Test Papers – ICSE – Class – X
Std-11, Science, Mathematics, Chapter-9, Straight Line
Good morning class, today we will study topic of straight lines. We all know what is a straight line.
Let us talk about the centroid of the triangle. See, centroid is point of intersection of medians of
triangle. The point where two medians meet what do we call that point, centroid. For example, say
this is a triangle in which the first vertex is meeting with the midpoint of the second side and the
second vertex is also meeting with the midpoint of another side. So the point of intersection is called
centroid. In general it is denoted by capital G. The third median also passes through G and the
coordinates of centroid are x1 plus x2 plus x3 by 3, y1 plus y2 plus y3 by 3. So add the three x
coordinates and divide it by 3. And add the three y coordinates and divide by 3.
Next thing is slope of straight line. We will just have a quick revision of basic concepts of straight
lines, slope of straight line. If a straight line makes angle theta with positive direction of x axis then
its slope is tan theta. If that theta lies from 0 to 90 degrees, the slope is positive. The line will be like
this. If theta is pie by 2, then how will the line be, vertical line, so we call the slope as infinite tan 90
not defined is infinite. And if the slope is negative then the angle will be pie from pie by 2, the angle
will be more than 90 degrees. Theta is greater than 90. If a straight line passes through two points,
x1y1 and x2 y2, if a straight line passes through two points, x1y1 and x2y2, then its slope is y2 minus
y1 upon x2 minus x1. These are all basic concepts, we have studied all this in class 11th.
Next, angle between two straight lines. If the slope of one straight line is m1 and the slope of the
other straight line is m2 and between them there is angle theta. Tan theta will be modulus of m1
minus m2 upon 1 plus m1m2, tan theta is modulus of m1 minus m2 upon 1 plus m1m2. If two lines
are parallel then m1 equals to m2. When will two lines be parallel, when both their slopes are equal.
Two lines are perpendicular when m1m2 is minus 1, that is slope of one line is negative times
reciprocal of slope of other line. If two lines are perpendicular slope of one line is negative times
reciprocal of slope of other line.
Let us move forward, equations of straight lines, if a line passes through point x1y1 and its slope is
m, if a line passes through x1y1 and its slope is m then its equation is y minus y1 equals to m into x
minus x1. If any line is passing through a point x1y1 and the slope is m so that equation will be y
minus y1 equals to m into x minus x1. All the students know this and we have been using this for a
long time. Is it clear to everyone? Yes. So to get an equation of a line we need two things, one is
slope, other is point from which line passes.
Next, intercept form of line, if any line is intersecting x axis at a,o and y axis at o,b. If a line is
intersecting x acis at a,o and y axis at o,b so the equation of that will be x by a plus y by b equals to
one, x by a plus y by b equals to 1. How did you get that? You took a stroke at two points and then
taking any point wrote the equation of the line. Are you clear on all this?
General equation of line: ax plus by plus c is equal to zero. A linear equation in xy is equation of a
straight line. What will the slope of this equation be minus times a by b, minus times coefficient of x
divided by coefficient of y.
Distance of a point from the line: if we have to take perpendicular distance from one point to a
straight line, how will we do that? We will put the point on the line ax1 plus by1 plus c. We will take
the modulus of whatever value we get upon square root of a square plus b square. Distance of point
x1y1 from the line ax plus by plus c equal to zero is modulus of ax1 plus by1 plus c upon square root
of a square plus b square.
Next is distance between two parallel lines. If there are two parallel lines like this. One is ax plus by
plus c1 equal to zero. And other is ax plus by plus c2 equal to zero. So the distance between these
two lines will be modulus of c2 minus c1 upon square root of a square plus b square. If the
coefficients of x and y of the two parallel lines are the same so the distance between them will be
the difference of constant terms upon square root of a square plus b square.
Children, these were the few concepts of straight lines which we have done a quick and fast revision.
On the basis of all these concepts now we will solve questions.
Let us talk about first question. If vertex of triangle is 1,1 and mid points of two sides are minus 1,2
and 3,2, we have to find the centroid of triangle.
Std-11, Science, Mathematics, Chapter-15, Limits Continuity and Derivability
Good evening class, today we will talk about Limits. Let us talk about few simple formulae of limits,
limit extending to a, x raised to par n minus a raised to par n upon x minus a is equal to n into a2 par
n minus 1. Limit extending to zero, e raised to par x minus 1 upon x is equal to 1. Put it as x to zero
and it is zero by zero format and the limiting value comes as 1. Similarly limit extending to zero, a
raised to par x minus 1 upon x is log a to the base e.
More formulae on limits, Limit extending to zero, log of 1 plus x upon x is 1. Limit extending to zero,
log of 1 plus x upon x, log to the base a is 1 upon log a to the base e.
