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Lesson Posted on 07 Feb Tuition/Class XI-XII Tuition (PUC)/Mathematics

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) =... read more

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)?

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.

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Lesson Posted on 07 Feb Tuition/Class XI-XII 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(x-1-7) (100+y-20) x-8 y+80x-8 = 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... read more

Rs. paise
x y
-
7 20

(x-1-7) (100+y-20)

x-8 y+80

x-8 = 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.

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Lesson Posted on 07 Feb Tuition/Class XI-XII Tuition (PUC)/Mathematics

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... read more

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 Tuition/Class XI-XII Tuition (PUC) Tuition/Class XI-XII 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)^2-4*1991a-4 (1991-a)^2-4 (1989-a)(1993-a) 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 a-b = -1 and... read more

Method 1:

x^2+(1991+a)x+1991a+1

Now for integers d should be either 0 or perfect square so,

(1991+a)^2-4*1991a-4

(1991-a)^2-4

(1989-a)(1993-a)

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

a-b = -1 and 1991 - b = 1

or

a-b=1 and 1991-b = -1

2 values of a and b

a = 1989 and 1993 Sum = 3982.

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Lesson Posted on 07 Feb Tuition/Class XI-XII 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 a-3d, a-d, a+d, a+3d 2a - 4d = 4 2a + 4d = 6 a= 5/2; d = 1/4 A = (a-3d)(a-d) = 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.,... read more

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 a-3d, a-d, a+d, a+3d

2a - 4d = 4

2a + 4d = 6

a= 5/2; d = 1/4

A = (a-3d)(a-d) = 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.

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Lesson Posted on 07 Feb Tuition/Class XI-XII Tuition (PUC)/Mathematics

Sujoy D.

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)(x-1)(x-2) x... read more

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

Method 1:

(x+2)(x+1)(x-1)(x-2)

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.

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Lesson Posted on 07 Feb Tuition/Class XI-XII Tuition (PUC)/Mathematics

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(1-f) 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... read more

Question:

If x = (9 + 4rt(5))^48 = [x] + f, where [x] is integral part of x and f is a fraction, then x(1-f) equals:

a) 1 b) less than 1 c) more than 1 d) between 1 and 2.

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(1-f) = 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 XI-XII 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: Best approach: Let f(x-1) = 1 and f(x) = 1So 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... read more

Method 1:

Best approach: Let f(x-1) = 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(x-1)=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

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## Class XI-XII Tuition (PUC) in:

Lesson Posted on 07 Feb Tuition/Class XI-XII Tuition (PUC)/Mathematics

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Question: If a+b+c=0, then the quadratic equation 3aX^2+2bX+c=0 has 1) at least one root in (0,1) 2) one root in (2,3)and the other on (-2,-1) 3) imaginary roots 4) none of these Answer: a = 1 ;b = -1 ; c = 03x^2 - 2x = 0x(3x - 2) = 0x = 0 ; 2/3 option 1 correct. read more