Test Papers – ICSE – Class – X
Std-11, Science, Mathematics, Application of Derivatives
Good afternoon students, today we will study maxima and minima increasing decreasing
functions and other applications of Derivatives. Let us start.
Derivatives are very widely used and they have many different applications. As we studied in
the last lecture, firstly they are used in finding equation of tangents, equation of normals.
Secondly, they are also used for checking whether the function is increasing or decreasing.
Their third application is maxima and minima and mean value of theorems and many other
different applications. We have talked about tangents and normals in last class, today we
will talk about increasing, decreasing functions, maxima and minima and mean value
theorems.
Function is increasing when derivative is positive. Function is decreasing when derivative is
negative. When derivative is positive function is increasing function and when derivative is
negative function is decreasing function. If derivative is positive function will increase and
derivative is negative function will decrease. Increasing function will look like this, going
from left to right, value of function increases, graph goes upwards and decreasing function is
like this, going from left to right and graph goes downwards. Decreasing function and
increasing function. This graph is made when derivative is negative and this graph is made
when derivative is positive.
Maxima and Minima – I hope everybody had studied maxima and minima for school, yes,
very good. This is exactly same as what you studied for boards. Exactly, same types of
questions are asked. Firstly we have to find critical points. Critical points are the points
where f dash x is zero or not defined. First of all we will find critical points and after that we
will check if critical point is maxima or minima. Second point is we check whether critical
point is maxima or minima. Critical point is that point where derivative is zero. Now, how to
check whether critical point is maxima point or minima point. Say, x equals to a is critical
point, now we find f double dash a. We will find double derivative, if it is negative then it is
maxima or f double dash a is positive then x equals to a is minima. I hope everybody got it.
Yes, very good, to check maxima and minima of functions, we find critical points. Critical
points are the points whether f dash is zero or not defined. After that we can apply two ways
firstly what I have taught you is second derivative test. You find double derivative and put
critical point if the answer is negative then critical point is maxima point or if it is positive
then it is minima point. But double derivative point many a times fail, when double
derivative doesn’t exists only or double derivative instead of being positive or negative is
zero.
So, we have one more test which is first derivative test. What do first derivative test says? It
says if sign of derivative changes from negative to positive, what is this – sign of f dash x, on
the both sides of critical point, we have to check the sign of f dash x, if it goes from negative
to positive, then critical point is, function has increased instead of decreasing. As soon as the
function increases instead of decreasing, critical point is point of minima. You can very
clearly say function is increasing instead of decreasing. Means the value decrease at a
minimum value and then increases, so the critical point is of minima. And if sign of f dash
becomes turns positive to negative on the both sides of critical point. Means f dash should
change from positive to negative, after the functions increasing it decreases so the critical
point is point of maxima. To find maxima and minima functions, we need to do two things,
first is we need to critical points, second thing is we should check whether critical point is of
maxima or minima, for this we have two tests. Second derivative test and first derivative
test. In second derivative test, in double derivative value we will put critical value and if it is
positive then minima and if it is negative then maxima. In first derivative test, on both the
sides of critical points, we will check sign of f dash, if first derivative sign changes from
negative to positive, so the function is from decreasing to increasing, means this point will
be of minima. And if first derivative sign changes from positive to negative, then the function
will go from increasing to decreasing. So this point will be of maxima. And third thing which
may happen is.
Std-11, Science, Mathematics, Probablity
Greetings for the day, today we will study topic Probability. Probability, in which we will talk
about chance of happening of an event, chance of happening of any event. Say, everybody
knows about classical definition of Probability. We had studied in 10th the classical definition
that probability of any event is equal to number of favourable cases divided by total number
of sample space. Say, we rolled the dice, these are the six possible sample space which can
happen. Question was asked, probability of getting number divisible by 6, how many sample
space is there totally – 6, how many favourable cases, getting number divisible by 6, how
many numbers are there which is divisible 6? Only 1. So what will be the probability? 1 by 6.
Say, let us take one more question. Dice is rolled, what is probability of getting an even
number? What is the sample space? 6, that is 1, 2, 3, 4, 5, 6, from this 6, any number can
come. How many even numbers are there? 2, 4, 6 – 3 cases are favourable hence the
answer will be half ( ½ ). These are very elementary questions that we people did in 10th
class.
Further moving on we did few questions based on Permutations and combinations in 11th
class and you studied more about conditional probability based theorem, total probability
theorem and binomial distribution in 12th class. Today, we will talk about all these theorems
and the questions based on these, majorly the questions which come in competitive
examination.
Let us start, first question of the assignment, an old man was dialling a 7 digit telephone
number, an old man was dialling a 7 digit telephone number, after dialling the first 5 he
forgets the next 2, but he remembers that both are different. What is the probability of
making a right call? What is the total work, to dial 2 numbers? How many ways are there to
dial first number – 10 ways. After dialling first, how many ways are there to dial next one – 9
because he knows that both numbers are not same. So, how many ways are there for
dialling the two numbers – 90 and how many favourable? So how ways are there for right
calls? Only 1. The correct number only will be dialled. Hence, probability is 1 by 90, answer is
1 by 90. To dial two numbers, how many ways are there? 10 into 9 and in favourable way 1,
so the answer is 1 by 90.
