuclei decreased by 75%
means 25% left
means 2 half lives......(because after first half life, 50%remains, and after next half life half of that)
= 2 x 5760
= 11520 years
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A) Activity is proportional to amount of nuclei.
If activity falls by 75% it means nuclei decreased by 75%
means 25% left
means 2 half lives......(because after first half life, 50%remains, and after next half life half of that)
= 2 x 5760
= 11520 years
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Q) The quadratic equation whose roots are twice those of 3x^2+5x-2=0 is ?
A) 3x^2+5x-2 = 0
= x^2 - (sum of roots)x + (product of roots)
If roots are multiplied by two, it should become:
x^2 - 2(sum of roots)x + (2*2*product of roots)
So answer is
3x^2 + 2*5x - 2*4 = 0
=> 3x^2 + 10x - 8 = 0
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A) 3x^2+5x-2 = 0
= x^2 - (sum of roots)x + (product of roots)
If roots are multiplied by two, it should become:
x^2 - 2(sum of roots)x + (2*2*product of roots)
So answer is
3x^2 + 2*5x - 2*4 = 0
=> 3x^2 + 10x - 8 = 0
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Q) Let z be a complex number ,then the minimum value of |z-2|+|z-3|+|2z-9| is ?
A) |z-2|+|z-3|+|2z-9|
Let z = a + ib
|z-2|+|z-3|+|2z-9|
= sqrt [ (a-2)^2 + b^2 ] + sqrt [ (a-3)^2 + b^2 ] + sqrt [ (2a - 9)^2 + 4b^2 ]
for minimising, b^2 will be 0.
= sqrt [ (a-2)^2 ] + sqrt [ (a-3)^2 ] + sqrt [ (2a - 9)^2 ]
= |a-2| + |a-3| + |2a - 9|
if seeing first two terms, a - 3> 0 then a > 3
then |2a - 9| will be 9 - 2a
= a - 2 + a - 3 + 9 - 2a
= 4
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A) |z-2|+|z-3|+|2z-9|
Let z = a + ib
|z-2|+|z-3|+|2z-9|
= sqrt [ (a-2)^2 + b^2 ] + sqrt [ (a-3)^2 + b^2 ] + sqrt [ (2a - 9)^2 + 4b^2 ]
for minimising, b^2 will be 0.
= sqrt [ (a-2)^2 ] + sqrt [ (a-3)^2 ] + sqrt [ (2a - 9)^2 ]
= |a-2| + |a-3| + |2a - 9|
if seeing first two terms, a - 3> 0 then a > 3
then |2a - 9| will be 9 - 2a
= a - 2 + a - 3 + 9 - 2a
= 4
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Q) In order to quadruple the resistance of a wire of uniform cross section, a fraction of its length was stretched uniformly till the final length of the wire was 1.5 times the original length. The value of the fraction elongated of the wire in comparison to original length of the wire is ?
A) let length of wire = L
let length of fraction = x
let fraction x be stretched to 't' times its length
R = 
L/A
L/A4R = (L-x)/A + tx/(A/t)
(L-x)/A + tx/(A/t)4(L/A) = L/A - x/A + t2x/A
L/A) = L/A - x/A + t2x/A3L/A + x/A = t2x/A
L/A + x/A = t2x/A3L + x = t2x
x/L = 3/ t2 - 1 ----------------------(1)
also given,
new length of wire/original length of wire = 1.5
L - x + tx / L = 1.5
0.5L = tx - x
x/L = 0.5 / t-1 ----------------------(2)
comparing (1) and (2)
0.5 / t-1 = 3 / t2-1
0.5(t+1) = 3
0.5t = 2.5
t = 5
now put this value of t in (1)
x/L = 3/ t2 - 1
x/L = 3 / 5^2 - 1
x/L = 3/24
x/L = 1/8
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Q) Let f(x)=(2x/2x^2+5x+2) AND g(x)=(1/1+x).Find the set of real values of x for which f(x)> g(x) .
