Q) Let f(x) = and f(0) = 0 . Then f(1) = ?

A)

=

=

By Applying partial fractions -

=

=

=

=

BY using the statement that f(0) = 0 ........We get C = 0

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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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Q) What is iota raised to the power iota? (i = sqrt(-1) )

A)
cos@ + i.sin@ = e^(i.@)
if we take @ = pi/2:
i = e^(i.pi*/2)
Raise both sides to exponent i:
i^i = e^ (i.pi/2 . i)
i^i = e^(-pi/2)

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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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Q) Solve for x:
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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