Q13/130) For real numbers a,b what is max(a,b)+min(a,b)

It is a + b

To prove there are 3 cases

1) a >b

In this case max(a,b) = a and min (a,b) = b and so sum = a +b

2) a= b

In this case max(a,b) = min (a,b) = a and so sum = 2a and a+ b= 2a

3) a < b

In this case max(a,b) = b and min (a,b) = a and so sum = a +b

so in call cases sum = a + b

Q13/129) What is the value of x in the following equation : (ab)^2 = (bc)^4 = (ca)^x = abc

From (ab)^2= abc we have ab = (abc)^(1/2)

Similarly (bc) = (abc)^(1/4)

ca = (abc)^(1/x)

multiplying all 3 we get (abc)^2 = (abc)^(1/2+1/4 + 1/x)

or 2 = ½ + ¼ + 1/x or x = 4/5

Q13/128) Consider n^2 + 20n + k = m^2 where k is an integer. As k varies, you can get cases of 0 or more solutions. What values of k causes 0 solutions?

if k is 2 mod 4 then there is no solution

Reason:
(n^2+20n + k) = (n+10)^2 + (k-100)

Now if n is odd then (n+10)^2 + (k-100) mod 4 = 1 + 2 = 3 so cannot be perfect square
if n is even then (n+10)^2 + (k-100) mod 4 = 2 so cannot be perfect square

so n is neither odd nor even .

so no solution

Q13/127) find integers in positive solutions to xyz + 20xy + yz + 5zx + 100x + 20y + 5z = 1913

Because we have xyz, xy, yz, xz, x, y, z terms in the left we should add some d and factor if possible

(x+a)(y+b)(z+c) – d

= xyz + ayz + bzx + cxy + bcx + abz + acy + abc – d

Comparing with given expression we get a = 1,  b = 5, c = 20

So we get (x+1)(y+5)(z+20) = xyz+ yz +20xy + 5xyz + 100 x + 5z + 20y + 100

Comparing with coefficient we get d = 100

So (x+1)(y+5)(z+20) = 1913 + 100 = 2013 = 3 * 11 * 61

So x = 2, y = 6, z= 41

Q13/126) If one of the roots of the equation 2x^2 - x -2 = 0 is a, prove that the other root is 4a^3 - 6a - 3/2

One solution is a so

2a^2 –a -2 = 0

Let other root be p = 4a^3 -6a – 3/2

So 4a^3 – 4 a = 2a^2

So 4a^3 -6a – 3/2 = 2a^2  – 2a – 3/2 = (2a^2 – a – 2) – a + ½ = ½ -- a

Now sum of roots = ½

Product of roots =  a(1/2- a) = ½a(1-2a) = 1/a – a^2 = - ½(2a^2- a)  = -1

So p is another root

Q13/125) For how many positive integer values of n does the equation 2x^2+689x+n=0 have an integer solution?

let (2x+ a)(x+b) = 2x^2+689x+n

as n is +ve both a and b positive or -ve, 

or 2x^2 + (a + 2b) = 2x^2+689x+n

a and b cannot - ve as sum is positive

0 < 2b < 689 and hence 0 < b < 344.5

so there are 344 values of b and also n

Q13/124) Let P(x) = x^3 - a x^2 + x - b Prove that there is a root between a and b (inclusive)

P(a) = a^3 - a^3 + a - b = a - b
P(b) = b^3 - a b^2 + b - b = b^2 (b - a)

If a = b then P(a) = P(b) = 0 so a and b are roots

If a and b are not same then P(a) and P(b) are of opposite signs so according to the intermediate value theorem, there exists c between a and b, such that

P(c) = 0.

Q13/123) Are there any repeating digit whole numbers that are perfect squares?

that is

Is there any number N that satisfies all of the following conditions?

(A) N is a whole number greater than 9.
(B) N is made of a single digit that repeats (1,111 or 777,777, e.g.)
(C) N is a perfect square.

Solution

The squares mod 10 can be ending with digits 1,4,9,6,5,0
so the numbers with all 2,3,7,8 cannot be perfect square

11 mod 4 , 55 mod 4, 99 mod 4 = 3, 66 mod 4 = 2

so the numbers with all 1,5,9,6

as numbers ending with all 4s ( all 11’s * 4) cannot be a perfect square

so there is no N.

