Sponsor Area
Comparing 
......................................
Now 
Comparing 
f(x) = x - 1, a = 0, b = 5
Now 
Comparing 
we get
f(x) = 2x +3 , a = 1, b = 3
f(a+2h) = f(1+2h) = 2(1+2h)+3 = 5+4h
... ... ... ... ... ...
Now, 
Evaluate 
as the limit of a sum.
Comparing 
..................................................................
Now 
Evaluate 
as limit of sums.
Comparing 
we get
....................................................................
Now 
= 
Evaluate 
as the limit of sums.
Comparing 
.......................................................................
Now, 
Evaluate 
as the limit of a sum.
Comparing 
.............................................................
Now, 
Evaluate 
as the limit of a sum.
Comparing 
we get
...................................................
Now, 
Evaluate 
as the limit of sums.
Comparing 
...........................................................................
Now, 
Evaluate 
as the limit of a sum.
Comparing 
we get,
............................................................................
Now 
Evaluate 
as the limit of sum.
Comparing 
f(a) = f(1) = 1+1
................................................................................
Now 
Evaluate as limit of sums 
Comparing 
............................................................................
Now, 
Evaluate as limit of sums 
Comparing 
.............................................................
Now, 
Sponsor Area
Evaluate as limit of sums: 
Comparing 
.................................................................
Now, 
Evaluate the following integrals as the limit of a sum
Comparing 
with 
,
........ ............. .......... ........... ........
Now 
]
Evaluate the following integrals as the limit of a sum
Comparing 
......... ........... ........... .......... .......... .............
Now,
Evaluate 
as the limit of a sum.
Comparing 
.......................................................
Now, 
Evaluate 
as a limit of a sum.
Comparing 
we get
.............................................
Evaluate 
as the limit of sums.
Comparing 
.....................................................
Now 
Evaluate the following definite integrals as limit of sums.
Let I = 
Since 
by second fundamental theorem,
I = F(3) - F(2) = 
Sponsor Area
Let I = 
Let 
Put 
by the second fundamental theorem,
I = F(9) - F(4) = 
Evaluate the following definite integral
Let I = 
Let 
Put sin 2t = y; 
By second fundamental theorem,
If f (x) is of the form f (x) = a+b x+cx2, show that
Here 
R.H.S. = 
∴ L.H.S. = R.H.S. Hence the result.
Sponsor Area
Prove the following:
Let I = 
Put 
Multiplying both sides by x2 (x + 1), we get,
1 ≡ A x (x + 1) + B (x + 1) + C x2 ...(1)
Putting x = 0 in (1), we get,
1 = B, ∴ B = 1
Putting x + 1 = 0 or x = – 1 in (1), we get,
1 = C (– 1 )2, ∴ C = 1
(1) can be written as
1 ≡ A (x2 + x) + B(x + 1) + C x2 ...(2)
Equating coeffs. in (2) of x2, we get,
A + C = 0 ⇒ A + 1 = 0 ⇒ A = – 1
I = 
Evaluate 
Let I = 
Put 
When x = 0, y = 4 + 3 sin 0 = 4 + 0 = 4
When x = 2 
, y = 4 + 3 sin 2 
= 4 + 3 x 0 = 4 + 0 = 4
Evaluate the following integral using substitution.
