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Lesson Posted on 08 Mar CBSE/Class 10/Mathematics CBSE/Class 12/Science/Mathematics
Honesty and Dedication helps in education and success in life.
Jagdish S.
Mathematics teaching for CBSE ICSE,PUC state CET,JEE IIT and Engineering maths ,CAT,MAT,International...
Honesty and Dedication helps in education and success in life.
As per my 22 years of experience of teaching mathematics i would like to say to all student that we should be Honesty and Dedicated in study then we will be successful in life. If we earn money or marks by wrong way then we will not feel happiness. Get less mark but with honesty that will make you feel happiness.
read lessAnswered on 24 Feb CBSE/Class 12/Science/Mathematics
Srinivas Rao
Math Guru
Lesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics
If A=(1+ 1/38)^38 And B=(1 + 1/37)^37 Then 1) A=B, 2) A>B, 3) A>4, 4) A
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Answer:
Method 1:
A = [(k + 1)/k ]^k
B = [(k/(k  1)]^(k  1)
Apply PMI now to get : A > B
Method 2:
Solution:
(1+ 1/n)^n = e ~ 2.78xx.. as n tends to infinity.
Now its an increasing function because just check for
A = (1 +1/2)^2 and B= (1 +1)^1 ==> A= 9/4= 2.25 and B=2. So clearly A>B here.
So, as the function is increasing so, it will also satisfy for our given A and B. i.e., A>B here.
f(x)= (1 + 1/x)^x is alwz an increasing function. and as it is increasing function so, if x> y then f(x) > f(y) alwz.
It also satisfies from here.
read lessLesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics
What Is The Value Of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Question:
A function f(x) is defined as f(x) = f(x – 2) + x(x – 2) for all the integer values of ‘x’. Given that f(1) + f(6) = 0. What is the value of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?
Answer:
Method:
f(x) = f(x – 2) + x(x – 2)
f(1) + f(6) = 0 => f(6) = f(1)
f(2) = f(0)
f(3) = f(1) + 3
f(4) = f(2) + 8
f(5) = f(1) + 18
f(6) = f(4) + 24 = f(2) + 32
f(2) = f1  32
f(4) = f1 24
f(1) + ... f(6) = f1 32 f1  24 + f1 + 3 + f1 + 18
= 56 + 21 = 35.
read lessLesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Answer:
Rs. paise
x y

7 20
(x17) (100+y20)
x8 y+80
x8 = 3y
y+80 = 3x
x = 29
y = 7
x+y = 36
I have actually made two columns one for Rupees one for Paise! Here messed up that format! Anyway, I divided it into Rupees and Paise separately instead of converting the entire thing into paise! That way we are getting two variable but only one eqation.
read lessLesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics
How Many Pairs Of Integers (x, y) Exist Such That x^2+ 4*y^2< = 100
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Question:
How many pairs of integers (x, y) exist such that x^2+ 4*y^2<= 100?
Solution:
Method 1:
Case and Fundamental counting:
y = 0, 1, 2 =>19 ways
y = 3 =>17ways
y = 4 =>13 ways
y = 5 => 1 way
Total = 19*3+17+13+1=88
Similarly for y negtve total ways =19*2+17+13+1=69
Again for y=10 x=0 and y=10 x=0
So total ways = 88+69+2=159
Method 2:
Ellipse: Geometrical approach:
x^2/10^2 + y^2/5^2 = 1 => Ellipse
Case 1: 4 Boundary points cutting x and y axis.
Case 2: x =+/ 9 => y = +/2 and 0 => 19*5 = 95 points
Case 3: y = +/ 3 => x^2 <= 64 => 2 * 17 = 34 points
Case 4: y = +/ 4 => x^2 <= 36 => 2 * 13 = 26 points
Total: 4 + 95 + 60 = 159 points
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Lesson Posted on 07 Feb CBSE/Class 12/Science/Mathematics
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Question:
The equation 3x+11y=k where k is a 2 digit number has exactly three solutions in which both x and y are positive integers.
Which of the following is not a value of x? 5,17,19,25
Answer:
In equation 3x+11y = k, x increases by 11 and y decreases by 3 as k is constant.
Now take the value of x= 5 and put y =1 we get k= 26(true) Next value of x = 5 + 11 = 16, k = 59(true).
Next value of x = 27, k = 92(true)
So solution set = {5, 16, 27}
Similarly check for x = 17.
For x = 19, we have k = 68 (true),
Next x = 30 then k = 108(false)
Even you take x = 19  11 = 8, k = 35 (true) but on x = 30 is false which is the third possible value of a solution set. and
At last x = 25, k = 86 (true)
Next value of x = 3, 14 having k = 20, 53 respectively.
So solution set {3, 14, 25}.
read lessAnswered on 05/06/2017 CBSE/Class 12/Science/Mathematics Tuition/Class XIXII Tuition (PUC)
Ca Prashanth Reddy
Tutor
Answered on 23/05/2017 CBSE/Class 12/Science/Mathematics Tuition/Class XIXII Tuition (PUC)
Ca Prashanth Reddy
Tutor
Looking for Class XIXII Tuition (PUC)
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Answered on 25 Feb CBSE/Class 12/Science/Mathematics Tuition/Class XIXII Tuition (PUC)
Salman Ahmad Siddiqui
Maths Tutor
UrbanPro.com helps you to connect with the best Class XIXII Tuition (PUC) in India. Post Your Requirement today and get connected.
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