Q3/075) Evaluate tan^2 20 + tan^2 40 + tan^2 80.

Observe that tan(3⋅20∘)=tan60=√3

-  tan(3⋅40)=- tan120=- tan(180−60)=tan60=√3  

and tan(3⋅80)=tan 240=tan(180+60)=tan60=√3

As tan3θ=(3tanθ−tan^3θ)/)1−3tan^2 θ)

the roots of the equation t^3−3√3t^2−3t+√3=0 ( Putting tan3θ=√3)

will be tan20,tan(−40)=−tan40∘,tan80∘

that is of f(t) =t^3−3√3t^2−3t+√3=0
and we shall construct an equation whose roots are
tan^2 20,tan^2 40 tan^2 80
shall be f(x^(1/2) = 0)
putting t = x^(1/2) we get
so x^(3/2) - 3√3x−3x^(1/2) +√3=0
or x^(3/2) - 3 x^(1/2) = 3√3x -√3
or
√x(x-3) = √3(3x -1)
square both sides to get
x(x-3)^3 = 3(3x-1)^2
or x(x^2-6x+ 9) = 3(9x^2 - 6x + 1)
or x^3 - 33 x^2 + 27x - 3 = 0
as it is cubic roots are tan^2 20,tan^2 40 tan^2 80
and sum of roots = - coefficient of x^2 or 33

Q13/074) solve 2cos(x)≤∣ √∣1+sin(2x) ∣-√∣1-sin(2x) ∣∣≤√2

one solution cos x <= 0

so x is between pi/2 and - 3pi/2

second case cos x > = 0 and sin x >= 0

now sin x < cos x gives
cos x < 1/2 | cos x + sin x - ( cos x - sin x) <= 1/ sqrt(2)

or cos x < sin x <= 1/sqrt(2)

cos x >= 1/sqrt(2) => sin x >= 1/ sqrt(2) so no solution

cos x > 0 and sin x >= cos x gives cos x <= sqrt(2)

siimiliarly ranges sin x < 0 2 ranges | sin x | < cos x and | sin x | > cos x need to be anlaysed

by symetry we get cos <= 1/ sqrt(2)

so solution set cos^1 (1/ 2sqrt(2) to 2pi - cos^1 (1/2sqrt(2))

Q13/073) Solve x^3−3x=√(x+2)

If we square both sides( I am not squaring to find the expression) then
we get x 6th order polynomial the coefficient of x^6 is 1 and constant term of is -2 as x+2 goes to left
so if it has a rational root it is factor of -2 that is 1/-1/2/-2
1 and -1 can be ruled out as they give give irrational so checking - 2 and 2 we get that x= 2 satisfies both sides
hence it is a root.
to find all solutions is a different matter ( others are irrational)

Q13/072) find the period of (sinx)^4+(cosx)^4

(sinx)^4+(cosx)^4 = ( sin ^2 x + cos^2 x) - 2 sin ^2 x cos ^2 x
= 1 - 2 sin ^2 x cos ^2 x
= 1- 1/2( 4 sin ^2 x cos ^2 x)
= 1 - 1/2( sin 2x)^2 
= 1/2( 2 - sin ^2 2x)
= 1/4( 4 - 2 sin ^2 2 x)
= 1/4( 3 + cos 4x)

now cos x has period 2pi so cos 4x has period pi/2 hence (sinx)^4+(cosx)^4 has period pi/2

Q13/071) Let x,y two natural numbers such that x > y, x-y=96,? and the greatest common divisor of x and y is 16. Then x= ?, and y=?

as such there is no unique solution that is there is more than one solution

now gcd(x,y ) = 16 so x = 16m, y = 16 n and gcd(m,n) = 1 and m-n = 6

so if we take m = 6k + 1, n = 6k - 5 then gcd(m,n) = 1 and we meet the critera

so x = 96 k + 16 and y = 96 k - 80 for random k

k = 1 gives 112, 16
k =2 gives 208, 112
so on

this is one set of solution and not exhaustive

Q13/070) Find the largest coefficient in the expansion of (1+x)^n given that the sum of the coefficients of the terms in the expansion is 4096

We get sum of the coefficient by putting x = 1

So (1+1)^n = 4096 or 2^n = 4096 = 2 ^12 => n = 12

The coefficients are 12Cn and as 12 is even largest coefficient is 12C6

Q13/069) given that a,b are roots of equation Ax^2 – 4x +1 = 0 and c , d are roots of equation Bx^2 – 6x + 1 = 0 find the values of A and B such that a,c, b,d are in HP.

a,c , b,d are in HP => 1/a, 1/c, 1/b and 1/d are in AP

A,b are roots of Ax^2 – 4x + 1 = 0 => 1/a, 1/b are roots of 1 -4x + A = 0

So 1/a + 1/b = 4

Similarly  1/c + 1/d =  6

As 1/c – 1/a = 1/d – 1/b we get 1/a = 1, 1/c = 2, 1/b = 3 and 1/d = 6.

using these we get A = 1/ac = 3 and B= 1/bd = 8

Q13/068) Solve xy(x+y)=30 , x^3+y^3=35

We have
 xy(x+y)=30… (1)

x^3+y^3=35 … (2)

x^3 + y^3 = (x+y)^3 – 3 xy (x+y) = (x+y)^3 – 3 * 30 = 35

 so (x+y)^3 = 125 => (x+y)= 5 .. (3)

from (1) and (3)

xy = 6 … (4)

solving (3) and (4)  (x,y) = (3,2) or (2,3)

Q13/067) Let x be a real number and let M=(3x−1)/(1+x) – ((√(∣x∣−2) +√(2−∣x∣))/ √∣2−x∣.

We are taking square root of |x| - 2 and its –ve so it has to be zero

So |x| - 2 = 0 or x = 2 or – 2

x cannot be 2 as 2-x is in denominator

so x = - 2 

hence putting the value x = -2 we get M = 7

as M^4 =  1 mod 10   

M^2000 = 1 mod 10 or M^2003 = M^3 mod 10 = 343 mod 10 or 3

3 is the unit digit

Q13/066) Prove that (a+b+c)/(a+b) + (a+b+c)/(b+c) + (a+b+c)/(c+a) > 9/2

a/(b+c) + b/(c+a)
= (ca + a^2 + b^2 + bc)/(b+c)(c+a)
>= (ca + 2ab + bc)/(b+c)(c+a) [since (a-b)^2 >= 0 implies a^2 + b^2 >= 2ab]
= {a(b+c) + b(c+a)}/(b+c)(c+a)

Thus, a/(b+c) + b/(c+a) >= a/(c+a) + b/(b+c)

Similarly, it can be proved that

b/(c+a) + c/(a+b) >= b/(a+b) + c/(c+a)
c/(a+b) + a/(b+c) >= c/(b+c) + a/(a+b)

Adding corresponding sides of the above three inequalities, we get

2{a/(b+c) + b/(c+a) + c/(a+b)} >= 3
i.e. a/(b+c) + b/(c+a) + c/(a+b) >= 3/2

adding one to each term on LHS and 3 on RHS

( 1 + a/(b+c)) + ( 1+ b/(c+a)) + ( 1+ c/(a+b)) >= 3/2 + 3

or (a+b+c)/(a+b) + (a+b+c)/(b+c) + (a+b+c)/(c+a) > 9/2

QED