Sponsor Area
The equation of curve is y2 = x
Required area = 
Find the area of the region bounded by y2 = 4 x, x = 1, x = 4 and the x-axis in the first quadrant.
Find the area of the region bounded by y2 = x - 2, x = 4, x = 6 and the x-axis in the first quadrant.
The equation of curve is y2 = x - 2. which is right handed parabola with vertex at (2, 0).
Two lines are x = 4 and x = 6
Required area = Area ABCD
(i) The given curve and lines are
y = x2, x = 1, x = 2 and x-axis
Required area = 
(ii) The given curve and lines are
y = x4, x = 1, x = 5 and x-axis
Required area = 
Find the area of the region bounded by x2 = 4 y, y = 2, y = 4 and the y-axis in the first quadrant.
The equation of curve is x2 = 4y, which is an upward parabola.
Lines are y = 2 and y = 4
Required area = Area ABCD
which is upward parabola. The shape of 
is shown in the figure. 
Find the area of the region bounded by x2 = 16 y, y = 1, y = 4 and the y-axis in the first quadrant.
The equation of curve is
Required area = 
and x-axis. Draw a rough sketch..
...(1)
Give the rough sketch of the curve y2 = x and the line x = 4 and find the area between the curve and the line.
Find the area bounded between the curve y2 = x and the line x = 3. Draw the rough sketch also.
The equation of parabola is
y2 = x ...(1)
The equation of line is
x = 3
Also. we know that parabola is symmetric about x-axis.
∴ required area = 2 (area ORP)
The equation of parabola is y2 = x
From the given condition, area OAP under y2 = x between x = 0 and x = a = area ABQP under y2 = x between x = a and x = 4.
Find the area of the region bounded by the curve y = x2 and the line y = 4.
The equation of parabola is
y = x2 ...(1)
The equation of line is
y = 4
This parabola is symmetrical about y-axis
Find the area of the region bounded by the curve y2 = 4x and the line x = 3.
The equation of parabola is
y2 = 4 x
The equation of line is
x = 3
Also, we know that parabola is symmetric about x-axis
∴ required area = 2 (area ORPO)
Sponsor Area
Using definite integrals, find the area of the circle x2 + y2 = a2.
The equation of circle is
x2 + y2 = a2 ...(1)
Its centre is a and radius a. We know that circle x2 + y2 = a2 is symmetrical about both axes.
required area = 
The equation of ellipse is
or 
or 
[In the first quadrant]
The ellipse is symmetrical about both the axes
∴ required area = 4 (area AOB)
...(1)
[
]
Draw a graph of 
and evaluate area bounded by it.
The equation of ellipse is
or 
or 
...(1)
(in the first quadrant)
The ellipse is symmetrical about both the axes.
∴ required area = 4 (area AOB)
The equation of the ellipse is
or 
or 
...(1)
(in the first quadrant)
The ellipse is symmetrical about both the axes,
∴ required area = 4 (area AOB)
Draw a graph of 
and evaluate area bounded by it.
...(1)
[
Using definite integrals, find the area of the ellipse 
.
The equation of the ellipse is
...(1)
The ellipse is symmetrical about the axes.
∴ required area = 4 (area OAB)
[
]
Sketch the region of the ellipse and find its area, using integration.
The equation of ellipse is
...(1)
The ellipse is symmetrical about both the axes.
∴ required area = 4 (area OAB)
and the circle x2 + y2 = 4.
The equation of circle is
...(1)
The equation of line is
...(2)
From (1) and (2), we get,
∴ point of intersection of circle (1) and line (2) is P 
.
From P, draw PM ⊥ x-axis.
Also OA = radius of circle = 2
∴ A is (2, 0)
Required area = Area OMAPO = Area of ∆OMP + area MAP
= A1 + A2 ..(1 )
where 
The equation of circle is
...(1)
The equation of line is
y = x ...(2)
From (1) and (2), we get
circle (1) and line (2) meet in P(4, 4)
Radius OA of circle = 
Required area = Area OAP
= Area of 
+ area MAP
...(1)
where 
Find the area bounded by the ellipse 
ordinates x = a e and x = 0 where b2 = a2 (1 - e2) and e < 1.
The equation of ellipse
or 
or 
or 
(in the first quadrant)
Ordinates are x = 0, x = a e.
Required area = 
[
ellipse is symmetrical about x-axis]
We are to find the area of the region bounded by the curves.
and x = 2
Now 
is an upward parabola with vertex A (0, 1), y =x is straight line passing through the origin, B(2, 2) and lies below the parabola.
Now area bounded by the parabola, above the x-axis and ordinates x = 0, x = 2.
Area bounded by the line y = x, above the x-axis and ordinates x = 0, x = 2.
We are to find the area of the region bounded by the curves
y = x2 + 2,
y = x,
x = 0
and x = 3
Now y = x2 + 2 is an upward parabola with vertex (0, 2).
y = x is a straight line passing through the origin,
(3, 3) and lies below the parabola.
Now area bounded by the parabola, above the x-axis and ordinates x = 0, x = 3
Area bounded by the line y = x, above the x-axis and ordinates x = 0, x = 3
is
Find the area bounded by the curve y = x2 and the line y = x.
OR
Find the area of the region {(x. y): x2 ≤ y ≤ x}.
...(1)
The given curve is
|x| + |y| = 1
or ± x ± y = 1
The given equation represents four lines
x + y = 1, x - y = 1,
- x + y = 1 and -x - y = 1
which enclose a square of diagonal 2 units length.
Required area is symmetrical in all the four quadrants.
∴ required area = 4 (area OAB)
Find the area of the region bounded by the line y = 3 x + 2, the x-axis and the ordinates x = - 1 and x = 1.
The equation of given line is
y = 3 x + 2 ...(1)
Consider the lines
x = -1 ...(2)
and x = 1 ...(3)
Line (1) meets x-axis where y = 0
putting y = 0 in (1), we get,
line (1) meets x-axis in 
Let line (1) meet lines (2) and (3) in B and D respectively. From B, draw BC ⊥ x-axis and from D, draw DE ⊥ x-axis.
Required area = Area of region ACBA + area of region ADEA
...(1)
...(2)
curves (1) and (2) intersect in points A (4, 12) and B(-2, 3).
The equation of curve is x2 = 4 y ...(1)
which is upward parabola with vertex O.
The equation of line is
x = 4 y - 2 ...(2)
Let us solve (1) and (2)
Putting x = 4y - 2 in (1), we get
From A, draw AM ⊥ x-axis and from B. draw BN ⊥ x-axis.
Required area = area AOB
= Area of trapezium BNMA - (area BNO + area OMA)
= 
The equation of parabola is
y2 = x ...(1)
The equation of line is
x + y = 2 ...(2)
From (2), y = 2 - x ...(3)
Putting this value of y in (1), we get,
(2 - x)2 = x
or x2 - 4 x + 4 = x or x2 - 5 x + 4 = 0
∴ (x - 1) (x - 4) = 0
∴ x = 1, 4
∴ from (3), y = 1, - 2
∴ parabola (1) and line (2) intersect in the points A (1, 1), B (4, - 1)
Also line (2) meets x-axis in C (2,0)
Required area is shaded.
Area above x-axis = area AOL + area ALC
Area below x-axis = Area OBM - area CBM
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and the x-axis.
OR
Draw the rough sketch and find the area of the region:
{(x, y): x2 < y < x + 2}
The equation of parabola is
x2 = y ...(1)
The equation of line is
y = x + 2 ...(2)
From (1) and (2), we get,
x2 = x + 2
∴ x2 - x - 2 = 0
⇒ (x - 2) (x + 1) = 0 ⇒ x = 2, -1
∴ from (2), y= 2 + 2, -1 +2 = 4, 1
∴ parabola (1) meets line (2) in two points A (2, 4) and B (-1. 1).
From A. draw AM ⊥ x-axis and from B. draw BN ⊥ x-axis.
Required area = Area AOB
= 
Draw a rough sketch of the curves y = sin x and y = cos x as x varies from 0 to 
and find the area of the region enclosed by them and the x-axis.
Let OB and CA represent the curves y = sin x and y = cos x as x varies from 0 to 
. The two curves intersect at D where
Required area = Area OAB + area OCA - area ODA
Sponsor Area
Find the area bounded by the curve y = cos x between x = 0 and x = 2 
.
The equation of curve is
y = cos x
Its rough sketch from x = 0 to x = 2 
is shown in figure.
Required area = area of the region OABO + area of the region BCDB + area of the region DEFD.
Find the area bounded by the curve y = sin x between x = 0 and x = 2 
.
is shown is the figure.
Using integration, find the area of the triangular region whose sides have the equations y = 2 x + 1, y = 3 x + 1 and x = 4.
The equations of the sides are
y = 2 x + 1 ...(1)
y = 3 x + 1 ...(2)
and x = 4. ...(3)
Subtracting (1) from (2), we get,
Putting x = 0 in (1), we get y = 0+1 = 1
line (1) and (2) intersect in A(0, 1)
From (1) and (3), we get,
line (1) and (3) intersect in B (4, 9)
From (2) and (3), we get,
x = 4, y = 12 + 1 = 13
∴ lines (2) and (3) intersect in C (4, 13)
∴ vertices of the triangle ABC are A(0, 1), B (4, 9), C (4, 13)
Required area = Area of ∆ ABC = Area of region AOMC - area of region AOMB
The equations of the sides are
2 x + y - 4 = 0 ...(1)
3 x - 2 y - 6 = 0 ...(2)
x - 3 y + 5 = 0 ...(3)
Solving (1) and (2), we get
Solving (2) and (3), we get,
Solving (1) and (3), we get,
or 
From B. draw BL ⊥ x-axis and from A. draw AM ⊥ x-axis.
Required area = Area of 
= Area of region BLMA - area of 
- area of 
Using integration, find the area of the region bounded by (2, 5), (4, 7) and (6, 2).
Let A(2, 5), B(4, 7), C(6, 2) be vertices of ∆ABC. from A, B, C draw AL. BM. CN ⊥ x-axis.
The equation of AB is
or 
or 
or 
The equation of BC is
or 
The equation of CA is
or 
Area of 
= Area ALMB+ area BMNC - area ALNC
The equation of AB is
or 
The equation of BC is
or 
The equation of CA is
From C, draw CD 
x-axis.
Required area = area of quad. ABCD - area of 
Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 1), (0, 5) and (3, 2).
Let A(-1, 1), B(0, 5), C(3, 2) be the vertices of the given triangle.
The equation of AB is
or y = 4x + 5 ...(1)
The equation of BC is
or 
...(2)
The equation of CA is
or 
Required area = Area of ∆ ABC
= Area of region AMOB + area of region BONC - area of region AMNC
The given vertices are A (3, 0), B (4, 6), C (6, 2). The equation of AB is
or 
or 
...(1)
The equation of BC is
or 
or 
or 
...(2)
The equation of CA is
or 
or 
...(3)
From B. C draw BM. CN ⊥ s on x-axis.