Let I = 
Put 
When x = 2, y = 5
When x = 3, y = 10
Evaluate the following definite integral:
Let I = 
Put log x = y; 
When x = 1, y = log 1 = 0
When x = 2, y = log 2
I = 
Evaluate the following integral using substitution
Let I = 
Put cos x = t, ∴ sin x dx = dt ⇒ sin x dx = – dt When x = 0, t = cos 0 = 1
When x = 
Evaluate 
Let I = 
Put x5 + 1 = y, ∴ 5 x4 dx = dy When x = –1 , y = – 1 + 1 = 0 When x = 1, y = 1 + 1 = 2
Evaluate the following integrals
Let I = 
Put 
When x = – 1, y = 1 + 1 = 2 When x = 1, y = 1 + 1 = 2
Evaluate the following integrals
Let I = 
Put 
When x = 0, y = 1 – 0 = 1 When x = 1, y = 1 – 1 = 0
I = 
Evaluate the following integrals:
Let I = 
Put 
When x = – 1, y = 4 + 1 = 5 When x = 1, y = 4 + 1 = 5
Prove that: 
Let I = 
Put x = a cos θ , ∴ dx = –a sin θ dθ
When x = a, a = a cos θ ⇒ 1 = cos θ ⇒ θ = 0
When x = –a, –a = a cos θ ⇒ – 1 = cos θ ⇒ θ = 
Evaluate 
Let I = 
Put sin x - cos x = y, ∴ (cos x + sin x) dx = dy When x = 0, y = sin 0 – cos 0 = 0 – 1 = – 1
When 
Evaluate
Let I = 
Put 
...(1)
and 
...(2)
Squaring and adding (1) and (2), we get
Dividing (2) by (1), 
Prove that:
Let I = 
Put tan2 θ = t, ∴ 2 tan θ sec2 θ dθ = dt When θ = 0, t = tan2 0 = 0
When 
Prove that:
Let I = 
Put tan2 ϕ = y , ∴ 2 tan ϕ sec2 ϕ dϕ = dy When ϕ = 0, y = tan2 0 = 0
When 
Prove that:
Let
Put sin x = t, ∴ cos x dx = dt
When x = 0, t = sin 0 = 0
When 
I = 
[Do not forget to change limits of integration]
Sponsor Area
Evaluate: 
Let I = 
Put cos 2x =y, ∴ –2 sin 2x dx = dy ⇒ 2 sin 2x dx = dy
When x = 0, y = cos 0 = 1
When 
I = -
Evaluate the following definite integral:
Let
I = 
Put tan 2 x = y, ∴ 2 tan x sec2 x dx = dy
When x = 0, y = tan2 0 = 0
When 
Evaluate the following definite integral:
Let 
Put sin x – cos x = y, ∴ (cos x + sin x) dx = dy
When 
When 
Evaluate the following definite integral:
Let I =
Put sin x – cos x = t, ∴ (cos x + sin x) dx = dt
Also sin2 x + cos2 x – 2 sin x cos x = t2 ⇒ 1 – sin 2 x = t2 When x = 0, t = sin 0 – cos 0 = 0 – 1= – 1
Evaluate the following definite integral:
Let I = 
Put sin x = t, ∴ cos x dx = dt
When x = 0. t = sin 0 = 0
When 
Let I = 
Put x = tan θ so that dx = sec2 θ dθ When x = 0, tan θ = 0 ⇒ θ = 0
When x = 1, 
Let I = 
Put x = tan θ so that dx = sec2 θ dθ When x = 0, tan θ = 1 ⇒ θ = 0
When 
Let I = 
Put x = tan θ so that dx = sec2 θ dθ
When x = 0, tan θ = 1 ⇒ θ = 0
When 
I = 
If 
cosx + x sin x
x sinx
x cosx
sinx + x cosx
B.
x sinx
f(x) = 
∴ f '(x) = x sin x [∴ of first fundamental theorem]
Let I = 
...(1)
...(2)
Adding (1) and (2), we get,
2 I = 
Let I = 
...(1)
Then 
...(2)
Adding (1) and (2), we get.