Q13/122) If k is a positive integer and x is an integer such that x - 2 is a multiple of 7 and x^6 - 1 is a multiple of 7^k, show that (x + 1)^6 - 1 is also a multiple of 7^k.

We have 

(a+b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + y^6

= a(a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) + y^ 6

If x -2 is divisible by 7 then taking a = 7k and b = 2

=> x = 7k+ 2 , b^6 mod 7 = 1

(a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) mod 7 = 6b^5 mod 7 not divisible by 7

now if a is divisible by 7^m and so (a^5 + 6a^4 b + 15a^3b^2 + 20a^2b^3 + 15ab^2 + 6b^5) is not divsible by 7

so x^6 – 1 mod 7^m is zero but x^6 – 1 mod 7^(m+1) is not zero

(x+1)^6 – 3^ 6 (as x+1 = 7k + 3) mod 7^(m+1) is is not zero

so if x^6 – 1 is divisible by 7^m <=> (x+1)^6 -1 is is divisible by 7^m

  I have proved that 
x^6 - 1 is a multiple of 7^k, iff (x + 1)^6 - 1 is also a multiple of 7^k.

  

Q13/121) Out of ( 2n + 1 ) tickets, consecutively numbered, 3 are drawn at random. Find the chance that numbers on them are in A.P.

The total number of cases (2n+1C3)

The lowest number can be 1 to 2n-1
if the lowest number is odd then there are even number of numbers and 2^nd number shall be in 1^st half.

So if lowest number id 2k +1 there are (n-k) choices for 2^nd and 3^rd number pairs

if the lowest number is even  then there are odd number of numbers and 2^nd number shall be in 1^st half and middle number not counting.

So if lowest number id 2k there are (n-k) choices for 2^nd and 3^rd number pairs

If lowest number is 1 total number of cases n( lower of the rest 2 from 2 to n+ 1
If lowest number is 2 total number of cases n-1
If lowest number is 3 total number of cases n-1
If lowest number is (2n-2) total number of cases 1( that is 2n-1 is the lower of the 2)
If lowest number is (2n-1) total number of cases 1( that is 2n is the lower of the 2)
So total number of cases = n(n-1)/2 *2 + n = n^2
So probability or chances= n^2/(2n+1C3)

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

Q13/110) Solve in positive integers (1+1/x) (1+ 1/y))(1 +1/z) = 2

Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3

(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2

(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)

Q13/109) Find all positive integer solutions of the equation 4x^3 – 12x^2 + 5x – 10y + 36y^2 – 18y^3 + 4x^2y + 6xy – 15xy^2 = 0

We put the number in decreasing order of power of x

4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−10y+36y^2−18y^3
Now factor the part independent of x
= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(5-18y+9y^2)
= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(3y-5)(3y-1)
Now I multiply by 2 and put 2x = z to get coefficient of x^3 as 1
= ½(8x^3−24x^2 +8x^2y +10x+12xy−30xy^2−4y(3y-5)(3y-1)
= 1/2(z^3−6z^2 +2 z^2y +5z+6zy−15zy^2−4y(3y-5)(3y-1)
= 1/2(z^3−z^2(6-2y) +z(5+6y+15y^2) −4y(3y-5)(3y-1)

Now we need to split - 4y(3y-5)(3y-1)into3 parts that the sum is 6- 2y

They are-( 3y-5),- (3y-1) , 4y ( it is easy to do so as 5 + 1 = 6

Now to check The coefficient of z is (5+6y+15y^2)

We see that – 4y(3y-5) – 4y(3y-1) + (3y-1) (3y-5) = -15y^2 + 6y + 5 which is true

So we get ½(( z- 4y) ( z+ 3y-5)(z+ 3y-1)) or (x-2y)(2x + 3y-5)(2x+3y-1)) = 0

Which gives us x = 2y or 2x + 3y = 5 or 2x + 3y = 1

X= 2y gives solution x= 2t, y = t;

2x + 3y = 5 gives x= 1, y= 1 and 2x + 3y = 1 has no solution ( as we are looking positive solution)

Q13/108) Evaluate the given sum k from 0 to n ∑ (nCk) /(k+1)

(nCk)/(k+1) = n!/(k! * (n-k)!) * (k+1) = 1/(n+1) ( n+ 1 C k+1)

So ∑ (nCk)/(k+1) = 1/(n+1) (∑ (n+1 C k) - (n+1 C 0))
= 1/(n+1) (∑(n+1Ck) - 1)
= 1/(n+1) ( 2^n+1) – 1)

Q13/107) What is the general formula for the equation 1*1!+2*2!+.....+n*n!