Required area = Area of ∆ AMB + Area of trap. MNCB - area of ∆ANC
The given vertices are A(3, 0), B(4, 5), C(5, 1).
The equation of AB is
or y = 5x - 5 ...(1)
The equation of BC is
or 
or 
or 
...(2)
The equation of CA is
or 
or 
...(3)
From B. C draw BM. CN ⊥s on x-axis
Required area = Area of ∆AMB + area MNCB — area of ∆ANC
Using integration, find the area of the region bounded by the triangle whose vertices are (1, 0), (2, 2) and (3, 1).
Using integration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).
Let A(-1, 0), B(1, 3), C(3, 2) be the vertices of the given triangle.
The equation of AB is
or 
...(1)
The equation of BC is
or 
or 
...(2)
The equation of CA is
or 
...(3)
From B, draw BM ⊥ x-axis and from C, draw CN ⊥ x-axis.
Required area = Area of ∆ABC
= Area of 
+ area of region BMNC - area of 
...(2)
from (1), 
the points of intersection of the given circles are 
The equation of two circles are
...(1)
and 
...(2)
Centre of circle (1) is O (0, 0) and radius OA = 1
Centre of circle (2) is A (1, 0) and radius AO = 1
Subtracting (1) from (2), we get,
Putting 
in (1), we get,
points of intersection of circles (1) and (2) are
Required area = Area of region OQAP
= 2 (area of region OMAP)
= 2 (area of region OMPO + area of region MAPM)
...(2)
and radius CB = 1
in (1), we get,
points of intersection of circles (1) and (2) are 
and 
The equations of curves are
...(1)
and 
...(2)
From (2), 
...(3)
Putting this value of y in (1),
or 
From P, draw PM 
x-axis.
Required area = Area OAPB
= Area OBPM - area OAPM
The equations of curves are
...(1)
and 
...(2)
From (1) and (2), we get
curves (1) and (2) intersect in O(0, 0), P(9, 9)
From P, draw 
x-axis.
Required area = Area OAPBO = Area OBPM - area OAPM
Calculate the area of the region bounded by the y = x2 and x = y2.
...(1)
...(2)
....(3)
x-axis
...(1)
...(2)
...(3)
The equations of curves are
...(1)
and 
...(2)
From (2), 
...(3)
Putting this value of y in (1),
or 
∴ x = 0, 4 a
∴ from (3), y = 0, 4 a
∴ curves (1) and (2) intersect in O (0, 0), P (4 a, 4 a).
From P. draw PM ⊥ x-axis.
Required area = Area OAPB
= Area OBPM - area OAPM
The equation of given curves are
y2 - x + 1 ...(1)
and y2 = -x + 1 ...(2)
From (1) and (2), we get,
x + 1 = -x + 1 ⇒ 2 x = 0 ⇒ x = 0
Putting x = 0 in (1). we get,
y2 = 1 or y = -1, 1
∴ points of intersection of given curves are (0, -1) and (0, 1).
Now (1) and (2) represent two parabolas having vertices at (-1, 0) and (1, 0).
Required area = 2 [area CAO + area OBC]
The given region is
{( x, y): 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
Thus region is the intersection of the following regions:
R1 = {(x, y) : 0 ≤ y ≤ x2 + 1}
R2 = { (x, y) : 0 ≤ y ≤ x + 1}
R3 = {(x, y) : 0 ≤ x ≤ 2}
The function with graph in the figure is
Consider the equations
y = x2 + 1 ...(1)
and y = x + 1 ...(2)
Putting y = x + 1 in (1), we get
x + 1 = x2 + 1, or x = x2 ⇒ x2 - x = 0 ⇒ x(x - 1) = 0
∴ x = 0, 1
∴ from (2), y = 1, 2
∴ curve (1) and (2) intersect in the points P (0, 1) and Q (1, 2).
The region considered is bounded by
y = f(x),
y = 0
x = 0
and x = 2
required area = 
Find the area of the region {(x, y): x2 + y2 ≤ 1 ≤ x + y}.
The given region is
Consider the equations
...(1)
and x + y = 1 ...(2)
From (2), y = 1 - x ...(3)
Putting this value of y in (1), we get,
Required area = Area of shaded region = 
Find the area of the region bounded by the circle x2 + y2 = 1 and x + y = 1. Also draw a rough sketch.
Consider the equations
...(1)
and 
...(2)
From (2), y = 1 - x ...(3)
Putting this value of y in (1), we get,
circle (1) and st . line x + y = 1 intersect in the points A(1, 0) and B(0, 1).
Required area = Area of shaded region = 
Find the area of the region {(x, y): x2 ≤ y ≤ |x|}.
Or
Find the area of the region bounded by the parabola y = x2 and y = |x|.
The given region is
This region is the intersection of the following regions
Consider the equations
...(1)
y = x ...(2)
and y = -x ...(3)
From (1) and (2), we get
from (2), y = 0, 1
∴ curve (1) and (2) intersect in the points O (0, 0) and A (1, 1).
Similarly, curves (1) and (3) intersect in the points O (0, 0) and B (-1, 1)
Required area = area of shaded region = 2 (area of region OAO)
Given region is
{(x, y): y2 ≤ 8 x, x2 +y2 ≤ 9}
Consider the equations
y2 = 8 x ...(1)
and x2 + y2 = 9 ...(2)
From (1) and (2). we get,
x2 + 8 x = 9 or x2 + 8 x-9 = 0⇒ (x + 9)(x - 1) = 0
⇒ x = - 9, 1
which gives the abscissa of the points of intersection P and Q.
Rejecting negative value of x, we get, x = 1
Required area = Area of shaded region
= 2 (area of region APOA)
= 2[area of region AMPA + area of region MOPM]
Given region is
Consider the equations
...(1)
and 
i.e., 
...(2)
From (1) and (2), we get,
curve (1) and (2) intersect in the points
From P, draw 
Here OA = 
Required area = Area of shaded region
= 2 (area of region OAPO)
= 2 [area of region OMPO + area of region MAPM]
The equation of circle is
x2 + y2 = 8 x ...(1)
The equation of parabola is
y2 = 4 x ...(2)
(1) can be written as
(x2 - 8 x) + y2 = 0 or (x2 - 8 x + 16) + y2 = 16
or (x - 4)2 + y2 = (4)2 ...(3)
which is a circle with centre C(4, 0) and radius = 4.
From (1) and (2), we get,
x2 + 4 x = 8 x or x2 - 4 x = 0 ⇒ x(x - 4) = 0
∴ x = 0, 4
∴ from (2), y =0, 4
∴ points of intersection of circle (1) and parabola (2) and 0(0, 0) and P(4, 4), above the x-axis.
Required area = area of region OPQCO
= (area of region OCPO) + (area of region PCQP)
We are able to find the area included between the curves
...(1)
and 
...(2)
From (1) and (2), we get,
circle (1) and line (2) intersect in points A (1, 0) and 
Required area is shaded.
Required area = 
The equation of circle is
...(1)
The equation of parabola is
...(2)
From (1) and (2),
or 
Rejecting x = -8 as parabola lies in 1st or 4th quadrant , we get x =2
When 
Required Area = Area of the circle - area of circle interior to the parabola 
Consider the equations
...(1)
and 
i.e., 
...(2)
From (1) and (2), we get,
From P, draw 
Here 
Required area = Area of shaded region
= 2 (area of region OAPO) = 2 [area of region OMPO + area of region MAPM]
Given region is
Consider the equations
...(1)
i.e. 
...(2)
From (1) and (2), we get,
Rejecting negative value of x, we get,
which gives the abscissa of the points of intersection P and Q.
Required area = Area of shaded region = 2 (area of region OAPO)
= 2[area of region OMPO + area of region MAPM]
where 
Given region is
Consider the equations
...(1)
i.e., 
...(2)
From (1) and (2), we get,
Rejecting negative values of x, we get
which gives the abscissa of the points of intersection P and Q.
Required area = Area of shaded region
= 2 [area of region APOA]
= 2 [area of region AMPA + area of region MOPM ]
= 
= 
Find the area of the smaller part of the circle x2 + y2 = a cut oil by the line 
The equation of circle is
...(1)
The equation of line is
...(2)
From (1) and (2), we get,
is 
and the straight line 
.
The equation of ellipse is
...(1)
The equation of line is
...(2)
Let line (2) meet ellipse (1) in A (3, 0) and B(0, 2).
Area of shaded region = Area of region OAB - area of 
Using integration, find the area of the smaller region bounded by the curve 
and the straight line 
The equation of ellipse is
...(1)
The equation of line is
...(2)
Let line (2) meet ellipse (1) in
A(4, 0) and B(0, 3)
Area of shaded region = Area of region OAB - area of 
Find the area of smaller region bounded by the ellipse 
and the straight line 
Consider the equations
...(1)
and 
...(2)
Line (2) meets ellipse (1) in A (a, 0) and B (0, b)
Area of shaded region = Area of region OAB area of ∆OAB
which is required area.
, where x is a fixed positive number, Hence, prove that the area of a circle of radius r is 
Let I = 
Let 
Now 
represents the area bounded by the circle 
the x-axis and the ordinate x = 0 and the ordinate x = r, i.e., it represents the area of the region in the first quadrant. 
Find the whole area of the circle x2 + y2 = a2.
A.
The equation of circle is
...(1)
We are to find the area of the circle lying between the lines x = 0 and x = 2 in the first quadrant.
Required area = 
2
B.
The equation of curve is
y2 = 4.x
We are to find the area bounded by the curve 
Required area = 
Sponsor Area
B.
The equation of circle is
...(1)
Its centre is O(0, 0) and radius OA = OB = 2
The equation of line is
x + y = 2 ....(2)
or 
line passes through A(2, 0), B(0, 2).
Required area = area of circle in first quadrant - area 
Area lying between the curves y2 = 4x and y = 2x is
B.
The equations of the curves are
...(1)
and 
...(2)
From (1) and (2),
or 
.
Required area = 
-9
D.
We want to find the area bounded by the curve y = x3, the x-axis and the ordinates x = - 2 and x = 1.
A rough sketch of the curve y = x3 is given in the figure.
Required area = 
0
C.
We want to find the area bounded by the curve 
x-axis and the ordinates x = -1 and x = 1
Required area = 
C.
The equation of circle is
...(1)
The equation of parabola is
...(2)
From (1) and (2),
or 
x = -8, 2
Rejecting x = -8 as parabola lies in 1st or 4th quadrant, we get x = 2
When x = 2, 
(1) and (2) intersect in 
Required Area = Area of the circle - area of circle interior to the parabola
The area bounded by the x-axis, y = cosx and y = sin x when 
B.
The two curves y = sin x and y = cos x meet where sin x = cos x , i.e., where x = 1
Required area (show shaded)
The equation of curves are
...(1)
and 
...(2)
From (2), 
...(3)
Putting this value of y in (1),
or 
From P. draw PM ⊥ x-axis.