Let I = 
...(1)
Then 
...(2)
Adding (1) and (2), we get,
By using the properties of definite integrals, evaluate the following integral:
Let I = 
...(1)
Then 
...(2)
Adding (1) and (2), we get,
By using the properties of definite integrals, evaluate
Let
[∴ cos x is an even function as cos (–x) = cos x]
Evaluate 
Comparing 
f (x) = sin5 x cos4 x
Now f (–x) = sin5 (–x) cos4 (–x) = – sin5 x cos4 x = – f (x)
∴ f (x) is an odd function
[
is an odd function]
By using the properties of definite integrals, evaluate the following integral:
Let I = 
For 
and for 
By using the properties of definite integrals, evaluate the following integral:
Let I = 
For 
By using the properties of definite integrals, evaluate the following integral:
LetI = 
For - 5 ≤ x ≤ – 2, x + 2 ≤ 0 ⇒ | x + 2 | = – (x + 2) and for – 2 ≤ x ≤ 5, x + 2 ≥ 0 ⇒ | x + 2 | = x + 2
By using the properties of definite integrals, evaluate the following integral:
Let I = 
For 2 ≤ r ≤ 5, x – 5 ≤ 0 ⇒ | x – 5 | = – (x – 5) and for 5 ≤ x ≤ 8, x – 5 ≥ 0 ⇒ | x – 5 | = x – 5
By using the properties of definite integrals, evaluate the following integral:
Let I = 
...(1)
By using the properties of definite integrals, evaluate the following integral:
Let I = 
...(1)
...(2)
Adding (1) and (2), we get,
Show that:
Let I = 
Adding (1) and (2), we get
Put cos x = t, ∴ – sin x dx = dt, or sin x dx = – dt When x = 0, t = cos 0 = 1 When x = 
, t = cos 
= – 1
Show that:
Let I = 
Put x = tan θ, ∴ dx = sec2 θ dθ
When x = 0, tan θ = 0 ⇒ θ = 0
When x = 1, 
...(1)
[
of (1)]
Show that:
Let I = 
...(1)
...(2)
Adding (1) and (2), we get,
Let 
Put 
When x = 0, t = tan 0 = 0
When 
Show that:
Let I = 
Put x = cos2 θ, θ dx = 2 cos θ (– sin θ) dθ = – 2 sin θ cos θ dθ = – sin 2θ dθ
When x = 0, 
When x = 1, cos2 θ = 1 ⇒ θ = 0
Show that:
Let I = 
Put cos x = t. ∴ sin x dx = – dt When x = 0, t = cos 0 = 1 When x = 
, t = cos 
= – 1
Let I = 
...(1)
Then I = 
...(2)
Adding (1) and (2), we get, 
Solution not provided.
Let I = 
........(1)
Put tan x = t, ∴ sec2 x dx = dt When x = 0, t = tan 0 = 0
When 
Show that:
Let I = 
Put x = a sin θ, ∴ dx = a cos θ dθ
When x = 0, a sin θ = 0 ⇒ sin θ = 0 ⇒ θ = 0
when x = a, 
If [ x ] stands for integral part of x, then show that 
Since 
has integral values at x = 0, 
is discontinuous at x = 0, 
Show that 
and g are defined as f(x) = f(a - x) and g(x) + g(a-x) = 4
We have
f(x) = f(a-x) ...(1)
and g (x) + g (a – x) = 4 ...(2)
Let I = 
...(3)
Evaluate 
Let I = 
...(1)
or 
...(2)
Adding (1) and (2), we get,
where 
and 
Put 2 x = t or 
When x = 0, t = 0
When 
Cor. 
Find the coordinates of the foot of perpendicular drawn from the point A
(-1,8,4) to the line joining the points B(0,-1,3) and C(2,-3,-1). Hence find the image of the point A in the line BC.
Suppose P be the foot of the perpendicular drawn from point A on the line joining
points B and C.
Let P’ (a, b, c) be the coordinates of the image of point A.
A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.
Given,
Bag X =4 white, 2 black
Bag Y=3 white, 3 black
Let A be the event of selecting one white and one black ball.
E1 = first bag selected
E2 = second bag selected
Three numbers are selected at random (without replacement) from first six positive integers. Let X denote the largest of the three numbers obtained. Find the probability distribution of X.Also, find the mean and variance of the distribution.
The first six positive integers are 1, 2, 3, 4, 5, 6.
We can select the two positive numbers in 6 × 5 = 30 different ways.
Out of this, 2 numbers are selected at random and let X denote the larger of the two numbers.
Since X is large of the two numbers, X can assume the value of 2, 3, 4, 5 or 6.
P (X =2) = P (larger number is 2) = {(1,2) and (2,1)} = 2/1
P (X = 3) = P (larger number is 3) = {(1,3), (3,1), (2,3), (3,2)} =4/3
P (X = 4) = P (larger number is 4) = {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3)} = 6/30
P (X = 5) = P (larger number is 5) = {(1,5), (51,), (2,5), (5,2), (3,5), (5,3), (4,5), (5.4)} = 8/30
P (X = 6) = P (larger number is 6) = {(1,6), (6,1), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5)} = 10/30
Find the distance of a point (2, 5, −3) from the plane 
Consider the vector equation of the plane.