We have k* k! = ((k+1)-1) * k! = (k+1)! – k!

So 1 * 1! = 2! – 1!

2 * 2 ! = 3! – 2!
n*n! = (n+1)! – n!

Adding we get 1*1!+2*2!+.....+n*n! = (n+1)! – 1! or (n+1)!- 1

Q13/106) Find three different prime numbers between 10 and 99, where the average of any two is a prime and the average of all three is a prime.

All the numbers have to be of the same form that is all 6n+1 or all 6n-1.
This is so because 6n+1 and 6m-1 have average 3(n+m) which is not a prime
this I took to reduce the number of test cases.
Now without loss of generality we can choose the 2 numbers and their mean to be taking
a, a + 6n, a + 12n, all 3 are primes ( a and a + 12n are numbers and a+6n is mean we choose this because average should be 6 or a multiple)
the next number has to be of the form a + 12m 
now (a+12n + a + 12m)/2 = a + 6(m+n)
and (a+ a + 12n + a + 12m)/3 = a + 4(m+n)
so the 7 prime numbers are a, a + 6n, a+ 12n, a+12m, a+ 6(n+m), a +6m, a + 4(m+n)
not necessarily in increasing order

a=11, n= 3, m= 5 gives the solution 11,29,47,71, 59, 41, 43.

Q13/105)if ab + bc +ca=0,then find 1/(a^2-bc) +1(/b^2-ca) +(1/c^2-ab)

ab + bc + ca = 0

so bc = - ab - ca = -a(b+c)

so a^2 - bc = a(a + b+ c)

so 1/ (a^2 - bc) = 1/(a(a+b+c))

similarly

1/ (b^2 - ca) = 1/(a(a+b+c))

1/ (c^2 - ab) = 1/(a(a+b+c))

adding we get (1/a^2-bc) +(1/b^2-ca) +(1/c^2-ab)= 1/(a+b+c) ( 1/a + 1/b+ 1/c)

= ( 1/a + 1/b+ 1/c)/(a+b+c)
=(bc + ac + ab)/((a+b+c)(abc)) = 0

Q13/104) Let a,b,c be the roots of x3−7x2−6x+5=0. Compute (a+b)(a+c)(b+c).

Let f(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37

Q13/103) Given f(x) = x -3 for x >= 1000 and f(f(x+5)) x < 1000 find f(84)

we evaluate f(999) through f(995)

f(999) = 998
f(998) = 997
f(997) = 998
f(996) = 997
f(995) = 998

now f(85)= f^184 (999) = f^183( 998) notation for f is applied 183 times

applying f twice gives 998 and so on applying 182 time gives 998 and then once more gives 997 which is the ans

for some discussions refer to http://mathhelpboards.com/challenge-questions-puzzles-28/find-f-84-a-6600.html#post30091

Q13/102) Given abc=1 Show that 1/(1+a+b^-1) + 1/(1+b+c^-1) + 1/(1+c+a^-1) = 1

we have 1/(1+ a + 1/b) = b/(b+ab+1) ...1

1/(1+b+ 1/c) = c/(c+bc+1) = c/(c+bc + abc) ( as 1 = abc)
= 1/( 1+ b+ ab) .. (2)

1/( 1+ c + 1/a) = a/( a+ ac + 1) = ab/( ab + abc + b) = ab/( ab+ 1 + b) ... (3)

adding (1) , (2), (3) we get the result

Q13/101) Solve : 36/x^(1/2) + 9/y^(1/2) = 42 – 9 x^(1/2) – y^(1/2)

36/x^(1/2) + 9x^(1/2) + 9 /y^(1/2) + y^(1/2) = 42

Hence

(36/x^(1/2) + 9x^(1/2) – 36) + (9 /y^(1/2) + y^(1/2)- 6) = 0

Hence 9( 2/x^*1/4 – x^(1/4))^2 + (3/y^(1/4)  – y^(1/4))^2 = 0
=> 2/x^(1/4) – x^(1/4) -= 0 or x = 2^2 = 4