Limit extending to zero sin x upon x is also 1. Limit extending to zero tan x upon x is also 1. Similarly
limit extending to zero sin inverse x upon x is also 1. Limit extending to zero tan inverse x upon x is
also 1. Good.
But how do we use this formulae, we use this formulae as limit extending to zero sin 2x upon x. Say
we have to find this. I will multiply it and divide it by 2. Sin zero upon same zero or what we can do is
we can put 2x equals to t limit t extends to zero, as x extends to zero, t also extends to zero. Sin t
upon t is 2 into 1 that is 2. Limit t extending to zero is sin t upon t is 1.
Next let us talk about L-Hospital’s rule, L-Hospital’s rule. It is applicable for the limits of the form zero
by zero or infinity by infinity. It is applicable for the limits of the form zero by zero or infinity by
infinity. It says that limit extending to a, fx upon gx is equal to limit extending to a f dash x upon g
dash x. Differentiate the above function separately and differentiate the below function separately.
And you get the value of limit.
For example, zero by zero format, differentiating numerator we get n into x to par n minus 1
denominator we get 1. Now put x equals to a, we will get the value of limit that is n into a to par n
minus 1.
One more question of L-Hospital say, find limit extending to zero, x minus sin x upon xq. Applying L-
Hospital rule we get differentiate the numerator 1 minus cos x denominator is 3x square. Format is
zero by zero, differentiate above separately and below separately. Put x as zero and see. Cos zero, 1
minus 1 zero, now also zero by zero format, and again L-Hospital’s rule it becomes, and limit
extending to zero, sin x upon x put 1, the answer becomes 1 by 6. Simple application of L-Hospital’s
rule.
Let us talk about questions, first question of your assignment. Put x as 1 and format is zero by zero.
Applying L-Hospital’s rule we get, now put, it is extending to 1, now put x to 1 we get 8 minus 2 upon
4 minus 2 that is 6 by 2 that is 3. Answer of this question is 3. Simple it was a zero by zero format.
We have applied L-Hospital’s rule, differentiate above separately and below separately and we got
the answer.
Second question of your assignment, put x as 1 and see, zero by zero format. So this limit L is equal
to, below there are factors 6 minus 1 into 3x plus 7, x minus 1 into 3x plus 7. x minus was again
factorized.
Std-11, Science, Mathematics, Chapter-14, Inverse Trigonometry
Good afternoon, students, welcome to the class. Today we will study inverse trigonometric functions
with respect to JEE Mains syllabus. We will solve those kinds of questions which come in JEE Mains.
Let us start with first very simple question. We have to find tan of cos inverse x.
See let cos inverse x is equal to theta, if cos inverse x is an angle equal to theta then x is equal to cos
theta, this means that x equals to cos theta. And we have to find tan of theta. If x is cos theta then
what will be tan theta is what we have to find out. If x is cos theta then what will tan theta be, we
have to find that out.
Simply speaking x is cos theta means if I take this as x and this as 1 then it will be cos theta, then this
will be 1 minus x square. So tan theta is equal to square root of 1 minus x square upon x. Using
simple triangles we can get the value of tan theta, square root of 1 minus x square upon x. I hope
everybody got it.
Second question is again very simple based on same thing, we have to find cos of 2 cos inverse 0.8. If
I take cos inverse 0.8 as theta then what will be cos theta, 0.8. I have to get cos 2 theta, what is cos 2
theta. 2 cos square theta minus 1, 2 into 0.8 into 0.8 minus 1. 2 into 0.64 minus 1 that is 0.64 into 2
is 1.28 minus 1 that is 0.28. Answer is 0.28 which is not in options so the answer is none of these.
Cos theta 0.8 is 2 cos square theta minus 1 that is 2 into 0.8 into 0.8 minus 1. Got it, very good.
Third question, similar questions are asked in school also. Let us apply tan inverse of x plus tan
inverse of y is equal to tan inverse of x plus y upon 1 minus xy. Because both these angles are less
than pi by 4, both are less than tan inverse, we can directly use this formula, this will become tan
inverse one by.
Std-11, Science, Mathematics, Chapter-13, Matrices and Determinants
Good morning students. Today we will talk about topic, Matrices and Determinants. We
know what is Matrix it is a rectangular arrangement of numbers and what is Determinant, it
is a square arrangement of numbers. Determinant is always associated with a value. So we
know many things about Matrices and Determinants, and we know that there are many
properties of both Matrices and Determinants. For example, we know that in Determinants,
if two rows or two columns are identical then value of Determinant is zero, if we interchange
two rows then value of Determinant gets multiplied by a negative sign. If we interchange
rows and columns then the value of Determinant remains same. We also know we can
apply operations like ith row goes to ith row minus some constant times and jth row. And by
doing this value of Determinant does not get changed. And there are many other properties
of Matrices and Determinants. Let us talk about few questions which can come in the
competition examinations. We will give you a quick glimpse of such type of questions.