Required area = Area OAPB = Area OBPM - area OAPM
Now, the area of the region OAQBO bounded by curves 
and 
Again, the area of the region OPQAO bounded by the curves x2 = 4 y, x = 0, x = 4 and x-axis
Similarly, the area of the region OBQRO bounded by the curve 
...(3)
From (1), (2) and (3), it is clear that the area of the region OAQBO = area of the region OPQAO = area of the region OBQRO, i.e., urea bounded by parabolas y2 = 4 x and x2 = 4 y divides the area of the square in three equal parts.
The given differential equation is
Its order is 2 and degree is 2.
.
so its order is one. It is a polynomial equation in 
and the highest power raised to 
is one, so its degree is one. 
So its order is two. It is a polynomial equation in 
and the highest power raised to 
is one, so its degree is one.
is one, so its degree is 1.
The given differential equation is
The highest order derivative present in the given differential equation is 
therefore, its order is 2. Since the given differential equation is not a polynomial in 
therefore, its degree is not defined.
and index of its highest power is 1.
The given differential equation is
y' + y = ex
The highest order derivative present in the given differential equation is y' and the index of its highest power is 1.
∴ the given differential equation is of order 1 and degree 1.
The given differential equation is
y'' + (y')2 + 2 y = 0
The highest order derivative present in the given differential equation is y' and the index of its highest power is 1.
∴ the given differential equation is of order 2 and degree 1.
The given differential equation is
y'' + 2 y' + sin y = 0
The highest order derivative present in the given differential equation, is y' and the index of its highest power is 1.
∴ the given differential equation is of order 2 and degree 1.
For each of the differential equations given below, indicate its order and degree (if defined).
is one, therefore, degree of the given differential equation is 1.
For each of the differential equations given below, indicate its order and degree (if defined).
therefore, order of the equation is 1. Also, index of the highest power of 
is 3. So degree is 3.
For each of the differential equations given below, indicate its order and degree (if defined).
therefore, order of the equation is 4. As 
occurs in the equation, therefore, degree of the given equation is not defined. 
state also if this is linear or non-linear.
The given differential equation is
or 
or 
It is of order 1, degree 2 and non-linear.
Find the order and degree of the equation y = px + 
when 
State also if this is linear or non-linear.
The given equation is
or 
or 
or 
It is of order 1, degree 2 and non-linear
Determine the order and degree of the differential equation:
Determine the order and degree of the differential equation:
The given differential equation is
Here highest differential coefficient is 
and its degree is one.
∴ order of given differential equation is 4 and degree 1.
is
3
2
1
not defined
D.
not defined
The given differential equation is
is
2
1
0
not defined
A.
2
The given differential equation is 
The highest order derivative present in the differential equation is 
.
∴ its order is 2
...(1)
is a solution of the differential equation
...(1)
Again differentiating w.r.t.x, we get,
is a solution of the differential equation
...(1)
...(2)
...(3)
L.H.S. = 
= 0 - R.H.S.
is a solution of the differential equation
.
Here 
...(1)
...(2)
Again differentiating w.r.t. x, we get,
...(3)
L.H.S. = 
is a solution of the differential equation
: 
y = ex + 1 ; ...(1)
∴ y' = ex ...(2)
and y'' = ex ...(3)
L.H.S. = y'' – y'
= ex – ex [∵ of (2), (3)]
= 0 = R.H.S.
∴ y = ex + 1 is a solution of y'' – y' = 0
: 
y = x2 + 2 x + C
∴ y' = 2 x + 2
⇒ y' – 2 x – 2 = 0
∴ y = x2 + 2 x + C is a solution of y' – 2 x – 2 = 0
y = cos x + C
∴ y' = – sin x
⇒ y' + sin x = 0
∴ y = cos x + C is a solution of y' + sin x = 0
: 
...(1)
...(2)
R.H.S. = 
is a solution of 
Sponsor Area
Differentiate w.r.t.x, we get
or 
: 
y – cos y = x ...(1)
Differentiate both sides w.r.t. x, we get,
y' + sin y . y' = 1
⇒ y'(1 + sin y) = 1 ...(2)
L.H.S. = (y sin y + cos y + x) y'
= (y sin y + y) y' [∵ of (1)]
= y (1 + sin y) y' = y. 1 [∵ of(2)]
= y
= R.H.S.
∴ y – cos y = x is a solution of (y sin y + cos y + x) y' = y.
: 
Here, 
is a solution of the given differential equation.
The given differential equation is
y = ex (a cos x + b sin x)
⇒ e–x y = a cos x + b sin x ...(1)
Differentiating (1) twice w.r.t. .x. we get
and 
or 
[
]
or 
or 
Hence the result.
: 
...(1)
: 
...(1)
0
2
3
4
D.
4
Since the number of arbitrary constants in the general solution of a differential equation of nth order is n.3
2
1
0
D.
0
Since there is no arbitrary constant in particular solution.The equation of the family of lines is
y = mx + c
Again differentiating w.r.t. 'x' we get, 
This is required differential equations.
Find the differential equations from y = k esin–1 x + 3.
The given equation is 
...(1)
which is required differential equation.
or 
or 
...(2)
The equation of given curve is
(x – a)2 – y2 = 1 ...(1)
Differentiating both sides w.r.t x, we get.
or 
Putting value of x – a in (1), we get,
which is required differential equation.
The equation of curve is
...(1)
Differentiating w.r.t., x successively two times, we get.
or 
...(2)
...(3)
and 
...(4)
From (3), (4), we get,
which is the required differential equation.
...(1)
...(2)
...(3)
The equation of given curve is
y2 = 4 a (x – b)
Differentiating w.r.t. x, we get,
or 
Again differentiating w.r.t x, we get.
or 
which is required differential equation..
The equation of given curve is
...(1)
Differentiating w.r.t.x, we get,
or 
....(2)
Again differentiating both sides w.r.t x we get,
or 
which is required differential equation.
The given equation is
...(1)
Differentiating both sides w.r.t. x, we get,
...(2)
Putting value of 2a from (2) in (1), we get,
or 
which is the required differential equation.
Find the differential equation that will represent the family of curves given by (a, b: parameter):
y = ax3
...(2)
or 
Find the differential equation that will represent the family of curves given by (a, b: parameter):
x2 + y2 = a x3
...(1)
...(2)
...(3)
...(4)
Find the differential equation that will represent the family of curves given by (a, b: parameter):
y = eax
...(1)
...(2)
or 
...(1)
...(2)
The given equation is 
...(1)
Differentiating w.r.t.x, we get, 
Again differentiating w.r.t. x, we get,
or 
which is required differential equation.
The given equation is
y = A sin mx + B cos mx. ...(1)
The given equation is 
...(1)
Differentiating w.r.t.x, we get,
or 
...(2)
Again differentiating w.r.t.x,
or 
or 
Which is the required differential equation.
...(1)
...(2)
...(3)
...(1)
...(2)
...(3)
Here, 
...(1)
Differentiating w.r.t.x, we get,
...(2)
Dividing (2) by (1), we get.
Hence the result.
The given equation is
y = 2 (x2 - 1) + c e-x2 or yex2 = 2 (x21) ex2 + c
Differentiating both sides w.r.t. x, we get.
or 
which is required differential equation.
The given equation is
...(1)
...(2)
...(3)
From (3), B = 
From (2), 
Putting value of A and B in (1), we get,
which is the required differential equation.
...(1)
...(2)
...(3)
which is the required differential equation. 
The given equation is
...(1)
which is required differential equation.
...(2)
...(3)
...(4)
...(5)
+
which is required differential equation.nThe equation of family of curves is
y = e2x (a + b x) ...(1)
Differentiating both sides w.r.t. x, we get,
y1 = e2x b + (a + b x). 2 e2x
or y1 = b e2x + 2 y [∵ of(1)]
∴ y1 – 2 y = be2x ...(2)
Again differentiating w.r.t. x, we get,
y2 – 2 y1 = 2 b e2x
or y2 – 2 y1 = 2 (y1 – 2 y) [∵ of (2)]
or y2 – 2 y1 = 2 y1 – 4 y
or y2 – 4 y1 + 4 y = 0, which is required differential equation.
...(2)
, which is required differential equation.
The given equation is
y = ae 2x + be– 3x ...(1)
Differentiating both sides w.r.t.x, we get
...(2)
Again differentiation both sides w.r.t.x, we get.
...(3)
Multiplying (2) by 2 and subtracting from (3), we get,
...(4)
Multiplying (2) by 3 and adding it to (3), we get,
...(5)
From (1), (4) and (5), we get,
or 
or 
which is required differential equation.
The given equation is
y = aex + be2x + ce3x ...(1)
∴ y1 = (aex + 2be2x – 3ce – 3x) ...(2)
Subtracting (1) from (2), we get,
y1 – y = b e2x – 4 c e–3x ...(3)
∴ y2 – y1 = 2 b e2x + 12 c e–3x ...(4)
Multiplying (3) by 2 and subtracting from (3), are get,
y2 – 3 y1 + 2 y = 20 c e–3x ...(5)
∴ y3 – 3 y2 + 2 y1 = – 60 c e –3x ...(6)
Multiplying (5) by 3 and adding to (6), we get ,
(y3 – 3 y2 + 2 y1) + 3 (y2 – 3 y1 + 2 y) = 0 or y3 – 7 y1 + 6 y = 0
which is required differential equation.
Obtain the differential equation from the equation y = ex (a cos 2x + b sin 2x). where a and b are arbitrary constants.
The given equation is
y = ex (a cos 2x + b sin 2x) ... (1)
Differentiating both sides w.r.t x, we get,
...(2)
Again differentiating w.r.t x,
or 
or 
or 
which is required differential equation.
...(2)
...(3)
...(4)
is the required differential equation.
Let the radius of circle = a
Since the circle has centre on x-axis and passes through the origin.
∴ its centre is (a, 0)
The equation of circle is
( x – a)2+( y – 0)2 = a2 or x2 –2 a x + a2 + y2 = a2
∴ x2 + y2 – 2 ax = 0 ...(1)
Differentiating both sides w.r.t. x, we get.
Putting this value of a in (1), we get.
which is required differential equation.
...(1)
or 
The equation of the family of circles having centre (0, b) on y-axis and radius 3 units is
x2 + (y – b)2 = 9 ...( 1)
Differentiating both sides w.r.t x, we get,
Putting this value of y - b in (1), we get,
or 
or 
which is required differential equation.
Let r be the radius of the circle whose centre is C and CM ⊥ x-axis, CN ⊥ y-axis.
∵ circle touches both the axes
∴ CM = CN = r
∴ C is (r, r)
∴ equation of circle is
(x - r)2 + (y – r)2 = r2 ...(1)
Differentiating both sides w.r.t.x,
or 
or 
Putting this value of r in (1), we get
or 
or 
or 
which is the required differential equation. 