And  3/y^(1/4)  – y^(1/4) = 0 or y = 9

Q13/100) 2 ^ sin x + 2 ^ cos x is greater than

A> 1/2

B> 2^(1/(2^(1/2)))

C> 2^(1/2)

D> 2^(1-(1/(2^(1/2))))

Solution

by AM GM inequality we have

(a+b) /2 >= sqrt(ab)

so 2^ sin x + 2^ cos x >= 2 sqrt(2^( sinx + cos x))

sin x + cos x = sqrt(2) ( sin x cos pi/4 + cos x sin pi/4) = sqrt(2) sin (x+pi/4)

lowest value is - sqrt(2)

so 2^ sin x + 2^ cos x >= 2 sqrt(2^-(sqrt(2)) = 2^(1-(1/2^(1/2))

hence D

 

Q13/099) If /(z - 5i)/ (z +5i)/ = 1 then prove that z is a real number

let z = x + iy

|z + 5i| = | z - 5i|

or | x + iy + 5i | = | x + iy - 5i|

or | x + iy + 5i |^2 = | x + iy - 5i|^2

or x^2 + (y + 5)^2 = x^2 + (y- 5)^2

or (y + 5)^2 = (y- 5)^2 or y^2 + 10y + 25 = y^2 - 10y + 25 => y =0 or z is real

Q13/098) If α and β are two different values of θ ( between 0 and 2π ) which satisfy the equation .................? . 6 cos θ + 8 sin θ = 9, Find the value of ..... sin ( α + β )

α and ß are the roots of the equation : 6 cosΘ + 8 sinΘ = 9
=> 6 cosα + 8 sinα = 9 ------------- (1) and
.....6 cos ß + 8 sin ß = 9 ------------- (2)
Subtracting (2) from (1), we have :
6(cosα - cos ß) + 8(sin α - sin ß) = 0
=> 4(sinα - sin ß) = 3(cos ß - cos α)
=> (sin α - sin ß)/(cos ß - cos α) = 3/4
=> [2 cos{(α + ß)/2} sin {(α - ß)/2}] / [2 sin {(α + ß)/2} sin {(α - ß)/2}] = 3/4
=> cot {(α + ß)/2} = 3/4
=> cot²{(α + ß)/2} = 9/16
=> 1 + cot²{(α + ß)/2} = 1 + 9/16 = 25/16
=> cosec²{(α + ß)/2} = 25/16
=> sin² {(α + ß)/2} = 16/25
=> sin {(α + ß)/2} = ± 4/5 --------- (3)
=> 1 - sin²{(α + ß)/2} = 1 - 16/25 = 9/25
=> cos²{(α + ß)/2} = 9/25
=> cos {(α + ß)/2} = ± 3/5 ---------- (4)
=> sin (α + ß) = 2 sin {(α + ß)/2} cos {(α + ß)/2} = ± 2(4/5)*(3/5) = ± 24/25

However as 6 cosα + 8 sinα = 9

so cosα and sinα both > 0 ( as if one is -ve then sum < 8) and so α is in 1st quadrant and similarly β

so ( α + β ) is either in 1st or 2nd quadrant so sin ( α + β )is positive and hence -24/25 need to be ruled out and hence it is 24/25.

Another solution

6 cos θ + 8 sin θ = 9
=> (3 / 5) cos θ + (4 / 5) sin θ = 9 / 10
=> sin (θ + φ) = 9/10

Clearly if
sin x = sin a
=> x = a or π - a
for given range

=> θ + φ = α + φ or π - (α + φ)
Clearly one of them is equal to ß + φ
=> π - (α + φ) = ß + φ
=> α + ß = π - 2φ
=> sin (α + ß) = sin (π - 2φ)
=> sin (α + ß) = sin 2φ
=> sin (α + ß) = 2 sin φ cos φ = 24 / 25

Q13/097) The LCM of 2numbers is 45 times their HCF.Sum of HCF and LCM is 1150, find the numbers?