First question of your worksheet, if a is greater than zero and discriminant of this quadratic is
negative then this Determinant is what? a is greater than zero and discriminant is negative
that is four times b square minus ac is less than zero. We also know if a is greater than zero
and discriminant is negative then this quadratic is greater than zero for all x. Then this
quadratic is always greater than zero. This is what we know from quadratic equations.
Let us talk about Determinant a, b, ax plus b, b, c, bx plus c, ax plus b, bx plus c, zero. By
applying R3 goes to R3 minus R1 into x minus R2. From row three multiply row one and
subtract it and also subtract row two, what do we get is, first two rows will remain same, this
will become zero, bx plus c minus bx plus c will again become zero and then what does this
become, ax square plus bx plus bx plus c that is ax square plus 2bx plus c with the negative
sign.
Now opening using the third row, if we expand from the third row, we will get value of
Determinant is equal to minus times ax square plus two bx plus c into ac minus b square.
That is, b square minus ac into ax square plus bx plus a square plus 2 bx plus c. Now let us
see what is the correct answer, we have opened this Determinant and it is this, this
Determinant is not equal to option b. Option b is incorrect because it is negative sign.
This zero is also wrong. Now is this positive or negative? b square minus ac is negative, and
ax square plus 2bx plus c is positive, negative into positive is negative, hence Determinant is
always negative. Answer is option c. Simply applying one property and after expanding it
we got the value, after that we saw that the sign was negative, and we got the answer.
Questions are generally based on simple properties of determinants. Let us move forward.
Now see the second question on your worksheet.
If l, m, n are pth, qth and rth term of GP, let us assume l is equal to A into K raised to the
power P minus one, m is equal to A into K raised to the power q minus one, n is equal to A
into K raised to the power r minus one.
GP where the first term is A and common ratio r, so write the values of l, m, n they are pth,
qth and rth term of GP. We have to talk about the value of the Determinant which is equal
to log l p 1, log m q 1, log n r 1. Let us put value of log l, what is log l? Log A into K raised to
power of p minus 1, log A into K raised to the power q minus 1, log A into K raised
to the power r minus one. This is value of Determinant, let us simplify this, log A plus p
minus 1 log k, value of Determinant becomes this. Now let us apply C one goes to C one
minus log A into C 3, column first goes to column first minus log A into column third.
Applying this value of.
Std-11, Science, Mathematics, Chapter-12, Vectors and 3D
Good afternoon class, today we will study the topic Vectors and 3D.
Vectors and 3D is a very important topic. Many questions come in JEE Mains, majorly of 3D. You
have to learn this topic very well. In this short lecture people talk about what kinds of questions
come in competition exams and how to solve them. So today we will start with Vectors and 3D,
Vectors and 3 dimensional geometry.
First I hope you people know about dot products of vectors. If there are two vectors, a vector and b
vector and angle between them is theta, then a dot b is equal to mod a into mod b into cos theta.
There are two vectors a vector and b vector then their dot product is mod a mod b cos theta. And
we also know that a plus b dot a plus b, what is this, this can also be written as a plus b whole square
which finally becomes mod a square plus mod b square plus twice of a dot b and many similar
properties of dot product, we know this and many similar properties of dot product.
Let us see how to use this in first question of your assignment. First question of your assignment, a
plus b plus c is equal to null vector. a vector plus b vector plus c vector whole square is also equal to
null whole square that is null. That is mod a square plus mod b square plus mod c square plus twice
of a dot b plus b dot c plus c dot a is equal to zero. Mod a square is 25, mod b square is 16, mod c
square is 9 plus twice of a dot b plus b dot c plus c dot a is equal to zero. Hence a dot b plus b dot c
plus c dot a is equal to minus 25, minus 15 upon 2 that is minus 25. Hence modulus of this is equal to
25, answer of first question of your assignment 25. Very simple question based on dot product. a
plus b plus c is equal to null vector. a plus b plus c whole square is zero. Squaring both sides mod a
square plus mod b square plus mod c square plus twice of a dot b plus b dot c plus c dot a is equal to
zero. Hence a dot b plus b dot c plus c dot a is equal to minus 25. Hence mod of a dot b plus b dot c
plus c dot a is equal to 25.
