Q13/120) The polynomial p(x)=x^3-3ax^2+bx-6 has factors (x-1) and (x-3). Find the value of the constants a and b?

This can be solved using  factor theorem and putting p(1) = 0 and p(3) = 0.

As below

p(1) = 1 – 3a + b – 6 = 0 or -3a + b = 5  … (1)

p(3) = 27- 27 a + 3b – 6 = 0 or 27 a -  3b = 21 … (2)

we can solve above 2 linear equations to get a = 2 and b = 11

But there is a shorter method

As (x-1) and (x-3) are factors so

P(x) = m(x-1)(x-3)(x-n) = x^3-3ax^2+bx-6

Now coefficient of x^3 = m = 1

Constant term = -3mn = -6 or n = 3

So f(x) = (x-1)(x-2)(x-3) = x^3 – 6x^2 + 11x – 6

comparing coefficients we get a = 2 and b = 11

Q13/119) Prove that : If .... Tan (x-y)/2 , Tan z and Tan (x+y)/2 are in Geometrical Progression then .cos x = cos y . cos 2 z

we have

tan A tan B + 1 = sin A/cos A sin B/cos B + 1 = ( sin A sin B + cos A cos B)/( cos A cos B)

= cos ( A + B)/ ( cos A cos B) ... (1)

tan A tan B - 1= sin A/cos A sin B/cos B - 1 = ( sin A sin B - cos A cos B)/( cos A cos B)
= - cos ( A - B)/ ( cos A cos B) ... (2)

from (1) and (2) (tan A tan B+ 1)/( tan A tan B- 1) = - cos ( A + B)/cos (A-B)
putting A= (x+y)/2, B= (x-y)/2 we get

(tan (x+y)/2 tan (x-y)/2 + 1)/tan (x+y)/2 tan (x-y)/2- 1) = - cos x / cos y ...(3)

from given condition

tan^2 z = Tan (x-y)2Tan (x+y)/2

or sin ^2 z/ cos ^2 z = Tan (x-y)2Tan (x+y)/2

using componendo dvidendo we get

( sin ^2 z + cos^2 z)/( sin ^2 z - cos^2 z) = (Tan (x-y)2Tan (x+y)/2 +1)/ (Tan (x-y)2Tan (x+y)/2-1)
or 1/(-cos 2z) = - cos x/ cos y

or cos x = cos y cos 2z

Q13/118) How many ordered quadruples of positive integers(w,x,y,z) are there such that w!=x!+y!+z!?

w cannot be 4 or more

as x = 3, y = 3, z =3 gives x! + y! + z! = 18(maximum) < 4 !

w = 3 ! =< x! + y!+z! = 6( x =2 , y = 2, z= 2) no more solution

w = 2! = 2 and so no solution as x = 1, z = 1, y = 1 give 3

w cannot be 1 then x,y,z cannot take any value

so one solution w = 3, x=y=z = 2

Q13/117) Find a such that limit of ((3x^2 + ax + a + 3) / (x^2 + x -2)) as x approaches -2.

(x^2+x-2) is zero at x = -2 so  of ((3x^2 + ax + a + 3) need to be zero else the ratio shall be infinite /zero form and it shall not exist

So f(x) = (3x^2 + ax + a + 3) should be zero at x = -2

f(x) = 12 -2a + a + 3 or a = 15

now ((3x^2 + ax + a + 3) = 3x^2 + 15x + 18 = 3(x+3)(x+2)

so ((3x^2 + ax + a + 3) / (x^2 + x -2)) = 3(x+3)(x+2)/((x-1)(x+2))

= 3(x+3)/(x-1) = 3 (1)/(-3) = - 1

Q13/116) To show that 1 = 1/2+1/4+ 1/8 + 1/16+ ….

We can prove from the RHS that

1/2+1/4+ 1/8 + 1/16+ .. is a GP with a= 1/2 and r = 1/2 and the GP converges to

1/2/(1-/12) = 1/2/(1/2) = 1

but beauty lies in expanding from LHS

1 = 1/2  + 1/2 

 = 1/2 + 1/4 + 1/4 ( take the last term and make into 2 halves)

=     1/2 + 1/4 + 1/8 + 1/8 and so on to get the result

Q13/115) If L.C.M of (a,b) is 432, and L.C.M of (b,c) is 72, and L.C.M of (c,a) is 432. Then the number of ordered pairs (a,b,c) is

LCM of  (a,b) is =432=2^4×3^3

LCM of  (b,c) is =72=2^3×3^2

 LCM of of (c,a) is =432=2^4×3^3

First find power of 2

Highest from LCM of (b,c) = 3 so  b cannot have power > 3 and c cannot power > 3

So power of 2 in a = 4

Now in b and c can be (3,0), (3,1), (3,2), (3.3), (0,3),(1,3),(2,3) as these combinations given power 3

Similarly you can find the power of 3

In a = 3 an in bc = (2,0),(2,1)(2,2),(1,2),(0,2)

So a = 432 and for bc there are 35 sets

for example one is (2^3*3^2,1)

Q13/114) Given f(x)=anx^n+an−1x^(n−1)+⋯+a1x+a0, where a0,aa,⋯,an are all smaller than 4 and not –ve Given that f(4)=2009, find f(1).

as no coefficient is >4 and we are given f(4) subtract the highest power of 4 as many times as it can go

f(4) =  1024 + 3 * 256 + 3 * 64 + 16 + 8 + 1 so

f(x) = x^5 + 3x^4 + 3x^3 + x^2 +2x +1
so f(1) = 11

Q13/113) If a=(√5+2)^101=b+p

where b is an integer, 0<p<1, evaluate ap

Solution 

If we take

a=(√5+2)^101

and b = (√5-2)^101

and expand both we see that the terms with odd power of (√5) shall be same in both and they shall be positive

so a-b =(√5+2)^101 - (√5-2)^101 is integer

now as (√5-2) < 1 so fractional part of (√5+2)^101 is (√5-2)^101 = p

so ap = (√5+2)^101 * (√5-2)^101 = (5-4) ^ 101 = 1

Q13/112) Show that if a > b then a^3 > b^3

We have (a^3-b^3) = (a-b)(a^2+ab+b^2)

We need to show that (a^2+ab+b^2) > 0

We have a^2 + ab + b ^2 = (a-b)^2 + 3ab … (1)

a^2 + ab + b ^2 = (a+b)^2 – ab … (2)

as a > b if a or b is zero from (1) or (2) a^2 + ab + b^2 > 0

if ab > 0 then from (1) a^2 + ab + b ^2 > 0

and if ab < 0 then from (2) a^2 + ab + b ^2 > 0

 from above we have the result

Q13/111) Given the system of equation below: a+2b+3c+4d=262 ..1 4a+b+2c+3d=123 ..2 3a+4b+c+2d=108 ..3 2a+3b+4c+d=137 .. 4 Evaluate 27a+28b+29c+30d.

Adding all (4) we get

10(a+b+c+d)=630 and this gives a+b+c+d=63…(5)

Mulitply (5) by 26 and add (1) to get

27a+28b+29c+30d = 63 * 26 + 262= 1900