Equation representing the family C is
(x + a)2+ (y – a)2 = a2 ...(1)
or x2 + y2 + 2 ax – 2 a y + a2 = 0 ...(2)
Differentiating equation (2) with respect to x, we get
or 
or 
Substituting the value of a in equation (1), we get
or 
or 
or 
which is the differential equation representing the given family of circles.
Let C denote the family of circles touching x-axis at origin. Let (0, a) be the coordinates of the centre of any member of the family where ‘a’ is an arbitrary constant.
∴ the equation of family C is
x2 + (y – a)2 = a2
∴ x2 + y2 + a2 – 2 a y = a2
or x2 + y2 – 2 a y = 0 ...(1)
a being arbitrary constant
Differentiating w.r.t. x, we get
...(2)
[Multiplying by y]
Multiplying (1) by 
...(3)
Subtracting (3) from (2), we get,
or 
or 
which is required differential equation.
The equation of family of circles touching y-axis at origin is
(x – h)2 + y2 = h2 where h is arbitrary constant ...(1)
[∵ if (h, 0) is centre of any member of family, then its radius = h]
Differentiating (1) w.r.t.x, 
⇒ x – h + y y1= 0 ⇒ h = x + y y1
Putting value of h in (1), we get,
(x – x – y y1)2 + y2 = (x + y y1)2 ⇒ y2 y12 + y2 = y2 + 2 x y y1 + y2 y12
⇒ x2 – y2 + 2 x y y1 = 0 , which is the required differential equation.
Let P denote the family of given parabolas and let (a, 0) be the 'focus of a member of the given family, where a is an arbitrary constant. Therefore, equation of family P is
y2 = 4 a x ...(1)
Differentiating both sides with respect to x, we get
...(2)
Substituting the value of 4 a from equation (2) in equation (1), we get
which is the differential equation of the given family of parabolas.
Let P denote the family of parabolas and let S (0. a) be the focus of a member of the given family, where a is an arbitrary constant.
∴ equation of family P is
a2 = 4 a y ...(1)
Differentiating both sides w.r.t. x, we get.
Putting value of 4a in (1), we get,
which is the required differential equation. 
...(1)
...(2)
...(3)
...(4)
...(1)
...(2)
...(3)
...(4)
...(1)
...(2)
...(3)
B.
The given equation is
...(1)
(B) is the correct answer.
C.
Here y = x
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Solve 
is required solution.
Solve 
The given differential equation is
Separating the variables, we get,
Integrating, 
or 
or 
is the solution. 
which is required solution.
is the solution.
Solve the differential equation
which is required solution.
The given differential equation is
Separating the variables, we get,
Integrating, 
This is required solution.
The given differential equation is
Separating the variables, we get, 
Integrating, we get, 
where c is an arbitrary constant.
or 
The given differential equation is
Separating the variables, we get,
or 
or 
is the required solution.
is an arbitrary constant. 
where c is an arbitrary constant.
is the required solution. The given differential equation is
(x y2 + 2x) dx + (x2 y + 2y) dy = 0
or 
Integrating, 
Solve:
The given differential equation is
Separating the variables, we get, 
is the required solution.
Solve:
x (1 + y2) dx + y (1 + x2)dy = 0
The given differential equation is
x(1 + y2) dx + y(1 + x2)dy = 0 
Integrating, 
which is required solution.
Solve:
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0.
For the following differential equation, find the general solution:
which is the required solution.
For the following differential equation, find the general solution:
which is required solution..
For the following differential equation, find the general solution:
sec2 x tan y dx – sec2 y tan x dy = 0.
For the following differential equation, find the general solution:
sec2 x tan y dx + sec2 y tan x dy = 0.
Solve:
3ex tan y dx + (1 – ex) sec2y dy = 0.
The given differential equation is
3ex tan y dx + (1 – ex) sec2y dy = 0
or 
Solve:
tan y dx + sec2 y tan x dy = 0.
Solve:
The given differential equation is
Separating the variables, we get,
Integrating, 
which is the required solution.
Solve:
...(1)
Solve:
Solve:
Solve:
Show that the general solution of the differential equation 
is given by (x + y + 1) = A (1 – x – y – 2 x y), where A is parameter.
The given differential equation is
Integrating, we get
Taking tangents on the two sides, we get
where 
Solve
The given differential equation is 
Separating the variables, we get,
or 
which is required solution.
Solve
The given differential equation is
Separting the variables, we get, 
Let 
Put cosx =t, 
Let 
Solve:
...(1)
Show that the given differential equation is homogeneous and solve it.
(x2 – y2) dx + 2xy dy = 0
given that y = 1 when x = 1.
Solve the differential equation:
...(1)
Solve the following initial value problem:
(1 + x y) y dx + (1 – x y) x dy = 0, y (1) = 1.
...(1)
from (1),
given that y = 1, when x = 0.
The given differential equation is
(1 + e2x ) dy + (1 + y2 ) ex dx = 0 or (1 + e2x ) dy = - (1 + y2 ) ex dx
Let I = 
Put 
Now, x = 0, y = 1
...(1)
Solve the differential equation;
x (1 + y2 ) dx – y (1 + x2 ) dy = 0 given that y = 0 when x = 1.
...(1)
given that y = 0 when x = 0.
The given differential equation is
Separating the variables, 
...(1)
when y = 0, x = 0, then from (1), we get,
from (1), 
which is required solution.
Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2 x2 + 1) dx (x ≠ 0).
...(1)
...(1)
For the differential equation 
find the solution curve passing through the point (1, -1).
...(1)
is slope of tangent to the curve at point (x, y).
...(1)
represents the required displacement. 
represents the required displacement. 
(i) Time-scalar
(ii) Volume-scalar
(iii) Force-vector
(iv) Speed-scalar
(v) Density-scalar
(iv) Velocity-vector
(i) Vectors 
are coinitial
(ii) Vectors 
are equal
(iiii) Vectors 
are collinear but not equal.
(i) Collinear vectors : 
(ii) Equal vectors: 
(iii) Coinitial vectors: 
are two non-collinear vectors having the same initial point. What vectorn is represented by 
?
Let 
Now, 
is the vector represented by the diagonal OC.
and 
respectively, show that 
Here P.V. of 
. P.V. of B = 
, P.V. of C = 
.
P.V. of D = 
Find 
.
Let O be origin of vectors. Then
Now, 
and 
form the three sides of a triangle. What are the other possibilities?
Let 
Now, 
Other possibilities are
Other possibilities are
then
?Here 
(i) In 
In 
(ii)
are the position vectors of the vertices P, Q and R respectively with reference to the origin of reference O ; find the position vector of S with reference to O.We have 
Now position vector of S with reference to O = 
If 
are the vectors determined by two adjacent sides of a regular hexagon, find the vectors determined by the other sides taken in order.
If 
be the vectors represented by the sides of a triangle, taken in order, then prove that 
Let ABC be a triangle such that
L.H.S. = 
If 
then prove that the points A, B and C are collinear.
We have 
( Triangle Law of Vectors)
are parallel vectors.
But B is the common point . 
are along the same line
∴ point A, B and C are collinear.
]
In 
...(1)
In 
...(2)
Adding (1) and (2), we get
Take A is origin. Let 
be position vectors of B, C respectively so that
L.H.S. = 
= (P.V. Of E - P.V. of B) + (P.V. of C - P.V. of D)
If 
are distinct non-zero vectors represented by directed lines from the origin to the points A, B, C and D respectively and if 
, then prove that ABCD is a parallelogram.
Here 
are the position vectors of A, B, C, D respectively with reference to O as origin
Now, 
.We have
...(1)
...(2)
...(3)
Adding (1), (2), (3), we get,
Let ABCD be the quadrilateral and O be the point of intersection of the diagonals AC and BD. Let 
. O is the midpoint of both AC and BD if and only if 
or if and only if 
i.e., if any only if 
We also know that a quadrilateral is a parallelogram if and only if its opposite sides are equal and parallel.
Hence the result follows.
are collinear, where 
are three non-coplanar vectors.
Let 
be position vectors of A, B and C respectively.
Now, 
Now, 
But these vectors have the same initial point
∴ point A, B and C are collinear.
are collinear, where 
are three non-coplanar vectors.
Let 
be the position vectors of A, B and C respectively.
Also, 
But the vectors have the same initial point point A, B and C are collinear.
If A and B are two points in space and if they become respectively A', B' on a translation, show that 
is the sum of 
show that ABQC is a || gm and Q. therefore a fixed point.
Here 
(given)
∴ ABQC is a parallelogram
But A. B, C are given to be fixed points and ABQC is a || gm
∴ Q is a fixed point.
then O is the point of intersection of lines joining the middle points of the opposite sides of ABCD.Let E. F, G, H be mid-points of the sides AB. BC. CD. DA respectively of quadrilateral ABCD.
∴ EFGH is a || gm [From Co-ordinate Geometry]
∴ diagonals EG and FH bisect each other at P (say)
Now, 
∴ O coincides with P which is the point of intersection of lines joining the middle points of the opposite sides of ABCD.
, prove that
(ii) 
(iii) 
are non-collinear vectors. 
are collinear vectors.
If 
and 
, show that
(i) 
have the same direction and 
(ii) 
have opposite direction and
Here 
and 
(i) 
Also, 
(ii) 
Also, 
...(1)
...(2)
∴ sides PQ and SR are equal and parallel.
Similarly sides PS and QR are equal and parallel.
∴ PQRS is a parallelogram.
We have
...(1)
and 
...(2)
Adding (1) and (2) we get,
Adding (3) and (4), we get,
Also, 
and 
Adding, 
be the position vectors of A, B, C, D respectively. Let P be the mid-point of diagonal AC and Q be mid-point of diagonal BD.
position vectors of 
respectively
in terms of 
.
Take A as the origin. Let 
be position vectors of B and D respectively such that
Now, 
Again 
...(1)
∵ O is centrc of regular hexagon
∵ O is the mid-point of each of the diagonals AD. BE. CF
In ∆ ABE, O is mid-point of BE
...(1)
In 
O is the mid-point of FC
...(2)
Also, 
...(3)
as O is the mid-point of AD.
Adding (1), (2) and (3), we get,
Let O be the centre of regular octagon ABCDEFGH.
O is mid-point of each of the diagonals AE, BF, CG and DH.
O is mid-point of AE.
...(1)
Similarly 
...(2)
...(3)
...(4)
Adding (1), (2), (3) and (4). we get,
Hence the result.
[Triangle Law of Vectors]
...(1)
...(2)
...(3)
Let 
as their corresponding components are not equal.
Let 
Direction ratios of 
are 1, 1, 1 and direction ratios of 
are 3, 3, 3.
∴ direction cosines of 
are
and 
i.e. 
are different vectors but have the same direction.
and 
are equal.
Here, 
and 
We know that two vectors are equal if and only if their corresponding components are equal
, where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.
A(1, 2, 3), B(-1, 2, 1) are given points. 
