Q13/045) If x is real show that x/x^2-5x+9 always lie in the interval [-1/11,1]?

let y = x/(x^2-5x+9)
so yx^2 - 5yx + 9y = x
or yx^2 - (5y+ 1) x + 9y = 0
for it to have real root x(as x is real)  discriminate > = 0 or
(5y+1) ^2 - 36y^2 >= 0
=> (5y + 1 + 6y)(5y+1 - 6y) >= 0
or (11y + 1)(1-y) >= 0 or y >= -1/11 and <= 1
so x/x^2-5x+9 always lie in the interval [-1/11,1]

Q13/044) N and M are positive integers such that N+M=21. The largest possible value of 1/N+1/M is a/b ( and b are co-primes) .Find a+b

1/N+1/M = (N+M)/NM = 21/(NM)
This is largest when NM is lowest.
clearly when N= 1 and M= 20 ( or N = 20, M = 1)
so NM = 20
so a= 21 and b = 20 hence a+b = 41

Q13/043) Let a,b,c,d be real numbers. Suppose that all the roots of the equation

z4+az3+bz2+cz+d=0 are complex numbers
lying on the circle ∣z∣=1 in the complex plane. The sum of the reciprocals of the roots is necessarily:

options

a) a
b) b
c) -c
d) d

Solution

Let α β,γ,δ be the roots of given Equation
Now all Roots are complex and lie on ∣z∣=1
and as coefficients are real so complex Roots are occur in pair
so Let α=x1+iy1 β=x1−iy1 and α.β=x1^2+y1^2=1
Similarly γ=x2+iy2 δ=x2−iy2 ] and γ.δ=x2^2+y2^2=1

Hence αβγδ = 1

Now  α β,γ,δ are roots of f(z) = z^44+az^3+bz^22+cz+1 = 0

So 1/α,1/β,1/γ,1/δ are roots of f(1/z)

= 1/ z^44+a/z^3+b/z^22+c/z+1 =0

Or = z^44+cz^3+bz^22+az+1 = 0

So sum of 1/α,1/β,1/γ,1/δ is –c (-ve  coefficient of z^3)

Q13/042) Solve ∣√(x−1)−2∣+∣√(x−1)−3∣=1

We know (√(x−1)−2)-(√(x−1)−3) = 1
so let (√(x−1)−2 = t

|t | + |t -1 | = 1

t cannot be < 0 then t-1 is less than -1 and  |t | + |t -1 | > 1

t cannot be > 1 then t-1 is less than -1 and  |t | + |t -1 | > 1

so we need to check 0 < = t < = 1

if t = 1 then |t | + |t -1 | = 1

if t = 0 then |t | + |t -1 | = 1

if 0 < t < 1 then t-1 < 0 so |t | + |t -1 | = t  + 1- t = 1

so 0 < = t < = 1

or   0 <= √(x−1)−2 <= 1

or 2 < = √(x−1) or <= 3

 5 < = x  <= 10

Q13/041)Find integral solution of (1-i)^n = 2^n

We have 1-i = (sqrt(2) cis (-pi/4)

So  (1-i)^n = 2^(n/2) cis (-npi/4) = 2^n

The modulo of LHS = 2^(n/2) and rhs = 2^n and both are same if

2^(n/2) = 2^n or n = 0

Then = 2^(n/2) cis (-npi/4) = 1 = RHS

So  n = 0  is the only ans

Q13/040)The fourth power of the common difference of an AP with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

proof

Without loss of generality we can take the 1^st term of the 4 consecutive terms to be a and let the difference be t

a and t are integers

We have the 4 terms = a, a+t, a+2t, a+ 3t and 4^th power of difference is t^4

Now a(a+t)(a+2t)(a+3t) + t^4

= (a(a+3t))((a+t)(a+2t)) + t^4

= (a^2+3ta)(a^2+3ta + 2t^2) + t^4

Letting a^2 + 3ta = p we get

= p(p+2 t^2) + t^ 4= (p^2+ 2pt^2+t^4) = (p+t^2)^2 = (a^2+3at+t^2)^2

Which is square of integer as a^2+3at+t^2 is integer

Q13/039) Show that 2^(1/2) 3^(1/2) and 5^(1/2) cannot be 3 terms of an AP

We know 2^(1/2) < 3^(1/2) < 5^(1/2)

Without loss of generality we can assume that 2^(1/2) is the 1^st term and < 3^(1/2) is the pth and 5^(1/2) is the qth term

If r is difference

So 2^(1/2) + (p-1) r = 3^(1/2)

2^(1/2) + (q-1) r = 5^(1/2)

Subtract to get (q-p) r = 5^(1/2) – 3^(1/2)

Or (q-p) r + 3^(1/2) = 5^(1/2)

Square both sides (q-p)^2 r + 3+ 2 sqrt(3)(q-p) = 5

Or sqrt(3) = (2-(q-p)^2)/2((q-p))

Right hand side is rational and LHS is irrational hence a contradiction

Hence proved