Second question of the assignment. There are 4 apples and 3 oranges. They are arranged
randomly. What is the probability that starting and ending position is occupied by oranges?
Let us consider all the apples different and all oranges different. We have to arrange, total
number of sample space 7 factorials. To arrange 7 things in 7 places, so how many ways — 7
factorials. Number of favourable cases. What do we have to arrange in the starting and
ending – orange? To arrange orange in the start is 3, after arranging orange in the start, so
how many ways are left to arrange orange in the end, 2 ways. After keeping 3, how many
left – 2. Now, after arranging these two oranges, how many fruits are left? 4 apples and 1
orange, to arrange them how many ways are there? There are 5 things which can be kept in
5 ways, after that 4, 3, 2, 1, total, so how many ways are there to arrange these – 6 factorials.
After multiplying all the numbers, you will get the answer as 6 factorials. Number of
favourable is 6 factorial. Hence probability is 6 factorial divided by 7 factorial, that is 1 by 7.
There are 4 apples and 3 oranges, they were arranged randomly, what is the probability that
oranges are at terminals, starting and ending position, oranges are there. Even if you solve
this question considering all apples to be same and all oranges to be same. Then also we will
get same answer, why. Total number of sample space, to arrange 7 things, how many ways –
7 factorials. Why we divided, because 4 apples are same and 3 oranges are also same. So,
how many favourable ways? In the starting 1 orange is kept and in the ending also 1 orange
is kept. So, 1 way and to arrange in 5 space, 5 fruits are left. So, to arrange these fruits, how
many ways are left? 5 factorial divided by 4 factorial is 5. Hence probability is 5 by 35 that is
1 by 7, got it. Starting orange in one way and at ending keep any orange, it is same. So, do it
by thinking oranges and apples are same or different, the answer will be same. But,
Std-11, Science , Mathematics , Permutation and Combination
Good afternoon, students, today we will start with next topic, Permutations and
Combinations. In permutations and combinations we will talk about counting. We will know
how to count fast. In this we will study how to count quickly.
For example, in this row 8 students are sitting. One way of counting is 1, 2, 3, 4, 5, 6, 7 and 8.
Other way of counting is 4 rows, 2 students in each row, 4 into 2 is 8. So what I used is I used
multiplication as tool. In the same way to count quickly we can use many different kinds of
tools by which counting will be very fast. And if we want to count in lakhs and crore which
will be difficult but counting by technique it will be easy. So, this is what we will learn in this
chapter.
Permutations and Combinations
Let us talk about first very important thing fundamental principle of multiplication,
fundamental principal of multiplication. If an event can occur in ‘m’ different ways, following
which another event can occur in ‘n’ different ways, then the total numbers of ways of
simultaneous occurrence of all these events in a definite order is m into n. If an event can
occur in ‘m’ different ways, following which another event can occur in ‘n’ different ways.
Then the total number of ways of simultaneous occurrence of all these events in a definite
order is m into n. If one work is done in ‘m’ way, following which means after that another
work can be done in ‘n’ way. Then simultaneously both, in same order means after one is
finished then second should be done. In this way, how many ways are there – m into n.
Simultaneously, means first one should be done and then second one also should be done.
We have to do not only one but both of them should be done. Total number of ways will be
m into n.
Let us take an example, say if from you house to reach Lakshya and you come by auto. You
come to Lakshya by auto. If there are 2 ways to reach from your house to the auto, say 1 and
2, and from the auto to reach Lakshya there are 3 different ways, a, b, c. So, to reach
Lakshya from your house, how many ways will be there? 1a, 1b, 2a, 2c. 1a, 1b, 1c, 2a, 2b, 2c.
To reach Lakshya from house how many different ways will be there – 6. So to arrive at this
6, how fast can we do this? What is the work, to come from house to Lakshya. It is divided in
2 different ways, from house to auto, from auto to Lakshya. To do the first work, there is 2
ways and from auto to reach Lakshya, 3 ways. To do both the work simultaneously how
many ways, 2 into 3 is 6. So, to reach from house to Lakshya how many ways are there? 6.