If 
and the coordinates of P are (1, – 1, – 2), find the coordinates of Q.
Here P = (1, -1, -2) 
P.V. of P = 
Let Q be 
Now, 
and 
are equal.
Since 
and 
are equal
their corresponding components are equal
Let 
and 
Is 
Are the vectors 
equal?
are not equal as their corresponding components are distinct.
which has magnitude 8 units.
Let 
Unit vector along 
is
Required vector = 
Required vector = 
Here, 
Let 
be parallel.
there exists a non-zero scalar 
such that 
Equating values of 
we get,
which is required condition.
are non-zero vectors, then they are parallel if and only if u2v2 – u2v1 = 0.
Let 
be parallel vectors. 
Similarly, if 
i.e., 
are parallel
Hence the result.
are parallel.
Here, 
Let 
be parallel..
there exists a non-zero scalar 
such that
Equating values of 
we get,
which is required condition.
Let 
be two units vectors such that 
is also a unit vector. 
are collinear whatever be 
Let 
be the position vectors of A, B, C respectively...
But A is their common point
∴ point A, B, C are collinear.
what are the vectors determined by its sides? Find the length of these vectors. Let A, B, C be vertices of given triangle with position vectors 
,
form the sides of right angled triangle.
respectively form the vertices of a right angled triangle.
are the vertices of a right angled triangle.
and 
are collinear.
Let 
be position vectors of A, B, C respectively.
respectively. Show that A, B and C are collinear.
Let 
be position vectors of A, B and C respectively.
respectively. Show that AB and CD are parallel.
Let 
be P.V s. of A, B, C and D respectively.
Let 
be the vector, where 
are the direction ratios of the vector.
Since vector is equally inclined to the axes
direction cosines of a vector equally inclined to axes are 
are 
direction cosines of 
i.e., coeffs. of 
are 
is equally inclined to the axes OX. OY and OZ.
Let 
where 
are angles made by vector with axes.
Let 
be unit vectors along OA, OC, OE respectively. Let OB, OD, OF be the diagonals of faces OABC, OCDE, OEFA respectively of the cube of side of length b.
Now 
unit vector along
Similarly unit vector along 
= 
and unit vector along 
Let 
be resultant of vectors 
magnitude of resultant vector = 
Let 
be a unit vector 
in XY - plane.
From P draw 
In rt. angled 
Also, as 
varies from 0 to 
, the point P traces the circle r2 + y2 = 1 counterclockwise, and this covers all possible directions.
∴ (1) gives every unit vector in the XY-plane.
Let 
be a unit vector OP
From P draw 
In rt. angled 
and 
be the position vector of B.
If 
, then is true that 
Justify your answer.
When 
, then 
form a triangle ABC where
Since sum of two sides of a triangle is always greater than the third side
C.
In ∆ABC, by triangle law of forces, we have
If 
and 
are two collinear vectors, then which of the following are incorrect:
the respective components of 
are proportional
both the vectors 
have same direction, but different magnitudes.
D.
both the vectors have same direction, but different magnitudes.
Since 
If 
is a non-zero vector of magnitude ‘a’ and λ a non-zero scalar, then 
is a unit vector if
D.
Since 
is a non-zero vector of magnitude a
Now, 
is a unit vector if 
i.e. if 
i.e. if 
i.e. if 
in the ratio 2 : 1 
ratio 2 : 1 is
externally in the ratio 1 :2. Also, show that P is the middle point of the line segment RQ.
externally in the ratio 1 : 2 i.e. internally in the ratio 1 : 2 is 
and 
respectively, in the ratio 2 : 1Let 
and 
be position vectors of P and Q respectively.
(i) The position vector of the point R dividing the join of P and Q internally in the ratio 2 : 1 is
i.e.. 
(ii) The position vector of the point R dividing the join of P and Q externally in the ratio 2 : 1 i.e.. internally in the ratio 2 : – 1 is 
If 
an 
are position vectors of points A and B respectively, then find the position vector of points of trisection of AB.
Let 
be position vectors of points P(2, 3, 4) and Q(4, 1, -2) respectively.
Position vector of mid-point of PQ is
i.e. 
i.e., 
Let 
be the position vectors of vertices A, B, C respectively.
Position vector 
of centroid is given by
Given points are A (1, –2, –8), B (5, 0, –2), C (11, 3, 7).
Let O be origin.
Let 
be the position vectors of vertices A, B, C respectively. Then position vector of centroid Q is 
L.H.S. = 
be the position vectors of B, C, D respectively. Let M be the point of intersection of diagonals of AC and BD.
are any four vectors in 3 - dimensional space with the same initial point and such that 
show that terminals, A, B, C, D of these vectors are coplanar. Find the point at which AC and BD meet. Find the ratio in which P divides AC and BD.
Since, 
∴ P.V. of point dividing AC in the ratio 1 : 3 is the same as the P.V. of mid-point of BD.
∴ AC and BD intersect at P whose P.V. is 
This point P divides AC in the ratio 3:1 and BD in the ratio 1:1.
respectively are such that 
. Show that the four points are coplanar. Also, find the position vector of the point of intersection of lines AC and BD.
Since, 
⇒ P.V. of point dividing AC in the ratio 2 : 3 is the same as the P.V. of point dividing BD in the ratio 4 : 1
∴ AC and BD intersect at P i.e., A, B, C, D are coplanar and P has P.V. as
Let 
be position vectors of vertices A, B, C, D repsectively.
Let E, F, G, H be mid-points of AB, CD, AC, BD respectively.
P.V. of E = 
P.V. of F = 
P.V. of G = 
P.V. of H = 
When P lies outside AB and is nearer to A than B. then AP<PB.
Also division is external
∴ – 1 < λ < 0
When P is nearer to B than A, the AP > PB
∴ λ < – I
Let 
be position vectors of vertices A, B, C, D of quadrilateral ABCD.
Let P, Q, R, S be mid-points of the sides AB, BC, CD, DA respectively of quadrilateral ABCD.
position vectors of P, Q, R, S are
respectively.
Position vectors of mid - point of PR is
Position vector of mid- point of QS is
Since position vector of mid-points of PR and QS is same.
∴ lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.
be the position vectors of A, B, C and D respectively with reference to O as the origin.
∴ P.V. of mid-point of diagonal BD = P.V. of mid-point of diagonal AC.
∴ diagonal AC and BD bisect each other.
(ii) Assume that the diagonals AC and BD of the quadrilateral bisect each other.
∴ P.V. of mid-point of diagonal BD = P.V. of mid-point of diagonal AC.
Let 
be the position vectors of the vertices A, B, C, D of the trapezium in which AB||CD
Now AB || CD
where 
is some scalar.
Let E be mid-point of AD and F be mid-point of BC.
∴ EF and DC are parallel.
Also EF is parallel to AB as AB is parallel to DC.
Now, 
Hence the result.
Let 
be position vectors of vertices A, B, C, D respectively.
Let E, F, G, H be mid-points of AB, CD, AC, BD respectively.
P.V. of E = 
P.V. of F = 
, where O is any arbitrary point.
be position vectors of vertices A, B, C respectively.
is parallel to 
where K divides AB in the ratio 1 : 3.
Let 
be position vectors of A, B, C respectively.
P.V. of D = 
+ (P.V. of F - P.V. of C)
and 
be position vectors of 
such that 
be position vectors of B and D respectively such that
position vector of Q is 
position vector of E is 
is 
. The straight lines FD and AE intersect at the point O. Find the ratio 
and 
be position vectors of B and D respectively.
and 
can form the sides of a triangle. Find the lengths of the medians of the triangle.
respectively.
We have
Draw a sphere with O as centre and AB as diameter. Take any point P in the interior of this sphere and join PA and PB. It is clear that P is an interior point of the sphere if and only if ∠APB is obtuse.
∴ from (1), it follows that the required set of points is the interior of the sphere with AB as a diameter.
If either vector 
But the converse need not be true. Justify your answer with an example.
Let 
be the two non-zero vectors
and 
are perpendicular to each other when 
Here, 
and 
Since 
are perpendicular to each other
there exist no real value of A.
with magnitude 1 and 2 respectively and such that 
Here,
a = 1, b = 2, 
Let 
be the angle between two vectors.
Now, 
Find the angle between two vectors 
with magnitudes 2 and 1 respectively, and such that 
Here a = 2, b = 1, 
Let 
be angle between 
Now,
Find the angle between two vectors 
having the same length (magnitude) 
and their scalar products is -1.
Let 
be angle between 
Now, 
having the same magnitude and such that the angle between them is 30° and their scalar product is 3.
Let 
be angle between 
.
...(1)
Now, 
Prove that
if and only if 
are orthogonal.
are orthogonal
...(1)
Let 
be the two non-zero vectors. Then, prove that 
are perpendicular if and only if 
(i) Assume that 
are perpendicular
...(i)
...(2)
Again, 
...(3)
Fom (2) and (3), we get.
(ii) Assume that
Prove that two vectors 
are perpendicular if
Tips: -
By writing the steps in the reverse order, we can prove that
only if 
If 
such that each is perpendicular to sum of the other two, find 
We have
...(1)
Since 
is perpendicular to 
Now, 
If 
, respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Deduce that AB and CD are parallel.
Let 
be position vectors of points A, B, C, D respectively.
Let 
be angle between AB and CD
If 
makes equal angles with 
and has magnitude 3, then prove that the angle between 
and each of 
is 
Let 
such that 
and O be the angle which 
makes with 
Again 
and 
Also, 
as sum of two vectors such that one is parallel to the vector 
and other is perpendicular to 
Let 
and 
Since 
is perpendicular to 
into vectors which are parallel and perpendicular to the vector 
Let 
be decomposed into two vectors 
.
where 
is parallel to 
and 
as the sum of a vector parallel, and a vector parallel, and a vector perpendicular to 
Let 
be expressed as
where 
is parallel to 
and B is perpendicular to 
form the vertices of a right angled triangle. Let 
be the position vectors of A. B, C respectively.
given position vectors 
form the vertices of a right angled triangle.
form a right angled triangle.Let 
be the position vectors of A, B, C respectively.
makes angle 
and an acute angle 
with 
, then find the component of 
and the angle 
.
.
makes an angle 
makes an angle 
is a unit vector
∴ = 60°, 120°
Rejecting 
= 120° as ө is given to be acute,
we have 
= 160°
Now scalar components of 
are 
and vector component of 
are 
Also 
Let 
be the position vectors of A, B, C respectively.
makes angles 
with 
and acute angle 
with 
, then find the components of 
and the angle 
.
