Q13/065) Show that ( sin x / sin 3x) + (sin 3x/ sin 9x) + ( sin 9x/ sin 27x) = ½( tan 27x – tanx)

we have tan 3x - tan x = sin 3x/ cos 3x - sin x/ cos x
= ( sin 3x cos x - cos 3x sin x)/ ( cos 3x sin x)
= sin 2x/( cos3x sin x)
= ( 2 sin x cos x)/(cos 3x sin x)
= 2 sin x / cos 3x
or sin x/ cos 3x = 1/2( tan 3x - tan x)  ..1

similarly s

sin 3x/ cos 9x = 1/2( tan 9x - tan 3x) ..2

sin 9x /cos 27 = 1/2( tan 27x - tan 9x) ..3

adding above 3 we get the result

Q13/064) Given x and α are non-negative real number and for number x, we have (x+1)^2≥α(α+1). Is x2≥α(α−1)?

we have (x+1)^2≥α(α+1)... (1)

Is x^2≥α(α−1)
we can chose x+1 to be t and have

t^2≥α(α+1)
now as t is positive so t > α

let t = α+h ( h >0)

so t^2 - α(α+1)
= (α+h)^2 - α(α+1)
= 2αh + h^2 - α >= 0 given (1)

we need to show that

(t-1)^2≥α(α-1)

(t-1)^2-α(α-1)
= (α+h-1)^2 - α(α-1)
= (α^2+h^2+1+ 2αh - 2α +2h) - α(α-1)
= h^2+1+ 2αh - α +2h
= ( h^2 + 2αh - α) + ( 1+ 2h)
> 0 as ( h^2 + 2αh - α) from (1)

hence proved that (t-1)^2≥α(α-1)

Q13/063) Find all values of x satisfying the equation 2x=|x|+|x-1|

2x=|x|+|x-1|

each term on the right >= 0 so LHS >= 0 or x >= 0

hence we get 2x = x + | x-1| or |x-1| = x

as x-1 cannot be x so (x-1) < 0 and hence - (x-1) = x or x = 1/2
it satisfies (x-1) < 0 and so x= 1/2 is the solution

Q13/062) prove cot6°cot42°cot66° cot78° =1

we know  cot x cot (60+ x) cot (60-x) = cot 3x  ..(1)

put x= 18 to get

cot 18  cot 42 cot 78  = cot 54

or cot 42 cot 78 = cot 54 tan 18 = cot 54 cot 72

so cot6°cot42°cot66° cot78 = cot 6 cot 54 cot 72 cot 66 … (2)

from (1) putting x = 6 we get

cot 6 cot 54 cot 66 = cot 18 or tan 72 …(3)

putting this on RHS of (2) we get

cot6°cot42°cot66° cot78 = (cot 6 cot 54 cot 66) cot 72

= tan 72 cot 72

= 1

Q13/061) Solve the equation cos^2(π/4(sinx+√2cos^2 x))−tan^2(x+π/4tan^2x)=1

as cos^2 t < = 1 and tan ^2 a >= 0 so the 1st term is 1 and 2nd term is zero.

So cos^2(π/4(sinx+√2 cos2x)) = 1 and tan^2(x+π/4tan^2x) = 0

So (x+π/4tan^2 x) = n π , x = 0 is a solution and other solutions are
= n π – π/4(this I found by guessing tan ^2x =1 and not rigorously)

cos^2(π/4(sin x+√2cos^2x)) = 1 when x = 2 n π – π/4

hence this is the solution

Q13/060) a,b,c,d are integers given :a+b=c, b+c=d , c+d=a now if b>0, find max(a+b+c+d)

by solving we get

a = -3b,
c = -2b
d = -b

so sum = a + b + c + d = - 5b

it is largest when b ls smallest

b = 1 as b > 0 and integer

so largest sum = - 5

Q13/059) If the sum of first p terms of an A.P. is q and the sum of first q terms is p, show that the sum of its first (p + q) terms is – (p + q).