Thus the Cartesian equation of the plane is
6x-3y+2z-4 = 0
Let d be the distance between the point (2, 5, -3)
Find the integrating factor for the following differential equation:
Consider the given differential equation:
Dividing the above equation by xlogx, we have, 
Evaluate:
In a set of 10 coins, 2 coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides
Let E1 E1 and A be the events defined as follows:
E1 = Selecting a coin having head on both the sides
E1= Selecting a coin not having head on both the sides
A = Getting all heads when a coin is tossed five times
We have to find P(E1/A).
There are 2 coins having heads on both the sides.
There are 8 coins not having heads on both the sides.
By Baye's Theorem, we have
How many times must a fair coin be tossed so that the probability of getting at least one head is more than 80%?
Let p denotes the probability of getting heads.
Let q denotes the probability of getting tails.
Suppose the coin is tossed n times.
Let X denote the number of times of getting heads in n trails. 
So the fair coin should be tossed for 3 or more times for getting the required probability.
Find x such that the four points A(4, 1, 2), B(5, x, 6) , C(5, 1, -1) and D(7, 4, 0) are coplanar.
A line passing through the point A with position vector 
is parallel to the vector 
Find the length of the perpendicular drawn on this line from a point p with vector 
Let the equation of the line be 
Now 
Length of the perpendicular drawn on the line from P = 
Three schools X, Y, and Z organized a fete (mela) for collecting funds for flood victims in which they sold hand-held fans, mats and toys made from recycled material, the sale price of each being Rs. 25, Rs. 100 andRs. 50 respectively. The following table shows the number of articles of each type sold:
School/Article School X School Y School Z
Hand-held fans 30 40 35
Mats 12 15 20
Toys 70 55 75
Using matrices, find the funds collected by each school by selling the above articles and the total funds collected. Also write any one value generated by the above situation.
Show that lines:
Also, find the equation of the plane containing these lines.
Two the numbers are selected at random (without replacement) from first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Find the mean and variance of this distribution.
First six positive integers are {1, 2, 3, 4, 5, 6}
No. of ways of selecting 2 numbers from 6 numbers without replacement = 
X denotes the larger of the two numbers, so X can take the values 2, 3, 4, 5, 6.
Probability distribution of X:
| X | 2 | 3 | 4 | 5 | 6 |
| P(x) | 1/15 | 2/15 | 3/15 | 4/15 | 5/15 |
| xi | P(X=xi) |  |
 |
| 2 | 1/15 | 2/15 | 4/15 |
| 3 | 2/15 | 6/15 | 18/15 |
| 4 | 3/15 | 12/15 | 48/15 |
| 5 | 4/15 | 20/15 | 100/15 |
| 6 | 5/15 | 30/15 | 180/15 |
 |
 |
Since differentiation operation is the inverse operation of integration, we have
Let us do this by integration by parts.
Therefore assume u = t; du = dt
Differentiating the above function with respect to x,
Evaluate:
We need to evaluate the integral
Comparing the coefficients, we have
2A = 1; 5A+B = 2
Solving the above equations, we have
Now consider I2:
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
The equation of the line 5x-25 =14-7y =35z can be rewritten as
Since the required line is parallel to the given line, so the direction ratio of the required line is proportional to 7,-5,1
The vector equation of the required line passing through the point (1,2-1) and having direction ratios proportional to 7,-5 1 is
Prove that if E and F are independent events, then the events E and F' are also independent.
Two events E and F are independent if
P(E ∩ F) = P(E).P(F)
Now,
P(E ∩ F') = P(E) – P(E ∩ F)
= P(E) – P(E). P(F)
= P(E)[1 – P(F)]
= P(E).P(F')
E and F' are independent events.
P(E ∩ F) = P(E).P(F)
Hence prove
Find 
⇒ 1 = X (y+2)2 + Y(y+1)(y+2)+C(y+1).....(2)
Putting y = -2 in (2)
1 = Z(-2+1)
⇒Z = -1
Putting y = -1 in (2)
1 = X(-1 +22)
⇒1 = A (1)
⇒A =1
Putting y =0 in (2)
1 = 4X + Y(2) +Z
⇒ 1=4 + 2Y-1
⇒ 1 =3 +2Y
⇒ -2 = 2Y
⇒ Y = -1
Evaluate: 
Adding (i) and (ii) we get
-sin xdx = dt
also, x = 0, t = cos 0 = 1 and x = π, t= cos π=-1
Evaluate:
For 0 < x < 1, x sinπx dx
For 1 < x <3/2, x sinπx< 0
Applying these in equation
Now applying the limits we get,
The random variable X can take only the values 0, 1, 2, 3. Given that P(X =0) = P(X = 1) = p and P(X = 2) = P(X = 3) such that Σpixi2 = 2Σpixi, find the value of p.
