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Lesson Posted on 07 Feb Tuition/Class XIXII Tuition (PUC)/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 Tuition/Class XIXII Tuition (PUC)/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 Tuition/Class XIXII Tuition (PUC)/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 Tuition/Class XIXII Tuition (PUC)/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
read lessLesson Posted on 07 Feb Tuition/Class XIXII Tuition (PUC) Tuition/Class XIXII Tuition (PUC)/Mathematics
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Answer:
Method 1:
x^2+(1991+a)x+1991a+1
Now for integers d should be either 0 or perfect square so,
(1991+a)^24*1991a4
(1991a)^24
(1989a)(1993a)
They both can be equal so for d = 0
a = 1989, a = 1993
Method 2:
b and c are roots
(b+a) (b+1991) +1 = 0
(b+a) (b+1991) = 1*1
ab = 1 and 1991  b = 1
or
ab=1 and 1991b = 1
2 values of a and b
a = 1989 and 1993 Sum = 3982.
read lessLesson Posted on 07 Feb Tuition/Class XIXII Tuition (PUC)/Mathematics
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Answer:
Method 1:
OA: 63/16
p, q, r, s are in HP
Put x>1/x then roots in AP
x^2  4x + A = 0 and
x^2  6x + B = 0
Let roots be a3d, ad, a+d, a+3d
2a  4d = 4
2a + 4d = 6
a= 5/2; d = 1/4
A = (a3d)(ad) = 63/16
B = (a+d)(a+3d) = 143/16
Method 2:
(p+q)/pq = 1/p+1/q = 4
Similarly 1/r+1/s = 6., 1/p, 1/q, 1/r, 1/s in Ap. So find p, q and find A using p+q = 4/A
p, q, r, s are in HP
Put x>1/x then roots in AP.
read lessLesson Posted on 07 Feb Tuition/Class XIXII Tuition (PUC)/Mathematics
The Largest Integer That Divides F(p) For All Values Of P Is : A) 72 B) 180 C) 240 D) 360 E) 720
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1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Question:
F(x) is a fourth order polynomial with integer coff and with no common factor.The roots of F(x) are 2,1,1,2.If p is a prime number greater than 97,then the largest integer that divides F(p) for all values of p is :
a) 72 b) 180 c) 240 d) 360 e) 720
Answer:
Method 1:
(x+2)(x+1)(x1)(x2)
x = 101 put
103*102 * 100*99. Largest integer hence = 360
All values of p, so 1st prime number after 97 consider it.
Method 2:
Every prime number can be expressed in 6k+1 form.
read lessLesson Posted on 07 Feb Tuition/Class XIXII Tuition (PUC)/Mathematics
x(1f) Equals: A) 1, B) Less Than, 1 C) More Than 1, D) Between 1 And 2
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Question:
If x = (9 + 4rt(5))^48 = [x] + f, where [x] is integral part of x and f is a fraction, then x(1f) equals:
a) 1 b) less than 1 c) more than 1 d) between 1 and 2.
Answer:
Method:
Take y = (9  4rt(5))^48 ; 0 < y < 1
=> xy = 1 and x+y = Integer
x + y = [x] + 1
[x] + f + y = [x] + 1
y = 1  f
So x(1f) = xy = 1
0 < y < 1
And x + y = Integer
When you add x + y, integer that would come would be [x] + 1.
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Lesson Posted on 07 Feb Tuition/Class XIXII Tuition (PUC)/Mathematics
If f Be A Function Such That f(x+1) + f(x1) = Root2 * f(x). Find f(x+4)/f(x)
Sujoy D.
1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
Answer:
Method 1:
Best approach: Let f(x1) = 1 and f(x) = 1
So the sequence becomes : 1, 1, (√2)  1, 1  (√2), 1, 1, 1  (√2), (√2)  1, 1, 1, ...
i.e. f(x) is a periodic function with period 8 and after every four terms, terms start repeating with negative sign as shown above.
Method 2:
Let f(x)=a and f(x1)=b then solve in terms of a and b we will get f(x+4)=a so ans is 1
f(x+2) + f(x) = rt2 . f(x+1)...... i
f(x+3) + f(x+1) = rt2 . f(x+2) ............ii
f(x+4) + f(x+2) = rt2 . f(x+3)
Or f(x+4) + f(x+2) = 2f(x+2)  rt2 . f(x+1)....... from ii
So f(x+4) = f(x+2)  rt2. f(x+1)
So f(x+4) =  f(x)..from i
So answer = 1
read lessLesson Posted on 07 Feb Tuition/Class XIXII Tuition (PUC)/Mathematics
If a+b+c=0, Then The Quadratic Equation 3aX^2+2bX+c=0
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1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....
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