Before starting with second question let us talk about in-center of the triangle, circle which is drawn
inside the triangle and which touches all three sides of triangle is called as in-circle, this circle is
called as in-circle. If position vectors of vertices of triangle are a vector, b vector and c vector, then
position vector of in center is I vector which is equal to BC times a vector plus AC times b vector plus
AB times c vector upon BC plus AC plus AB. In-center of triangle, if I is in-center of triangle and ABC
are vertices, position vectors of points ABC are a vector, b vector and c vector and position vector of
I is ivector, in-center is I vector, then I vector can be found by opposite sides, from each vector the
position vector multiply it with the length of the opposite side and then add it upon the length of all
the three sides, upon perimeter of triangle then we get the in-centre. Is this clear? Very good.
See, question number second is based on this thing only. If I is center of circle inscribed in a triangle
ABC, then we have been asked the value of this. Let us see, we know I vector is BC times a vector
plus AC times b vector plus AB times c vector divided by BC plus AC plus AB. Multiplying this and
simplifying we can see that it becomes BC times I vector minus a vector plus AC times I vector minus
b vector plus AB times I vector minus c vector is equal to zero. What is this? BC times AI vector plus
AC times BI vector plus AB times CI vector is equal to zero. We have taken minus as common from all
three. What I write here is BC times IA vector plus AC times IB vector plus AB times IC vector is equal
to zero. Hence in this question it is written that the sum is equal to what and the answer is the sum
is zero that means equal to null vector. Please understand this again. We know position vector of in-
center can be determined by this formula, just multiplying taking BC common AC common and AB
common we got this, and it becomes BC times IA vector and AC times IB vector and AB times IC
vector is equal to zero or null vector. The value that we have to find out we got that directly, answer
is null vector, option a. Very good.
Let us move on to third question of your worksheet, okay read third question. AB vector and AC
vector are given we have to find length of median. AB vector and AC vector are given we have to find
length of median.
Std-11, Science, Mathematics, Chapter-6, Complex Number
Good morning students, today, we will study Complex Numbers. Those numbers which are not real,
what do we call those numbers, Complex Numbers. Everybody of you know that the solution to this
type of equation is not real. So, this solution is called complex numbers. There are two solutions of
this equation, one is square root of minus 1 other is minus time square root of minus 1. Square root
of minus 1 is called iota. What is square root of minus 1? Iota square is minus 1, iota cube is iota
square into iota that is minus one into iota that is minus iota. And iota raised to par 4 minus 1 whole
square, iota whole square that is 1. So, what is iota, let me repeat again. Square root of minus 1, iota
square is minus 1, iota cube is minus iota, iota raised to par 4 is 1. Iota raised to par 5 again becomes
iota. Because iota raised to par 5 is iota raised to par 4 into iota and iota raised to par 4 is 1, hence
iota raised to par 5 is iota. So, you know what is iota? It is square root of minus 1. You have studied
briefly about iota in 11th standard. Now, we will use this and study about complex numbers. All
numbers which can be written in the form x plus iota y where x and y are real numbers are complex
numbers. All numbers which can be written in the form x plus iota y, where x and y are real numbers
are complex numbers. Where x is called as real part of complex number and y is called as imaginary
part of complex numbers. x is real part and y is imaginary part. Real part is also a real number and
imaginary part, which is y, it is also a real number. The real and imaginary part of complex numbers
both are real numbers.
Let us talk about Equality of two complex numbers. So, there are two complex numbers z1 which is
equal to x1 plus iota y1 and z2 which is equal to x2 plus iota y2, then z1 equals to z2 implies, z1 will
be equal to z2 when x1 is equal to x2 and y1 is equal to y2. When will two complex numbers be
equal when their real parts is also equal and their imaginary parts is also equal. Say, x minus 3 plus
iota y is equal to 4 plus 7 iota, when this type of equation is there, then we can directly say, x minus
3 is equal to 4 and y is equal to 7, that is x is equal to 7 and y is also equal to 7. We will equate real
parts separately and will also equate imaginary parts separately. This was example of equations of
complex number. In complex numbers, equations means real parts are equal as well as imaginary
parts are also equal.
Next thing that you know is Conjugate of a complex number. Say z is a complex number which is x
plus iota y, then z conjugate will be x minus iota y. Change the sign of imaginary part, whatever
complex number you get, it is called z conjugate. It is to be denoted by z bar, how do you denote it,
by z bar. Z conjugate is x minus iota y. Replace minus iota in the place of iota.
Let us solve few easy questions. Find the conjugate of z equals to 3 plus 4 iota. What will be the iota
of 3 and 4, 3 minus 4 iota. In place iota what you should replace? minus iota. Second question, find
the conjugate of z equals to 1 upon 1 plus iota. Z conjugate will be 1 upon 1 minus iota. What you
will replace in place of iota, minus iota. We get real value of conjugate. So similar question was
asked in AIEEE examination few years back, it was – find the conjugate of z equals to 1 upon iota
minus 1. That is first question of your worksheet. Very easy, z conjugate will be,
