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let the quadratic eq ax^2 +bx + c be such that a,b,c are distinct and each of a,b,c belong to {1,2,3.....n} such that x+1 divides ax^2+ bx +c.If such quad.polynomials are < 99,then max(n)=?

a)14  b)15   c)16   d)18   e)none  of these

Since, x+1 satisfies the equation ax^2+ bx+ c=0. Therefore, a(-1)^2+b(-1)+c=0. So, a - b+ c=0.
Now, a + c=b. Minimum value of b is 3. Number of quadratic equations will exceed 99 when b=12. So, the answer is none of these.

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38 Answers
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Find the condition for the equation ax^2 + bx + c = 0 for one root is n times the other

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X=2+2^2/3+2^1/3, then the value of x^3-6x^2+6x?

Hello Tarishi 

 

(x - 2) = 2⅔ + 2⅓ 

Cubing both sides 

(x - 2)³ =( 2⅔ + 2⅓ )³

 

 

= (2⅔)³ + ( 2⅓)³ + 3•2⅔• 2⅓( 2⅔ + 2⅓ )

 

[( a + b)³ = a³ + b³ + 3ab ( a + b )]

 

 

6 + 3 × 2(2⅔ + 2⅓) = 6 + 6(x - 2)

 

x³ - 8 - 6x² + 12x = 6 + 6x - 12

 

x³ - 6x²  + 6x = 6 - 12 + 8 = 2. 

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Solve the equation:x^4-2x^3+4x^2+6x-21=0 if two of its roots are equal in magnitude but opposite in sign .

Hello Tarishi , 

PFA the solution 

Solution
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If the constant term of (2x-1/x)^n is -160 ! Find n

Hello Tarishi , 

 

General term of (2x – 1/x) n   = nCr (2x)r (-1/x)(n-r)

= nCr (2r) (-1)(n-r) xr(1/x)(n-r)

Now ,
for constant term , power of x = 0

r = n – r

n = 2r

Constant Term = 2rCr2r (-1)r= - 160

r = 3 satisfies , Hence n = 2r = 6 .

0

Find a four digit perfect square such that its first two digits are the same and its last two digits are the same.

Hello Samyak , 

Let the four digit number be aabb 

1100a + 11b = x²

Clearly , x is a multiple of 11 

so , 1100a + 11b = (11y)²

100a+ b = 11y²

a + b mod 11 = 0 

b = 11-a

100a +11 -a = 11y² 

99a +11 = 11y² 

9a +1 = y² 
Only possibility , a = 7, 

So , b = 4. 

aabb = 7744

 

Alternate Approach : 

We know 122 ends with __44, i.e last 2 digits repeating

So the square must be (50k - 12)2 or (50k + 12)2

122 = 144 
382 = 1444
622 = 3844
882 = 7744 

Hence , 7744. 

0
20180826 005752

p² + q² = ( p+ q)² - 2pq 

 

We know , sum of the roots = -  ( Coefficient of x) / ( Coefficient of x²) 

 

So , p + q = ( a -2) 

 

and the product of the roots = ( constant term )/ ( coefficient of x² )

 

=> pq = - ( a +1)

 

p² + q² = (a -2)² + 2( a + 1) 

 

= a² + 4 - 4a + 2a + 2 

 

= a² -2a + 6 

 

= ( a -1)² + 5 . 

 

minimum value of ( a -1)² is zero. Hence the minimum possible value of p² + q² is 5. 

 

Option (D) 

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In the expansion of (x - √x - 7)^7, what is the coefficient of x^(5/2)?

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Hello sir , Please provide the solution to this problem

IMG 20180914 163833 EDIT 1 EDIT 1

 answer 101?

open all the brackets with + sign and solve it will give you x<=36

open all the brackets with -ve sign and solve it will give you x>=(-64)

total integer values =36+64 +1              1 is for 0

=101 (answr)

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Lord of Mordor distributed some magical rings among his Nazguls. The first Nazgul received 100 rings and one tenth of the remaining rings. The second Nazgul received 200 rings and one tenth of the remaining rings. The third Nazgul received 300 rings and one tenth of the remaining rings... and so on. After distributing the rings in this way, the Lord found out that he had given equal number of rings to all the Nazguls. How many Nazguls were there?

A. 6

B.9

C. 12

D. 13

9 is answer?

Let the total rings be n
1st  Nazgul recieved => 100 + (n - 100)/10 rings.  

2nd Nazgul recieved=> 200 + 1/10(n - 300 - (n- 100)/10) rings . 

since these both are equal, equate it.

n = 8100

so 1st Nazgul recieved = 100 + 8000/10 = 900

Since all person received same amount Hence , 8100/900 = 9 Nazguls

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A tells B , when I was 3/5th of your present age you were 5/7th of my present age . Present age of A is 28 years . B's Present age is ? 

Hello Richa, 

 

Let the present age of B be x years.  

 

So ,when A was 3/5 of x. That time B was 5/7 of 28 = 20. 

 

Now , 

 

28 - 3/5 x = x -20 

{ Difference should be constant } 

x = 30 . 

 

 

0
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TG FORUM SOl
0
15392741636701643210798

Sir,please provide the solution for both questions.

27th

save

Solution to the 27th Question : 

-2 , -1 , 1 and 2 are the roots of the equation so 

f(x)=k•(x+2)(x+1)(x-1)(x-2).

f(p)=k(p+2)(p+1)(p)(p-1)(p-2)/p

=>  f(p) × p is a product of 5 consecutive numbers.

So,must be divisible by 5.

Now p, p+1, p+2 are 3 consecutive numbers,so should be divisible by 3,will say the same for p, p-1, p-2. So, divisible by 3² i.e 9. 

p-1, p+1 are consecutive even numbers,must be divisible by 8(Because 2 × 2²).

Hence , divisible by 5, 9 and 8

Largest Integer =5 × 8 × 9 = 360.

 

Option (4) 

28th

x² + 2y² + 4z² + 3 ( xy + yz + zx) = 16 ... ( 1) 

y² + 2z² + 4x² + 3 ( xy + yz + zx) = 16 ... (2) 

 

z² + 2x² + 4y² + 3 ( xy + yz + zx) = 16 ... (3) 

On adding (1) , (2) and (3) 

 

7 ( x² + y² + z²) + 9 ( xy + yz + zx) = 48

 

x² + y² + z² =3

xy+yz+zx=3

x=y=z=+1 or-1

so only 2 solutions are possible 

hence option 2

0
F74A5512 D269 4571 9D26 67803D016D91

explanation please

for a cubic polynomial ax3 + bx2+cx+d=0 

three roots can be described as 

p +q+r=-b/a    

pq+qr+pr= c/a

pqr=-d/a

so in given equation 

p+q+r=0.................(1)

pq+qr+pr=a.................(2)

squaring (1) 

p2 +q2+ r2  +2(pq+qr+rp)= 0

p2 +q2+ r2  =-2(pq+pr+qr)

from (2)

p2 +q2+ r2  =-2a

a=-(p2 +q2+ r2  )/2

since roots are integers and squaring them will always give a positive value 

hence a will always be negative integer

so 

answer is option (3)

0
0ED2E302 D3A9 4378 ACA3 8B018D581A25

solution please 

answer is 4th option?

0
SERIES

Hi Richa , 

1540477531257 1701218473

 

thankyou so much. this helped

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