let 1st term be a and common difference be d

so pth term = a + (p-1) d and sum of 1st p terms = ( 2a + (p-1)d)* (p/2) = q

hence 2a + (p-1) d = 2q/p ..1

and q th term a + (q-1) p and sum of 1st q terms = ( 2a + (p-1)q)* (p/2) = p

hence 2a + (q-1) d = 2p/q ..2

subtract (2) from (1)
(p-q) d = 2(q/p-p/q) = 2(q^2-p^2)/(pq)

so d = -2 (p+q)/(pq)

now sum of 1st (p+q) terms

now (p+q)th term = ( a + (p+q-1) d)

and sum of its first (p + q) terms = ( 2a + (p+q -1) d) (p+q)/2
= ( 2a + (p-1) d + qd) (p+q)/2
= ( 2 q/p + qd) (p+ q)/2 as 2a + (p-1)d = 2q/p
= ( 2q/p + q (-2(p+q)/pq) * (p+q)/2 putting value of d
= ( 2q/p - 2 - 2q/p) * (p+q)/2
= - (p+ q)

Q13/058) For what values of b do the equation 1988x^2+bx+8891=0 and 8891x^2+bx+1988=0 have a common root?

we are given

 1988x^2+bx+8891=0 .. (1)

8891x^2+bx+1988=0.. (2)

We can eliminate x^2 from both to get

8891(1988x2+bx+8891) – 1988(8891x2+bx+1988) = 0

Or bx(8891-1988) = 1988^2- 9981 => bx = - (8891 + 1988)  = - 10879 ..3

Subtracting  2^nd one from 1^st we get (1988-8891) x^2 + (8891 – 1988) = 0 or x^2 = 1 +> x = 1 or – 1

From (3) if x = 1 then b = - 10879 and if x = - 1 then b = 10879

Q13/057) How many ordered pairs of integers (a,b) satisfy (1/a) + (1/b)= (1/6)

1/a + 1/b = 1/6
=> 6 * (b + a) = ab
=> 6b + 6a = ab
=> 6b = ab - 6a
=> a = 6b / (b - 6) = a So a, b and b-6 cannot be zero ( b-6 = 0 => b = 6 and a = infinite
letting b - 6 = u
we get a = 6 * (u + 6) / (u)
=> a = 6u/u + 36/u = a
=> a =6 + 36/u and b = u + 6
for a to be integer u has to be  integer and factor or 36 so we get the values as

u = 36( b = 42, a = 7), u = 18( b = 24, a =8  ), u = 12 ( b= 18, a = 9), u = 9( b= 15, a= 10), u = 6 ( b = 12, a = 12)

u = 4( b = 10, a = 15),u =3( b = 9, a = 18), u = 2( b= 8,  a = 24),u = 1 ( b = 7, a = 42)

u = -1 ( b = 5, a = - 30) ,u = -2 ( b = 4, a = - 12),u = -3 ( b = 3, a = - 6),u = - 4 ( b = 2, a = - 3),u = - 6( b = 0, a 0) inadmissible,u = - 9( b= -3, a = 2),u =- 12(b = 0 6, a = 3),u = - 18( b= - 12, a = 4),u = 36 (b = - 30, a= 6)

the above are solution there are 17 of them

Q13/056) Find Integer value of (a,b) which satisfy 4^a+4a^2+4=b^2

We have LHS is multiple of 4 so RHS must be multiple of 4 or b must be even and > 2^ a

For solution to exist we must have

 4^a+4a^2+4 >= (2^a + 2)^2

Or  4^a+4a^2+4 >= 4^a + 2^(a+2) + 4

Or 4a^2 >= 4 * 2^a or a^2 >=2^ a

We see from it a <= 4. SO we need to try for

A = 0 => b^2 = 5 no solution

A= 1 => b^2 = 2 no solution

A =2 => b = 6

A = 3=> b^2 = 64 + 36 +4 no solution

A= 4=> b^2 = 256+ 64 + 4 = 324 or b = 18

So solution set (2,6) and (4,18)

Q13/055) Let a,b,c be roots of equation x^3−6x^2+kx+k=0 and (a−1)^3+(b−2)^3+(c−3)^3=0. Find a, b, c and k.

