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