LCM = 45 *HCF
LCM + HCF = 46 * HCF = 1150

so HCF = 25 and LCM = 25 * 45

now both x and y ( the 2 numbers are) multiple of 25 and coprime

so we x = 25m, y = 25n and LCM(m,n) = 45, HCF(m,n) = 1

without loss of generality let m >= n

mn = 45 = 1 * 45 = 5 * 9
so m = 9, n =5 or m = 45 , n= 1

numbers are (225,125) or (1125,25)

Q13/096) Find sum of n terms ---- 1 + ( 1 + x ) + ( 1 + x + x² ) + ( 1 + x + x² + x³ ) + ..... + up to nth term.?

y = 1 + (1+ x) + (1+ x + x^2) + ...
y ( 1-x) = (1-x) + 1(1-x^2) + .. ( 1- x^(n))
= n - ( x + x^2 + .. x^(n)) as
= n - x ( 1 + x ... x^(n-1))
= n - x ( 1- x^(n))/(1-x)

or y = n/(1-x) - x (1-x^(n)/(1-x)^2

the above ans is correct if x is not 1

if x = 1 then we have 1 + 2 + 3 + .. n = n(n+1)/2

Q13/095) Find coefficient of xⁿ in the expansion of ( 1 + 2x + 3x² + 4x³ + ........ inf. )^(1/2)....?

let y = ( 1 + 2x + 3x² + 4x³ + ........ inf. ...(1)
y converges for |x| < 1

now xy = x + 2x^2 + 3x^3 ....(2)

subtract (2) from (1)

y - xy = 1 + x + x^2 ... = 1/(1-x)

so y(1-x) = 1/(1-x)

or y = 1/(1-x)^2

so ( 1 + 2x + 3x² + 4x³ + ........ inf. )^(1/2)..= 1/(1-x) = 1 + x + x^2 ...

so coefficient of x^n = 1 for all n

Q13/094) Solve |x+1|+|x-2| < 5

As -1 < 2 and in a line the distance between -1 and 2 is 3 so all the points from -1 to 2 satisfy the relation as the sum is the distance from -1 to 2

Now if it to the left of -1 the distance from -1 shall be added twice once for |x+1| and another time for |x-2| the sum < 2 and hence x > - 2

For the point to the right of 2 similarly x < 3

So -2 < x < 3

Q13/093) Find twin primes p and q(=p + 2) such that pq-2 is also a prime

P q are twin primes so they are 3 and 5 or p = 6n+ 1 and q = 6n – 1

If p = 6n- 1 and q = 6n + 1 then pq - 2 = 6n^2 – 1 – 2 = 36n^2 – 3 = 3(12n^2-1)

it is > 3 (minimum value = 33)  and 3 is a factor so cannot be prime

That leaves us with p = 3 , q = 5 and pq-2 = 13 meeting the condition

Q13/092) Prove 4 cos 36 sin 18 = 1

We have

sin 72 
= 2 cos 36 sin 36 using sin 2A formula
= 2 cos 36 ( 2 sin 18 cos 18) using sin 2A formula again
= 4 cos 36 sin 18 sin 72 as cos18 = sin(90-18) = sin 72

or

4 cos 36 sin 18 = 1

Q13/091) Prove that for every prime p > 7, p^6 - 1 is divisible by 504.

we have 504 = 9 * 56 = 7 * 8 * 9

we need to show that it is divisible by 7 , 8 and 9 as they are pairwise coprime

now as p is prime and 7 is also prime

so p^6 = 1 mod 7 by Fermat's Little theorem so p^6-1 is divisible by 7

p^6-1= (p^3-1)(p^3+1)

as these 2 are 2 successive even numbers Reason p is odd so p^3 is odd

so one is divisible by 4 and another by 2 so product is divisible by 8

one of p^3-1 and p^3 + 1 is divisible by 9

as p > 7 so p mod 3 = 1 or -1

to prove is if p = 1 mod 3 then p = (3n+1)

so p^3 = 27n^3 + 27n^3 + 9n + 1 or p^3-1 = 9(3n^3+3n^2 + 1) or dibsible by 9

if p = 3n-1 then p^3-1= 9(3n^3-3n^2 + 1) or divisible by 9

so p^6 -1 is divisible by 9

hence is divisible by 7 * 8 * 9 or 504

 

Q13/090) if M={x,xy,log(xy)}, N={0,∣x∣ ,y} given: M=N find : ( x + 1/y) + (x^2 + 1/y^2) + (x^3 + 1/y^3) …. + (x^2001 + 1/y^2001)

x and y cannot be zero as we have log(xy) and log zero is undefined .

so we have xy = 1 as log(xy) = 0

as x and y both are positive or –ve.