We have sum of the roots

a+b+c = 6

so (a + b+ c- 6) = 0

or (a-1)+(b-2)+(c-3) = 0

so (a-1)^3 + (b-2)^3 + (c-3)^3= 3(a-1)(b-2)(c-3) ( as x+ y + z = 0 => x^3+y^3 + z^3 = 3xyz

so 3(a-1)(b-2)(c-3) = 0 => a= 1 or b= 2 or c = 3

f(x) = x^3−6x^2+kx+k

  taking  1 as a root we get

f(x)=  1-6+ k+ k = 0 or k = 5/2

so we get x^3-6x^2 + 5/2 x + 5/2 = 0

or 2x^3 – 12x^2 + 5 x + 5 = 0

factoring we get (x-1)(2x^2 – 10 x - 5) = 0

we can solve (2x^2 – 10 x - 5)= 0 to get (10+/-√(140))/4 so b = (5+ √35)/2 and c = (5+ √35)/2 or b = (5-  √35)/2 and c = (5+ √35)/2  

similarly other 4 set of roots can be found

Q13/054) prove that:sin5x-2sinx(cos2x+cos4x) = sin x

we have 2 sin A cos B = sin ( A + B ) + sin (A- B) ..1

Putting A = x and B = 2x we get

2 sin x cos 2x = sin 3x – sin x ..2

Putting A = x and B = 4x we get

2 sin x cos 4x = sin 5x – sin 3x ..3

Add (2) and (3) to get 2 sin x cos 2x + 2 sin x cos 4x = sin 5x – sin x

Or sin 5x – (2 sin x cos 2x + 2 sin x cos 4x)= sin x

Or sin 5x – 2 sin x (cos 2x + cos 4x)= sin x

Proved

Q13/053) If the (m+1)th,(n+1)th,& (r+1)the terms of an A.P r in G.P & m,n,r are in H.P .Show that the ratio of the common difference to the first term in the A.P is -2/n.

Let 1^st term be a and common difference be t

(m+1)st term = a + mt

(n+1) st term = a + nt

(r+1^s)st term = a + rt

As they are in GP

So (a+ mt)(a+rt) = (a+nt)^2

=> a^2 + a (m+r) t + mrt^2 = a^2 + 2ant + n^2 t^2

=> a (m+r) t + mrt^2 = 2ant + n^2 t^2

 => a (m+r)  + mrt  = 2an  + n^2 t

=> a(m+r -2n) = t(n^2- mr)

=>t/a = (n^2 – mr)/(m+r -2n) ..(1)

As m .n and r are in HP 1/m + 1/r = 2/n ot (m+r ) = 2mr/n

So m+r – 2n = (2mr/n- 2n) = 2/n(rm-n^2) …(2)

From (1) and (2) we get the result

Q13/052) x,y are negative integers given : y=10x/(100−x) find all the possible solutions of y

y(100-x) = 10 x
or xy + 10x - 100y = 0
or x(y+10) - 100(y+10) = - 1000
or (x-100)( y+ 10) = - 1000 ..1

y < 0 so y + 10 < 10
x < 0 so x - 100 < -100

- 1000 = 1 * (- 1000) => y = - 9 , x = -900
similarly you can find other solutions
= 2 * (-500) => y = - 8, x = - 400
= 5 * (-200) => y = -5
= 4 * (-250) => y = - 6
= 8 * (-125) => y = -2

y + 10 has to be positive(as x-100) is –ve  and < 10 so these are only solutions