Let 
Since 
makes an angle 
with 
Again 
makes an angle 
Since 
is a unit vector
Now 
is the angle between 
Let 
Find a vector 
which is perpendicular to both 
is perpendicular to 
...(2)
...(3)
...(4) 
...(5)
Let 
Find a vector 
which is perpendicular to both 
Here 
Let 
Since 
is perpendicular to 
...(1)
and 
...(2)
Again 
...(3)
Multiplying (1) by 7 and (2) by -2, we get,
Adding, 
...(4)
Multiplying (1) by 2 and (3) by -1, we get,
Adding, 
From (4), 
From (1), 
Let 
Find a vector 
which is perpendicular to both 
and 
Here 
Let 
Since 
is perpendicular to 
and 3d2 – d3 = 0 ...(2)
Also, 
...(3)
Multiplying (1) by 3 and (2) by 1, we get,
3d1 – 3d2 = 0
3d2 – d3 = 0
Adding, 3d1 – d3 = 0 ...(4)
Subtracting (4) from (3), we get,
From (1), 
From (2), 
are respectively 9, 7 and 6. Find the vector.
Let 
be required vector.
From given conditions,
...(1)
...(2)
...(3)
Multiplying (1) by 7 and (3) by 1, we get,
14x + 21 y + 7z = 63
x – 3y – 7z = 6
Adding, we get.
15x +18y = 69 or 5x + 6y = 23 …(4)
Multiplying (2) by 6 and (4) by 1, we get,
24x + 6y = 42 ...(5)
5x + 6y = 23 ...(6)
Subtracting (6) from (5), we get,
19x = 19 or x = 1
Putting x = 1 in (2), we get,
4 + y = 7 or y = 3
Putting x = 1, y = 3 in (1), we get,
2 + 9 + z = 9 or z = – 2
are respectively – 1, 6 and 5. Find the vector.
Let 
be required vector.
From given conditions,
...(1)
...(2)
...(3)
...(4)
Multiplying (2) by 7 and (4) by 5, we get,
21 x - 35 z = -7
25 x+35 z = 145
Adding these equation, 46x = 138 
Let 
...(1)
be the unit vector which is perpendicular to the vectors
...(2)
...(3)
...(4)
Multiplying (3) by 1 and (4) by 2, we get,
x + 2y – z = 0
6x – 2y + 4z = 0
Adding, 
Multiplying (3) by 2 and (4) by 1, we get,
2x + 4y – 2z = 0
3x – y + 2z = 0
Adding, 
Putting 
Let 
be three vectors of magnitude 5, 3, 1 respectively. If each one is perpendicular to the sum of other two vectors, prove that 
...(1)
is perpendicular to 
, 
is perpendicular to 
is perpendicular to 
...(2)
...(3)
...(4)
If 
are three mutually perpendicular vectors of equal magnitude find the angle between 
We have
...(1)
Now 
Let 
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#7 {main}</pre>](/application/zrc/images/qvar/MAEN12067056-6.png)
If 
are mutually perpendicular vectors of equal magnitude, show that they are equally inclined to the vector 
We have
...(1)
Let 
be angle between ![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/5a/9e/e6cfd017e47b71817b27babff0d8.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/5a/9e/e6cfd017e47b71817b27babff0d8.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/MAEN12067063-5.png)
Let 
be the angle between
Let 
be angle between 
If 
are unit vectors such that 
, then find the value of 
Here
...(1)
Again from (1), 
Adding (2), (3) and (4), we get
satisfy the condition 
Evaluate the quantity 
Here
...(1)
Adding (2), (3) and (4), we get
Let 
be three vectors such that 
and each one of them being perpendicular to the sum of the other two, find 
We have
...(1)
Since 
is perpendicular to 
is perpendicular 
and 
is perpendicular to 
.
Now, 
then express 
in the form 
is perpendicular to 
Here
Since 
is parallel to 
Now, 
Since 
is perpendicular to 
.
and from (2), 
or 
If 
is the angle between two vectors 
, then 
only when
B.
Let 
be two unit vectors and 
be the angle between them. Then 
is a unit vector if choose the correct answer.
D.
Since 
are unit vectors.
Also, 
is angle between 
if 
i.e. if 
i.e. if 
i.e. if 
i.e if 
i.e. if 
i.e. if 
i.e. if 
Let 
so that BC = a, CA = b, AB = c
Now, 
position vector of E is 
.
Let ABC be the given triangle and O be the point of intersection of perpendicular bisectors OD and OE of sides BC and CA respectively.
Let F be mid-point of AB. Join O to F.
Take O as origin.
Let 
be the position vectors of A, B, C respectively.
The position vectors of D, E, F are 
respectively.
Since 
Again, 
Adding (1) and (2), we get,
perpendicular bisectors meet in a point.
be position vectors of B and C respectively with respect to origin A such that 
Then position vector of D w.r.t A is 
be position vectors of B and C so that 
.
position vector of D is 
(i) Draw a circle with centre at O and radius = 1.
Let P, Q be two points on the circle such that
∠POX = α and ∠QQX = β.
∴ ∠POQ = α + β
Co-ordinates of P, Q are (cos α, sin α) and (cos β, – sin β) respectively.
(ii) Draw a circle with centre at O and radius = 1.
Let P, Q be two points on the circle such that
∠POX = α, ∠QOX = β
∴ ∠POQ = α – β
Co-ordinates of P, Q are (cos α, sin α) and (cos β, sin β) respectively.
.
is perpendicular to 
Let ABCD be given rectangle and P, Q. R, S be mid-points of sides AB, BC, CD, DA respectively.
Take A as origin.
Let 
The position vectors of P, Q, R, S are 
PQRS is a rhombus.
be angle between AL and BM.
Prove, using vectors, that the altitudes of a triangle are concurrent.
OR
Prove that the perpendicular from the vertices to the opposite sides of a triangle are concurrent.
Let ABC be given triangle and H be the point of intersection of altitudes AL and BM. Join CH and produce it to meet BA in N.
Take H as origin.
Let 
be the position vectors of A, B, C respectively.
Since 
Again, 
...(2)
Adding (1) and (2), we get,
be the position vectors of B and D respectively so that
BC is equal and parallel to AD]
∴ position vector of mid-point of diagonal AC is same as position vector of mid-point of diagonals BD.
∴ diagonals AC and BD bisect each other.
Also, 
[
AD = AB as all sides of rhombus are equal]
diagonals AC and BD are perpendicular to each other.
Hence the result.
Take O as origin, 
Let P divide AB in the ratio 1 : 2
Let Q divide AB in the ratio 2 : 1
L.H.S. = 
R.H.S. = 
are position vectors of A, B, C respectively with D as origin. Let DA and CB be perpendicular.
if 
Here, 
which is a vector perpendicular to both 
a unit vector perpendicular to 
and 
, where 
.
Find a unit vector perpendicular to each of the vector 
where 
and 
Here 
which is a vector perpendicular to 
as well as 
which is a vector perpendicular to 
a unit vector perpendicular to 
is
and 
are (i) parallel (ii) perpendicular to each other.
Let 
(i) 
Since 
(ii) Since 
are perpendiuclar
Find the value of 
so that the two vectors 
are (i) parallel (ii) perpendicular to each other.
Let 
(i) 
Since 
(ii) Since 
are perpendicular 
Find the value of 
so that the vectors 
and 
are (i) parallel (ii) perpendicular to each other.
Let 
(ii) Since 
are perpendicular
Here 
which is vector perpendicular to 
.
a unit vector perpendicular to 
is 
a vector of magnitude 9 and perp. to 
is
Required vector can be = 
Let 
which is vector perpendicular to 
.
a unit vector perpendicular to 
is
a vector of magnitude of 19 and perpendicular to 
is
where 
and 
Here, 
and 
which is a vector perpendicular to 
a unit vector perpendicular to 
vector of magnitude 5 units and perpendicular to 
is
. Also find a unit vector perpendicular to both the vectors. 
and 
Let 
be P.Vs. of A, B and C respectively.
Scalar area of 
= 
be the position vectors of A, B, C respectively.
Let 
be position vectors of A, B, C repsectively.
Now, 
Area of 
be position vectors of A, B, C respectively.
is 
where 
are the position vectors of the vertices A. B and C respectively. Find the condition of collinearity of these points.
= 
, respectively are collinear if and only if 
are position vectors of A, B, C respectively.
If 
are the position vectors of the non-collinear points A, B, C respectively in space, show that 
is perpendicular to plane ABC.
We have,
Vector perpendicular to plane ABC
= Vector perpendicular to 
respectively. Show that the perpendicular distance from C to the straight line through A and B is 
Let 
be position vectors of A, B, C respectively.
From C, draw 
.
In rt. 
Given that 
what can you conclude about the vectors 
?
We have
...(1)
and 
...(2)
From (1), it is clear that
either 
From (2), it is clear that
either 
Now, 
are perpendicular and 
are parallel cannot hold simultaneously.
either 
Is the converse true? Justify your answer with an example.
If 
But the converse is not true, i.e., if 
then 
may not hold.
For example, consider the vectors 
Then 
also, 
but 
If 
show that 
. Interpret the result geometrically.
Now, 
...(1)
Similarly, 
...(2)
From (1) and (2), we get, 
Let 
If 
are mutually perpendicular unit vectors and 
then 
Since, 
are mutually perpendicular unit vectors
.
and give a geometrical interpretation of it.
L.H.S. = 
Gemetrical Interpretation
Let O be the point of intersection of diagonals AC and BD of || gm ABCD.
Let 
implies that the area of || gm ABCD is twice the area of the || gm whose adjacent sides are semi-diagonals of the first parallelogram.
If 
are three vectors such that 
show that the three vectors 
are orthogonal in pairs and 
.
...(2)
If 
and 
Find a vector 
such that 
and 
Here , 
Let 
Now, 
...(1)
x - z = 1 ...(2)
x - y = 1 ...(3)
Also, 
...(4)
From (1), y = z
from (4), we get,
...(5)
Subtracting (3) from (5), we get,
Let 
and 
. Find a vector 
which is perpendicular to both 
is perpendicular to both 
in the form 
in the form 
Let 
Let 
Since 
is parallel to 
From (1), 
are unit vectors, Suppose that 
and the angle between B and C is 
Prove that 
Since 
Let 
...(1)
where 
is some scalar
where 
is perpendicular to 
being a right-handed orthogonal system of unit vectors in space. Show that 
in another such system.
are mutually orthogonal vectors.
Now,
from a right handed orthogonal system of unit vectors.
Let vectors 
be such that 
then 
is a unit vector, if the angle betweeen 
is
B.
Here, 
Now, 
is a unit vector
i.e. if 
i.e. if 
where 
is angle between 
.
i.e. if 
i.e. if 
i.e. if 
respectively is
1
2
4
C.
2
Position vectors of vertices A, B, C, D of rectangle are
respectively 
Dividing both sides by a, b, c we get,
Similarly, 
...(1)
From (1) and (2), we have
be unit vectors along OX and OY respectively.
be two vectors such that 
so that
be the unit vector along z-axis.
...(2)
respectively.
position vectors of D, E, F are 
Find the differential equation of the family of lines passing through the origin.
Consider the equation, y = mx, where m is the parameter.