We have |x| in N . now x = |x| and xy = y => x = y = 1

but x and xy have to be different and |x| and y has to be different

so y = x = - 1 giving M={- 1,1,0 } and N={0, 1 ,0-1}

So x = y =-1 and sum = x^n + 1/y^n = -2 for n = odd and 2 for n = even

So sum = -2 as 1st 1000 pairs cancel leaving with x^2001 + 1/y^2001 = - 2

Q13/089) Find the real solution(s) to the equation 36/x^(1/2) +9/y^(1/4) = 42−9x^(1/2) −y^(1/2)

We have

36/x^(1/2) + 9x^(1/2) +     9/y^(1/4)+ y^(1/2) = 42

Complete the square for x and y terms giving

36/x^(1/2) + 9x^(1/2) - 36  +    9/y^(1/4)+ y^(1/2) = 42

Hence 9( 2/x^1/4 – x^(1/4))^2 + (3/y^(1/4) – y^(1/4))^2 = 0
so  2/x^(1/4) – x^(1/4) = 0 or x = 2^2 = 4
And 3/y^(1/4) – y^(1/4) = 0 or y = 9

Q13/ 088) Determine the smallest integer that is square and starts with the first four digits 3005.

we can have even number of digits or odd number of digits and they have to be treated separately

if even then we have

sqrt (3005 * 10^2n) = 54.8177 * 10^n
sqrt (3006* 10^2n) = 54.8270 * 10^n

if odd digits then

sqrt (30050 * 10^2n) = 173.34 * 10^n

sqrt (30060 * 10^2n)= 173.37 * 10 ^n

from the 1st set we get square root  5482 as smallest where a digit is different and from the second at least

17335

so it is 5482^2 = 30052324

Q13/087) Prove that if x = log_a(bc),y = log_b(ca),z = log_c(ab) then the value of xyz - x - y - z is 2?

we get

bc = a^x ..1
ca = b^y...2
ab = c ^z ...3

from (3)

c^(xyz) = (ab)^(xy)
= a^(xy) * b^(xy)
= (a^x)^y) * (b^y)^x)
= (bc)^y ( ac)^x ( from 1 and 2)
= c^y c^x ( b^y)(a^x)
= c^y c^x (ac) (bc)
= c^y c^x c^2(ab)
= c^ x c^y c^2 c^z
= c^(x+y+z+2)

hence xyz = x + y + z + 2 or
xyz - x - y - z = 2

Alternatively we can solve as below

\

multiply (1) by a

abc = a^(x+1) => a = (abc)^(1/(x+1) ..4

similarly

b = (abc)^(1/(y+1) ..5

c = (abc)^(1/(z+1) .. 6

multiply to get

abc = (abc)^(1/(x+1) + 1/(y+1) + 1(z+1))

or (1/(x+1) + 1/(y+1) + 1(z+1)) = 1

or (y+1)(z+1) + (x+1)(z+1) + (x+1)(y+1) = ( x+1)(y+1) (z+1)

or yz + y + z + 1 + xz + z + x + 1 + xy + x + y + 1 = xyz + xy + yz + xz + x + y+ z + 1

or x+y + z + 2 = xyz

Q13/086) Solve x- 2/(x-1) = 1 – 2/(x-1)

X cannot be 1 as x-1 is in denominator

Canceling 2/(x-1) from both sides x = 1

As x cannot be 1 so no solution

Q13/085) The sum of all real root of the equation |x-2|^2 + |x-2| - 2= 0 is

Let |x -2 | = t

So t > = 0

t^2 + t – 2 = 0 or (t-1)(t+2) = 0

so t = 1 as – 2 is not possible

|x-2| = 1 or x -2 = 1or -1

So x = 3 or 1 so sum of real roots = 4

Q13/084) The sum of the rational terms in the expansion of (2^(1/2) +3^(1/5))^10 is

For it the power of 2 and of 3 both should be integer

So the terms should be (10C 2p)  (2^(1/2)^2p * (3^(1/5)^ 5q) where 2p + 5q = 10

Solution of 2p + 5q =1 0 are p = 0 q =2 or q = 0 p = 5

P = 0 q = 2 give (10c0) 3^2 = 9

P = 5 q = 0 give (10c10) 2^5 = 32

So the rational terms are 32 and 9 and sum is 41