Let us take one more example, say from house to reach classroom. You come by auto. From
house to reach the auto 2 ways, from auto to reach Lakshya – 3 ways and from Lakshya to
reach classroom there are 2 ways. So, how many ways will be there to reach classroom from
your house? 1ap, 1aq, 1bp, 1bq, 1cp, how, 1cp, if you follow the first way from home to
auto, then auto came by c way and after reaching Lakshya you took p way to reach
classroom and in the same way 1cq. Now how much this is? 6. Similarly, 2ap, 2aq, 2bp, 2bq,
2cp, 2cq, total is 12. Now, to count this it took some time. Suppose if I use the formula. Total
work is divided in 3 parts. To do 1st work 2 ways, after doing 1st to do second work 3 ways
and after 2nd work to do 3rd work 2 ways. Hence total number of ways is 2 into 3 into 2 that is
12. I hope you people got it, fundamental principle of multiplication. If to do first work there
is ‘m’ ways, for 2nd work ‘n’ and for 3rd work ‘p’ then to do all the 3 simultaneously, one after
then other is m into n into p.
Let us take few more examples. In how many ways 5 people can be seated in a 5 seater car?
5 people, 5 seater car. In how many ways 5 people can be seated in a 5 seater car? How
many place is there for a to sit, 5. After doing first work, after making place for a, so, how
many seats are there for b to sit, a must have sat in one seat, 1 place has been used from 5.
Now how many seats are left for b? 4. How many seats for b to sit – 4. Now two seats are
used. So after making place for a and b, how many seats are left c? 3. After making a, b, c,
sit, now only 2 seats are left. For d 2 seats and for e 1 seat. To do all the work
simultaneously, what we will do, multiple. So, to make 5 people sit in a 5 seater car, there
are 120 ways. For a there are 5 places, after making place for a, for b there are 4 places. That
means if I had to make 5 people to sit one by then first made a to sit, then b, then c, then d
and e. What you always do is you break whole of the work into small jobs. To do all the
works you write it down and then you multiple them. When will it be applied, when you do
all the work simultaneously, one by one.
Let us take one more example. How many different words can be formed using letters of
word ‘DELHI’? How many different words can be formed after arranging word ‘DELHI?’ It will
be 5 letter words, because there is 5 letters. We should think we have to fill five places. To
fill the first place 5 ways, after the first place is filled any one letter will be used. How many
letters are left – 4, so to fill the second place, 4 ways because letters can’t be repeated, to fill
second place 4. After filling first 2 places, in 3 place 3, then 2 and then 1. So, what will be the
total after filling all the places? 120. We have to arrange them to make a word. We can’t add
new letter and also can’t repeat the existing letters. So, to fill first place 5 ways, after 1 letter
is used, so 4 letters are left. To fill the second place, 4 ways, now 2 letters have been used.
So, for the third place 3 ways, then 2, then 1.
Let us solve few more questions, question first of your sheet. How many 5 lettered words
can be formed by rearranging letters of word ‘knife’ such that they start and end with
vowels? Where all is it constant, in the beginning and at end, what will come here, vowels?
So, from where it is constant, start question from there. From where it is constant, start
question from there? In the first place, vowel will only come. How many ways are there to
fill first place? 2, i or e, first place.
State Board Semi-English Sample Test Paper (2) Std 10 Sanskrit Composite
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State Board Semi-English Test Paper (1) Std 10 Sanskrit Composite
One of the ways to make sure that you are prepared for your exams is to solve question papers. Prepared by our experts in accordance with the guidelines given by STATE BOARD SEMI-ENGLISH, These Sample Papers for class 10th Sanskrit Composite will help students gain confidence and make them ready to face their school examinations This way you know how strong your preparations are and can realize your strengths and weaknesses. Download the STATE BOARD SEMI-ENGLISH Sanskrit Composite Sample Papers and test yourself. Take full advantage of the STATE BOARD SEMI-ENGLISH Sanskrit Composite Sample Papers. Good Luck!
Why should you refer these Sample Papers?
- These sample papers help you in revising your learnt concepts.
- Since they are based on the STATE BOARD SEMI-ENGLISHpapers, so they help you in giving a feel of the final exam. This actually helps in improving your performance on the final day.
- These papers aid in finding out your weak points so that you can further brush up your skills.
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State Board Semi-English Sample Test Paper (2) Std 10 Sanskrit Entire
One of the ways to make sure that you are prepared for your exams is to solve question papers. Prepared by our experts in accordance with the guidelines given by STATE BOARD SEMI-ENGLISH, These Sample Papers for class 10th Sanskrit Entire will help students gain confidence and make them ready to face their school examinations This way you know how strong your preparations are and can realize your strengths and weaknesses. Download the STATE BOARD SEMI-ENGLISH Sanskrit Entire Sample Papers and test yourself. Take full advantage of the STATE BOARD SEMI-ENGLISH Sanskrit Entire Sample Papers. Good Luck!
Why should you refer these Sample Papers?
- These sample papers help you in revising your learnt concepts.
- Since they are based on the STATE BOARD SEMI-ENGLISHpapers, so they help you in giving a feel of the final exam. This actually helps in improving your performance on the final day.
- These papers aid in finding out your weak points so that you can further brush up your skills.
So, what are you waiting for? Just click here to download the STATE BOARD SEMI-ENGLISH Sanskrit Entire sample paper