A) f(x) > g(x)
f(x) - g(x) > 0
2x/ (2x^2 + 5x+ 2) - 1 / (1+x) > 0
2x / (2x+1)(x+2) - 1/1+x > 0
-3x - 2 / (2x-1)(x+2)(x+1) > 0
3x+2 / (2x-1)(x+2)(x+1) <>
A) f(x) > g(x)
f(x) - g(x) > 0
2x/ (2x^2 + 5x+ 2) - 1 / (1+x) > 0
2x / (2x+1)(x+2) - 1/1+x > 0
-3x - 2 / (2x-1)(x+2)(x+1) > 0
3x+2 / (2x-1)(x+2)(x+1) <>
x not equal to -1, -2, -1/2 as we get 0 in denominator... and x not equal to -2/3 as numerator becomes 0
Now apply wave curve method.
x E (-2, -1) U (-2/3, -1/2)
Q) LET 
be defined on the interval [0 ,1] . The odd extension of F(x) on the interval [ -1 , 1 ] is ?
A) G(x) = { - F(-x) -1<= x < 0
{ F(x) 0<= x <= 1
where G(x) is the odd extension.
WHERE
- F(-x) = - [ -4sinx + 3cosx + log (|x| + sqrt ( 1 + x^2 ) ]
= 4 sin x - 3 cos x - log (|x| + sqrt (1 + x^2) ]
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be defined on the interval [0 ,1] . The odd extension of F(x) on the interval [ -1 , 1 ] is ?A) G(x) = { - F(-x) -1<= x < 0
{ F(x) 0<= x <= 1
where G(x) is the odd extension.
WHERE
- F(-x) = - [ -4sinx + 3cosx + log (|x| + sqrt ( 1 + x^2 ) ]
= 4 sin x - 3 cos x - log (|x| + sqrt (1 + x^2) ]
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Q) One bag contains 4 white and 5 black balls. Another bag contains 6 white and 7 black balls. A ball is transferred from the first bag to the second bag. Find the probability that the ball drawn is white.
Case 1. white transferred :
Case 2. black transferred :
= 29/63
A) Prob. of white ball being transferred = 4/9
Prob. of black ball being transferred = 5/9Prob. of white ball being drawn from 2nd bag ---->
Case 1. white transferred :
7 white 7 black so 1/2
Case 2. black transferred :
6 white 8 black so 6/14 = 3/7
Total Prob. = (4/9) (1/2) + (5/9)( 3/7)
= 2/9 + 5/21 = 14 + 15 / 63= 29/63
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Q) An unbiased die is rolled until a number greater than 4 appears on the die . Find the probability that an even number of tosses are needed.
A) Prob. that a number greater than 4 appears = 2/6 = 1/3
Prob. that a number less than or equal to 4 appears = 2/3
Prob. that an even no. of tosses are needed =
(2/3)*1/3 + (2/3)^3*1/3 + (2/3)^5 * 1/3 ......
= 2/9 (1 + (2/3)^2 + (2/3)^4 + (2/3)^6 + ....)
= 2/9 [ 1 / ( 5/9) ]
= (2/9)*(9/5)
= 2/5
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A) Prob. that a number greater than 4 appears = 2/6 = 1/3
Prob. that a number less than or equal to 4 appears = 2/3
Prob. that an even no. of tosses are needed =
(2/3)*1/3 + (2/3)^3*1/3 + (2/3)^5 * 1/3 ......
= 2/9 (1 + (2/3)^2 + (2/3)^4 + (2/3)^6 + ....)
= 2/9 [ 1 / ( 5/9) ]
= (2/9)*(9/5)
= 2/5
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Q) An electric kettle takes 4A current at 220V. How much time will it take to boil 1 kg of water from temperature 20 degrees C ? (The temperature of boiling water is 100 degrees C).
A) P = IV = 880 Watts
P = work / time = 880 joules per sec
Q = m.c.del T
c is 4200 J per kg per K for water
Q = 1*4200* 80 = 336000
Time = Q / W = 336000 / 880 = 381.81 seconds
A) P = IV = 880 Watts
P = work / time = 880 joules per sec
Q = m.c.del T
c is 4200 J per kg per K for water
Q = 1*4200* 80 = 336000
Time = Q / W = 336000 / 880 = 381.81 seconds
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