Thus, the above equation represents the family of lines which pass through the origin.
y = mx ....(1)
Differentiating the above equation (1) which respect to x,
Thus we have eliminated the constant, m.
The required differential equation is
Sketch the region bounded by the curves 
and find its area using integration.
Consider the given equation. 
This equation represents a semicircle with centre at the origin and radius = 
Given that the region is bounded by the above semicircle and the line 
Let us find the point of intersection of the given curve meets the line 
Squaring both the sides, we have,
Consider the following figure
Thus the intersection points are 1,2 and 2,1 ( ) ( ) Consider the following sketch of the bounded region.
Required Area, 
Find the particulars solution of the differential equation 
given that y =1 when x = 1.
Find the particulars solution of the differential equation 
Find the value of 'p' for which the vectors 
are parallel.
Since the vectors are parallel, we have
If the cartesian equations of a line are 
write the vector equation for the line.
Given that the cartesian equation of the line as 
Any point on the line is of the form:
Thus, the vector equation is of the form:
Given that 
are two perpendiculars vectors.
Thus, 
Also given that, 
We need to find the value of 
Consider 
Show that the four points A, B, C and D with position vectors
respectively are coplanar.
Given position vectors of four points A, B, C and D are:
These points are coplanar, if the vectors, 
are coplanar. 
These vectors are coplanar if and only if, they can be expressed as a linear combination of other two.
So Let
Comparing the coefficients, we have, 
These values of x and y satisfy the equation 3x + 3y = -2.
Hence the vectors are coplanar.
The scalar product of the vector 
with a unit vector along the sum of vectors 
is equal to one. Find the value of 
and hence find the unit vector along 
Given that
Now consider the sum of vectors 
Thus, n is:
Find the distance of the point (2, 12, 5) from the point of intersection of the line 
Any point in the line is
The vector equation of the plane is given as
Thus, the point of intersection of the line and the plane is:
Distance between (2, 12, 5) and (14, 12, 10) is:
If a unit vector 
and 
are two equal vectors, then write the value of x+y+z.
Given that 
and 
are equal vectors.
If 
are two vectors such that 
then prove that vector 
is perpendicular to vector 
[Using (1)]
is perpendicular to 
Find the coordinates of the point, where the line 
intersects the plane 
. Also find the angle between the line and the plane.
The equation of the given line is 
Any point on the given line is 
If this point lies on the given plane x - y + z - 5 = 0, then
Putting 
in 
we get the point of intersection of the given line and the plane is (2, -1, 2).
Let 
be the angle between the given line and the plane. 
Thus, the angle between the given line and the given plane is 
Find the vector equation of the plane which contains the line of intersection of the planes. 
and 
and which is perpendicular to the plane 
The equation of the given planes are
The equation of the plane passing through the intersection of the planes (1) and (2) is
Given that plane (3) is perpendicular to the plane 
This is the equation of the required line.
Find the vector equation of the plane passing through three points with position vectors 
Also, find the coordinates of the point of intersection of this plane and the line 
Let the position vectors of the three points be, 
So, the equation of the plane passing through the point 
So, the vector equation of the required plane is 
The equation of the given line is 
Position vector of any point on the given line is
....(2)
The point (2) lies on plane (1) if, 
Putting 
in (2), we have
Thus, the position vector of the point of intersection of the given line and plane (1) is 
and its co-ordinates are (1, 1, -2).
Prove that x2 – y2 = c(x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2)dx = (y3 – 3x2y) dy, where C is a parameter.
Given equation is
(x3 – 3xy2)dx = (y3 – 3x2y) dy
which is a homogeneous equation.
Therefore substituting y = vx
We have,
Integrating the equation both sides, 
thus substituting the values of I1 and I2, 
If 
are mutually perpendicular vectors of equal magnitudes, show that the vector 
is equally 
inclined to Also, find the angle which 
with 
.
Since 
are mutually perpendicular vectors, we have
It is given that
coplanar.
coplanar.
Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).
the vertices of the triangle be A(–2, 1), B(0, 4) and C(2, 3).
Equation of line segment AB is
AL, BO and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ABOL) + Area (BOMCB) – Area (ALMCA)
Find the magnitude of each of the two vectors , having the same magnitude such that the angle between them is 60o and their scalar product is 9/2.
Given that,
As, we know that,
Hence, the magnitude of each of the vectors is 3.
Let and . Find a vector which perpendicular to both
Since is perpendicular to both and , therefore, if is parallel to
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are respectively, externally in the ratio 1:2. Also, show that P is the midpoint of the line segment R.
Point R divides the lines segment PQ externally in a ratio of 1 : 2 .
Now, we need to show that P is the mid-point of RQ.
Hence proved.
Find a unit vector perpendicular to each of the vector , where
So, the unit vector, perpendicular to each of the vectors is given by
If a line has direction ratios 2,-1,-2 then what are its direction cosines?
The direction cosines are
If are three vectors such that and find the value of
Considering dot product on both sides,
Solve the following differential equation:
Hence the given equation is an homogeneous equation.
Let y = v x
How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?
Let the man toss the coin n times. then n tosses are n Bernoulli trials
Probability ( P ) of getting a head at the toss of a coin is
It is given that
P ( getting at least one head )
The minimum value of n which satisfies given inequality is 3.
Thus, the man should toss the coin 3 or more than 3 times.
Evaluate:
Equating the coefficients of x2, x and constant term, we obtain
A - B = 0
B - C = 0
A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1
Find the equation of the line passing through the point (-1,3,-2) and
perpendicular to the lines
We know that, equation of a line passing through x1, y1, z1 with direction ratios a, b, c
So, the required equation of a line passing through ( - 1, 3, - 2 ) is:
Solving equation ( ii ) and ( iii ) by cross multiplication,
Putting the value of a, b and c in ( i ) gives
Find the particular solution of the following differential equation:
Integrating both sides, we get:
substituting this value in equation ( i ), we get:
Now, at x = 0 and y = 0, equation ( ii ) becomes:
This is the required particular solution of the given differential equation.
A manufacturer produces nuts and bolts. It takes 1 hours of work on machine A and 3 hours on machine B to product a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of `17.50 per package on nuts and `7 per package of bolts. How many packages of each should be produced each day so as to maximize his profits if he operates his machines for at the most 12 hours a day? From the above as a linear programming problem and solve it graphically.
Let the manufacturer produce x packages of nuts and y packages of bolts.
The given information can be complied in a table as follows.
| Nuts | Bolts | Availability | |
| Machine A ( h ) | 1 | 3 | 12 |
| Machine B ( h ) | 3 | 1 | 12 |
The profit on a package o nuts is Rs. 17.50 and on a package of bolts is Rs. 7.
Therefore, the constraints are
Total profit, Z = 17.5 x + 7 y
The mathematical formulation of the given problem is
Maximise Z = 17.5 x + 7 y ..............( i )
Subject to the constrain,
The feasible region determined by the system of constraints is as follows:
The corner points are A ( 4, 0 ), B ( 3, 3 ) and C ( 0, 4 ).
The values of Z at these corner points are as follows:
| corner point | Z = 17.5 x + 7 y | |
| O ( 0, 0 ) | 0 | |
| A ( 4, 0 ) | 70 | |
| B ( 3, 3 ) | 73.5 | Maximum |
| C ( 0, 4 ) | 28 |
The maximum values of Z is Rs. 73.50 at (3, 3 ).
Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs. 73.50.
Find the equation of the plane determined by the point A( 3, - 1, 2 ), B( 5, 2, 4 ) and C( -1, -1, 6 ) and hence find the distance between the plane and the point P( 6, 5, 9 ).
We know that, equation of a plane passing through 3 points,
Also, perpendicular distance of P ( 6, 5, 9 ) to the plane 3 x - 4 y + 3 z - 19 = 0
A girl throws a die. If she get a 5 OR 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 OR 4, she tosses a coin two times and notes the number of heads obtained. If she obtained exactly two heads, what is the probability that she threw 1, 2, 3 OR 4 with the die?
Consider the following events:
E1 = Getting 5 OR 6 in a single throw of the die
E2 = Getting 1, 2, 3 OR 4 in a single throw of the die
A = Getting exactly 2 heads
We have to find, .
Using the method of method of integration, find the area of the region bounded by the following lines:
3x – y – 3 = 0,
2x + y – 12 = 0,
x – 2y – 1 = 0
Given equations are:
3 x - y = 3 ........( i )
2 x + y = 12 ........( ii )
x - 2 y = 1 .......( iii )
To solve ( i ) and ( ii ),
( i ) + ( ii ) 5 x = 15 x = 3
( ii ) y = 12 - 6 = 6
Thus ( i ) and ( ii ) intersect at C ( 3, 6 ).
To solve ( ii ) and ( iii ),
( ii ) - 2 ( iii ) 5 y = 10 y = 2
( ii ) 2 x = 12 - 2 = 10 x = 5
Thus, ( ii ) and ( iii ) intersect at B ( 5, 2 ).
To solve ( iii ) and ( i ),
2 ( i ) - ( iii ) 5 x = 5 x = 1
( iii ) 1 - 2 y = 1 y = 0
Thus ( iii ) and ( i ) intersect at A ( 1, 0 ).
= 6 + 8 - 4
= 10 sq. unit
If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is:
5
10
B.
Given equation of a circle is x2 + y2 -4x +6y -12 = 0, whose centre is (2,-3) and radius
Now, according to given information, we have the following figure.
x2+y2-4x +6y-12 =0
Clearly, AO perpendicular to BC, as O is mid-point of the chord.
Now in ΔAOB,. we have
Let P be the point on the parabola, y2=8x which is at a minimum distance from the centre C of the circle, x2+(y+6)2=1. Then the equation of the circle, passing through C and having its centre at P is:
x2+y2−4x+8y+12=0
x2+y2−x+4y−12=0
x2+y2− 4 x +2y−24=0
x2+y2−4x+9y+18=0
A.
x2+y2−4x+8y+12=0
Centre of circle x2 + (y+6)2 = 1 is C (0,6)
Let the coordinates of point P be (2t2, 4t)
Now, let 
The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is:
4/3
C.
We have, 2b2/a = 8 and 2b = ae
⇒ b2 = 4a and 2b = ae
Consider, 2b = ae
Consider, 2b = ae
⇒ 4b2 = a2e2
⇒ 4a2 (e2-1)=a2e2
⇒ 4e2-4 = e2
[e>0]
If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?
3a2−26a+55=0
3a2−32a+84=0
3a2−34a+91=0
3a2−23a+44=0
B.
3a2−32a+84=0
We know that, if x1, x2..... xn are n observations, then their standard deviation is given by
The Boolean Expression (p∧~q)∨q∨(~p∧q) is equivalent to:
~p ∧ q
p ∧ q
p ∨ q
p ∨ ~ q
C.
p ∨ q
Consider, (p ∧~q) ∨ q ∨(~p ∧ q)
≡ [(p ∧~q) ∨ q] ∨ (~p ∧ q)
≡[(p ∨~q) ∧ t] ∨ (~p ∧ q)
≡((p ∨ q) ∨ (~p ∧ q)
≡(p ∨ q ∨ ~p) ∧ (p ∨ q ∨ q)
≡(q ∨ t) ∧ (p ∨ q)
≡ t ∧ (p ∨ q)
≡ p ∨ q
The area (in sq units) of the region described by {x,y): y2 ≤ 2x and y ≥ 4x-1} is
7/32
5/64
15/64
9/32
D.