It is given that the random variable X can take only the values 0,1,2,3.
Given:
P(X = 0) = P( X =1) =p
P(X =2) =P(X =3)
Let P (X =2) = P(X =3) =q
Now,
P(X =0) + P(X =1) +P(X =2) +P(X =3) =1
⇒ p +p+(q+q) = 1
since Σpixi2 = 2Σpixi,
Often it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a six.
Find the probability that it is actually a six.
Do you also agree that the value of truthfulness leads to more respect in the society?
Let H1 be the event that 6 appears on throwing a die
H2 be the event that 6 does not appear on throwing a die
E be the event that he reports it is six
Yes ruthless leads to more respect in the society.
Find the area bounded by the circle x2 + y2 = 16 and the line √3y=x in the first quadrant, using integration.
The area of the region bounded by the circle, x2+y2=16, x=√3y, and the x-axis is the area OAB.
Solving x2+y2=16 and x=√3y, we have
(√3y)2+y2=16
⇒3y2+y2=16
⇒4y2=16
⇒y2=4
⇒y=2
(In the first quadrant, y is positive)
When y = 2, x = 2√3
So, the point of intersection of the given line and circle in the first quadrant is (2√3,2).
The graph of the given line and circle is shown below:
Required area = Area of the shaded region = Area OABO = Area OCAO + Area ACB
Area OCAO = 12×2√3×2=2√3 sq units
Evaluate:
Reducing it to proper rational fraction gives
Equating the coefficients we get, A = 2 and B = 3
Substituting in equation (i), we get
Evaluate:
On substituting the values of A and B, we get
Substituting I1 and I2 in ( i ), we get
Two sides of a rhombus are along the lines, x−y+1=0 and 7x−y−5=0. If its diagonals intersect at (−1, −2), then which one of the following is a vertex of this rhombus?
(−3, −9)
(−3, −8)
(1/3, -8/3)
(-10/3, -7/3)
C.
(1/3, -8/3)
As the given lines x-y +1 =0 and 7x-y-5 = 0 are not parallel, therefore they represent the adjacent sides of the rhombus.
On solving x-y+1 = 0 adn 7x - y -5 = 0. we get x =1 and y =2
Thus, one of the vertex is A(1,2)
Let the coordinate of point C be (x,y)
Then, 
⇒ x+1 =- 2 and y =-4-2
⇒ x=-3 and y =-6
Hence, coordinates of C = (-3,-6)
Note that, vertices B and D will satisfy x-y +1 =0 and 7x - y-5 = 0, therefore the coordinate of vertex D is (1/3, -8/3)
The distance of the point (1, −5, 9) from the plane x−y+z=5 measured along the line x=y=z is:
B.
the equation of the line passing through the point (1,5-9 and parallel to x =y=z is
Thus, any point on this line is of the form
(λ +1, λ-5 ,λ+9)
Now, if P (λ +1, λ-5, λ+9) is the point of intersection of line and plane, then
(λ+1) - (λ-5) +λ+9 = 5
λ +15 = 5
λ = -10
therefore coordinates of point P are (-9, -15,-1)
Hence, required distance
=
If the line 
lies in the plane lx +my -z = 9, then l2 +m2 is equal to
26
18
5
2
D.
2
Since the line
lies in the plane lx my-z =9, therefore we have 2l-m-3 = 0
[∴ normal will be perpendicular to the line]
⇒ 2l-m = 3
and 3l -2m +4 = 9
[∴ point (-3,-2-4) lies on the plane
⇒ 3l-2m = 5
On solving eqs (i) adn (ii), we get
l =1 and m=-1
therefore, l2 +m2 = 2
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls, is
A.