9/32
Given region is {x,y): y2 ≤ 2x and y ≥ 4x-1} y2 ≤ 2x represents a region inside the parabola,
y2 = 2x .. (i)
and y ≥ 4x-1 represents a region to the left of the line
y = 4x-1 .... (ii)
The point of intersection of the curve (i) and (ii) is
(4x-1)2 = 2x
⇒ 16x2 + 1-8x = 2x
16x2-10x+1 = 0
x = 1/2, 1/8
therefore, the points where these curves intersects, are 
Hence, required area,
Let y(x) be the solution of the differential equation 
(x ≥1). Then, y (e) is equal to
e
0
2
2e
C.
2
Given differential equation is 
This is a linear differential equation.
therefore,
Now, the solution of given differential equation is given by
The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse
27/4
18
27/2
27
D.
27
given equation of ellipse is 
Therefore, Extremities of one of latus rectum are
therefore, area of quadrilateral = 4 x Area of ΔPOQ
= 
= 27 sq units
Let O be the vertex and Q be nay point on the parabola x2 = 8y. If the point P divides the line segment OQ internally in the ratio 1:3 then the locus of P is
x2= y
y2 =x
y2 =2x
x2 = 2y
D.
x2 = 2y
Any point on the parabola x2 = 8y is (4t, 2t2). Point P divides the line segment joining of O (0,0) and Q (4t,2t2) in the ratio 1:3 Apply the section formula for the internal division.
the equation of the parabola is
x2 = 8y
Let any Q on the parabola (i) is (4t, 2t2).
Let P (h,k) be the point which divides the line segment joining (0,0) and (4t, 2t2) in the ratio 1:3.
If =-1 and x =2 are extreme points of f(x) =α log|x| + βx2 +x, then
α = -6, β = 1/2
α = -6, β = -1/2
α = 2, β = -1/2
α = 2, β = 1/2
C.
α = 2, β = -1/2
Here, x =-1 and x = 2 are extreme points of f(x) = α log|x| +βx2 +x then,
f'(x) = α/x +2βx + 1
f'(-1) = -α -2β +1 = 0 .... (i)
[At extreme point f'(x) = 0]
f'(2) = α/x +4βx + 1 = 0 .. (ii)
On solving Eqs (i) and (ii), we get
α = 2 and β = -1/2
The area of the region described by A = {(x,y): x2 +y2 ≤ 1 and y2 ≤1-x} is
A.
Given, {(x,y): x2 +y2 ≤ 1 and y2 ≤1-x}
Required area = 
Let the population of rabbits surviving at a time t be governed by the differential equation.
If p(0) = 100 then p(t) is equal to
A.
Given differential equation 
is a linear differential equation
Here, p(t) = 
Hence, solution is
p(t), IF = ∫Q(t)IF dt
The intercepts on X- axis made by tangents to the curve, 
which are parallel to the line y =2x, are equal to
±1
±2
±3
±4
A.
±1
Let 
be two unit vectors. If the vectors 
and 
are perpendicular to each other, then the angle between 
is
π/6
π/2
π/3
π/4
C.
π/3
For x ∈, (0, 5π/2) define f(x). Then f (x) = 
has
local maximum at π and 2π.
local minimum at π and 2π
local minimum at π and the local maximum at 2π.
local maximum at π and local minimum at 2π.
D.
local maximum at π and local minimum at 2π.
local maximum at π
and local minimum at 2π
The area of the region enclosed by the curves y = x, x = e, y =1/x and the positive x-axis is
1/2 square units
1 square units
3/2 square units
5/2 square units
C.
3/2 square units
Required area = OAB + ACDB
Solution of the differential equation
cos x dy = y (sin x - y) dx, 0 < x < π/2, is
sec x = (tan x + C ) y
y sec x = tan x + C
y tan x = sec x + C
tan x = (sec x + C)y
A.
sec x = (tan x + C ) y
since cos xdy = y sin x dx - ydx
This is linear differential equation,
Therefore,
The circle x2+ y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if
-85 < m < -35
-35 < m < 15
15 < m < 65
35 < m < 85
B.
-35 < m < 15
x2+ y2– 4x – 8y – 5 = 0
Centre (2, 4)
Let 
and 
. Then, the vector b satisfying a x b + c = 0 and a.b = 3
A.
a x b +c = 0
⇒ a x (a x b) + a x c = 0
⇒ (a.b)a-(a.a)b +a x c = 0
⇒ 3a - 2b + a x c = 0
⇒ 2b = 3a +a x c
If the vectors 
are mutually orthogonal, then (λ,μ) is equal to
(-3,2)
(2,-3)
(-2,3)
(3,-2)
A.
(-3,2)
since, the given vectors mutually orthogonal, therefore
a.b = 2-4+2 = 0
a.c = λ-1 + 2μ = 0 ....(i)
b.c = 2λ + 4 +μ = 0 ... (ii)
On solving Eqs. (i) and (ii), we get
μ = 2 and λ = - 3
Hence, (λ,μ) = (-3,2)
Let In ∫tan x dx,(n> 1) . I4 + I6 = a tan5x + bx5 + C, where C is a constant of integration, then the ordered pair (a, b) is equal to
C.
I4 +I6 = ∫(tan4x + tan6x) dx
= ∫tan4 x sec2 x dx
= 1/5 tan5 x + c
⇒ a = 1/5, b = 0
Let a, b and c be distinct non-negative numbers. If the vectors 
the Geometric Mean of a and b
the Arithmetic Mean of a and b
equal to zero
the Harmonic Mean of a and b
A.
the Geometric Mean of a and b
depends on
only y
only x
both x and y
neither x nor y
D.
neither x nor y
are non-coplanar vectors and λ is a real number, then the vectors 
are non-coplanar for
all values of λ
all except one value of λ
all except two values of λ
no value of λ
C.
all except two values of λ
Condition for given three vectors to be coplanar is 
=0
⇒ λ = 0, 1/2.
Hence given vectors will be non coplanar for all real values of λ except 0, 1/2.
The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is:
2/3
1/3
A.
Normal to the plane x + y +z = 7 is
Sponsor Area
Sponsor Area
: 
...(1)
is a solution of 
...(1)
which is the required solution.
which is the required solution.
,
which is the required solution.
are collinear.
then what can you say about 
?
. Find the vector 
.
Find scalars a and b such that 
.
is a unit vector. 
is a unit vector![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/c2/da/7ff59b0191e721985443ce2d419b.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/c2/da/7ff59b0191e721985443ce2d419b.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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is vector 
, find the coordinates of B.
unit vector in the direction of 
find a unit vector in the direction of 
.
find the unit vector in the direction of the vector 
.
, 
and 
, find a vector parallel to the vector 
.
and 
are collinear.
are collinear.
?
and hence calculate its direction cosines.
and 
, find the direction ratios of 
when
if for a unit vector 
is the unit vector and 
then find 
is a unit vector
...(1)
and 
then what can be concluded about the vector 
?
be the angle between vectors
and 
if 
and 
show that 
are 
are perpendicular to each other. 
then show that the vectors 
perpendicular to each other?
find the value of 
so that 
is perpendicular to 
.
is perpendicular to 
and 
Find 
such that 
and 
are orthogonal .
are orthogonal
if the vectors 
and 
are orthogonal.
such that 
is perpendicular to 
, then find the value of 
is perpendicular to 
find the angle between 
be angle between 
are two vectors such that 
find the angle between 
.
be angle between 
.
with magnitudes 
be the angle between and 
are two vectors such that 
find the angle between 
be the angle between 
.
such that 
and 
.
= 
be the angle between 
having the same magnitude and such that the angle between then is 
...(1)
and 
if 
show that 
prove that either 
or 
or 
is perpendicular to 
.
is orthogonal to 
, for any vectors 
are perpendicular, then 
which is true. 
if two vectors 
are such that 
if 
...(1)
if 
...(1)
are unit vectors, and 
is the angle between them, find 
in terms of 
.
we always have 
.
, we always have 
is any vector, then show that
...(1)
on the vector 
.
on the vector 
.
on the vector 
in the direction of 
in the direction of 
.
on 
where
= 
and 
be angle between the vectors
and 
. Also find the projection of 
be angle between 
form a right angled triangle. 
forms the sides of a right-angled triangle. Also, find the remaining angles of the triangle.
are mutually 
with a unit 
and 
is equal to 1. Find the value of 
show that the angle 
between the vectors 
is given by
...(1)
show that the angle between 
is 
be angle between 
, 
if 
if 
and 
then show that 
= 
are two vectors such that 
and 
find the angle between 
.
if 
is an angle between 
and 
and 
and 
.
be any three vectors, then 
then prove that 
where m is a scalar.
and 
then 
, where 
is a scalar.
(iii) 
is perpendicular to 
and 
then 
is parallel to 
will be parallel to 
then 
when 
is equal to 
is
makes angle 
then find the value of 
.
be three unit vectors such that 
then the angle between 
is
if θ is the angle between vectors b and c, then a value of sin θ is
and 
are the sides of a ΔABC, then the length of the median through A is 
2y-x+3 =0, X -axis and lying in the first quadrant is 
and ∠BAD be an acute angle. If 
is the vector that coincides with the altitude directed from the vertex B the side AD, then 
is given byLet ABCD be a parallelogram such that AB = q,AD = p and ∠BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by (1)
are not perpendicular and 
are two vectors satisfying: 
The vector 
is equal to
is equal to
are non–coplanar vectors and p, q are real numbers, then the equality 
holds for
, then the maximum value of |Z| is equal to
lies in the plane of the vectors 
and bisects the angle between 
.Then which one of the following gives possible values of α and β?
are related by 
.Then the angle between 
is
are unit vectors and θ is the acute angle between them, then is a 
unit vector for 
respectively is the force along 
where D is the foot of the perpendicular from A onto BC. The magnitude of the resultant is
are any three vectors such that 
respectively are the vertices of a right-angled triangle with C =π/2 are
acting along IA, IB and IC respectively are in equilibrium, where I is incentre of ∆ABC. Then P : Q : R is
the value of 
is equal to
are non -coplanar vector λ is a real number then
be three non-zero vectors such that no two of these are collinear. If the vector 
is collinear with
is collinear with 
(λ being some non-zero scalar) then 
equals
which displace it from a point 
The work done in standard units by the forces is given by 
be such that 
If the projection 
is equal to that of 
are perpendicular to each other then 
equals
be non-zero vectors such that 
If θ is the acute angle between the vectors 
then sin θ equals