There seems to be ambiguity in this question. It should be mentioned that boxes are different and one particular box has 3 balls.
Then, number of ways = 
The number of points having both coordinates as integers that lie in the interior of the triangle with vertices (0,0), (0,41) and (41,0) is
901
861
820
780
D.
780
Required point (x,y) is such that it satisfies
x +y < 41
and x> 0 and y>0
Number of positive integral solutions of the equation x +y+ k = 41 will be number of intergral coordinates in the bounded region.
therefore, the total number of integral coordinates,
= 
Locus the image of the point (2,3) in the line (2x - 3y +4) + k (x-2y+3) = 0, k ε R is a
straight line parallel to X - axis
a straight line parallel to Y- axis
circle of radius 
circle of radius 
C.
circle of radius
(2x-3y +4) +k (x-2y+3) = 0 is family of lines passing through (1,2). By congruency of triangles, we can prove that mirror image (h,k) and the point (2,3) will be equidistant from (1,2).
Therefore, Locus of (h,k) is PR = PQ
⇒ (h-1)2 + (k-2)2 = (2-1)2 + (3-2)2
(x-1)2 + (y-2)2 = 2
Locus is a circle of radius = 
The distance of the point (1,0,2) from the point of intersection of the line 
and the plane x-y +z = 16 is
8
13
D.
13
Given equation of line 
and the equation of the plane is
x-y+z = 16 ... (ii)
Any point on the line (i) is
(3λ +2, 4λ-1, 12λ +2) = 16
11λ + 5 = 16
11λ =11
λ =1
therefore, Point of intersection is (5,3,14)
Now, distance between the points (1,0,2) and (5,3,2)
The equation of the plane containing the line 2x-5y +z = 3, x +y+4z = 5 and parallel to the plane x +3y +6z =1 is
2x + 6y + 12z = 13
x+3y+6z = -7
x+3y +6z = 7
2x+ 6y+12z = - 13
C.
x+3y +6z = 7
Let equation of plane containing the lines 2x- 5y +z = 3 and x+y+4z = 5 be
(2x-5y+z-3) + λ(x+y+4z-5) = 0
⇒ (2+λ)x + (λ-5)y + (4λ + 1)z -3 -5λ =0... (i)
This plane is parallel to the plane x +3y +6z = 1
On taking first two equalities, we get
6λ -30 = 3 + 12λ
-6λ = 33
λ = - 11/2
So, the equation of required plane is
The Integral 
is equal to
π-4
D.
By using the formula, 
It breaks given integral in two parts and then integrates separately.
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
3/2
5/2
7/2
9/2
C.
7/2
2x+y+2z-8 =0
2x+y+2z+5/2 = 0
Distance between two parallel planes
If the lines
are coplanar, then k can have
any value
exactly one value
exactly two values
exactly three values
C.
exactly two values
If ∫f x dx =Ψ (x),then ∫x5 f(x3)dx is equal to
C.
Given, ∫f(x) dx = Ψ (x)
Let l = ∫x5f(x3)dx
Put x3 = t
⇒ x2dx = dt/3 ...... (i) 
Statement I - The value of the integral 
is equal to π/6.
Statement II-
Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
Statement -I is True; Statement -II is False.
Statement -I is False; Statement -II is True
D.
Statement -I is False; Statement -II is True
Let
a true statement by property definite integrals.
If the integral 
then an equal to
-1
-2
1
2
D.
2
| ⇒ |
If g(x) = 
then g(x +π) equals
g(x)/g(π)
g(x) +g(π)
g (x) - g(π)
g(x). g(π)
B.
g(x) +g(π)
C.
g (x) - g(π)
Integral 
To find g(x+π) in terms of g(x) of g(π)
The value of 
is
log 2
π log 2
D.
π log 2
If the angle between the line x =
and the plane x + 2y + 3z = 4 is cos-1 
then λ equal
2/3
3/2
2/5
3/5
A.
2/3
Angle between the line and plane is
But given that angle between line and plane is
Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3
Statement-2: The number of ways of choosing any 3 places from 9 different places is 9C3 .
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is false, Statement-2 is true.
B.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement - 1 :
B1 + B2 + B3 + B4 = 10 = coefficient of x
10in (x1 + x2 + .....+ x7)4
= coefficient of x6 in (1 - x7)4 (1 - x)-4
= 4+6-1C6 =9C3
Statement -2 :Obviously 9C3
Statement-1 : The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line: 
Statement-2: The line:
bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is false, Statement-2 is true.
B.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Mid- point of AB ≡ M(1,3,5)
M lies on line
Direction ratios of AB is < 0, 6, - 4 >
Direction ratios of given line is < 1, 2, 3 >
As AB is perpendicular to line
∴ 0.1 + 6.2 - 4.3 = 0
There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is
3
36
66
108
D.
108
The number of ways in which two balls from urn A and two balls from urn B can be selected
= 3C2 x 9C2 = 3x 36 = 108
A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle θ with the positive z-axis, then θ
30°
45°
60°
75°
C.
60°
cos2α +cos2β + cos2γ = 1
α = 45°,β = 120°, γ = θ
An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is
1/3
2/7
1/21
2/23
B.
2/7
Number of ways to select exactly one ball
=3C1 × 4C1 × 2C1
Number of ways to select 3 balls out of 9 is
9C3
Required probability
The line L given by 
passes through the point (13, 32). The line K is parallel to L and has the equation 
. Then the distance
between L and K is
D.
Since, the line L is passing through the point (13,32)
Therefore,
The line K is parallel to the line L then its equation must be
Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ..., 20}.
Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is 1 85.
Statement-2: If the four chosen numbers form an AP, then the set of all possible values of common difference is {±1, ±2, ±3, ±4, ±5}.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is false, Statement-2 is true.
C.
Statement-1 is true, Statement-2 is false.
d = 1, x = 17
d = 2, x = 14
d = 3, x = 11
d = 4, x = 8
d = 5, x = 5
d = 6, x = 2
Clearly statement is wrong
Hence (3) is the correct answer
Let k be an integer such that triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point
A.
We have,
⇒ 5k2 + 13k – 46 = 0
or
5k2 + 13k + 66 = 0 (no real solution exist)
∴ k =–23/5
or k = 2
k is an integer, so k =2
As 
Therefore (2, 1/2)
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends,3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is
484
485
468
469
B.
485
Total number of ways
4C0 · 3C3 · 3C3 · 4C0 + 4C1 · 3C2 · 3C2 · 4C1 + 4C2 · 3C1 · 3C1 · 4C2 + 4C3 · 3C0 · 3C0 · 4C3= 485
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point
C.
y'= 1 at point (0, 1)
∴ Slope of normal is –1
Hence equation of normal is x + y = 1
∴ (1/2, 1/2)satisfy it.
A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one–by–one, with replacement, then the variance of the number of green balls drawn is
6/25
12/5
6
4
B.
12/5
We can apply binomial probability distribution Variance = npq
5
then which one is true?
I > 2/ 3 and J > 2
I < 2/ 3 and J < 2
I < 2/ 3 and J > 2
I > 2/ 3 and J < 2
B.
I < 2/ 3 and J < 2
Sponsor Area
Sponsor Area
as the limit of a sum.
as the limit of sum.
with 
as the limit of a sum.
we get,
as limit of sum. 
as the limit of a sum.
as the limit of a sum.
as the limit of a sum.
as the limit of a sum.
as the limit of a sum.
we get,
as the limit of a sum.
by second fundamental theorem,
find both a and b.
a = -1
we have a = -1, b = 1
equals
I = 
I = 
I = 
.
...(1)
from (1), 
or 
or 
I = 
I = 
[integrating by parts]
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
...(1) 
...(2)
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
...(1)
....(2)
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
...(1)
...(2)
.
....(1)
...(2)
...(1)
...(2)
is an even function as 
]
is an odd function of x.
...(1)
...(2)
...(1)
...(2)
...(1)
...(1)
[
]
...(2)
...(1)
...(1)
[
]
...(1)
...(1)
...(1)
...(2)
is
...(1)
is equal to
...(1)
is 
...(1)
on the three axes.
, find the value of a.
, we have
.
is equal to 
is equal to 
dx is equal to 
. If the firm employs 25 more workers, then
is equal to
is
Then F(e) equals
is
is equal to
Then f (π/2) is
is equal to
then the value of I2/I1 is