Q13/051) Prove that for integers n,m and p if √ (n) + √(m) = p then √(n) and √ (m) are integers

√ (n) + √(m) = p

ð     √ (n) = p -  √(m)

ð     n = p^2 – 2p √(m) + m^2

ð     √(m) = (p^2 +m^2 –n)/ (2p)

ð     √(m) is rational

Now √(m) is rational so there exists integers such that m = s/t

So m = s^2/t^2 or s^2 = mt^2

As left hand side is a square and so RHS is s square and hence m has to be s square then √(m) is rational and √ (n) = p - √(m) is also rational.  

Q13/050) Given m,n integers satisfying : √(m+2007)+√(m−325) = n find Max(n) and min (n)

we have

(m+ 2007) – (m- 325) = 2332 ..1

√(m+2007)+√(m−325) = n … (2) (given)

Dividing (1) by (2) we get

√(m+2007)-√(m−325) = 2332/n … (3)

Add (2) and (3) to get 2√(m+2007)= (2332/n +n)

Now RHS has to be even so n is even factor of 2332/2 or 1166

Hence n has to be largest even factor of 1166  or 1166

Now to find the lowest n

put m- 325 = p (which is >=0)

we get √(p+2332)+√p which is >= sqrt(2332) >= 49

is has to be even smallest even factor of 2232 but not < 49

now 2332 = 2 * 2 * 11 *53 hence minimum n = 2 * 53 = 106

Q13/049) Solve: sin^−1(x)+cos^−1(2x)= π/6

We have cos ^-1 x = sin ^-1 ( 1-4x^2)^(1/2)

now given equation

 sin ^-1(x) = π/6 – cos ^-1(2x)

take sin  of both sides

x = sin (π/6) cos cos ^-1(2x) - cos (π/6) sin  cos ^-1(2x)

or x = ½ * 2x – sqrt(3)/2 * (1- 4x^2)^(1/2) = x - sqrt(3)/2 * (1- 4x^2)^(1/2

or * (1- 4x^2)^(1/2) = 0 => x = +/- ½

One can check by putting the values x = ½ satisfies and -1/2 does not.

So solution = x =1/2

Q13/048) Without using a calculator find which is greater √1001 + √+999 or 2 √1000

We have √ (n+1) - √n = 1/(√(n+1) + √n)…  (1)

And √(n) - √(n-1)  = 1/(√(n + √(n-1)) …2

As

1/(√(n+1) + √n) < 1/(√(n + √(n-1)) ( as denominator of LHS is larger)

So √ (n+1) - √n < √(n) - √(n-1)

So √ (n+1) +  √(n-1)  < 2√(n)

Putting n = 1000 we get

√1001 + √+999 < 2 √1000

or 2 √1000 is larger

Q13/047) Find limx→1 1/(2(1−(x)^(1/2)))−1/(3(1−(x)^(1/3))).

Put x = u^3 to get 1/(2(1-u^3) – 1/(3(1-u^2)

= (3(1-u^2) – 2 (1 – u^3))/(6(1-u^2)(1-u^3))

Numerator = 3(1-u)(1+u) – 2(1-u)(1+u+u^3))

= (1-u)( 3+ 3u – 2 -2u – 2u^2)

= (1-u)( 1 + u- 2u^2)

= (1-u)(1-u)(1 + 2u)

= (1-u)^2 (1+ 2u)

Denominator = 6(1-u)^2(1+u)(1+u+u^2)

So ratio = ( 1+ 2u)/(6(1+u)(1+u+u^2))

And at x= 1 or u= 1 we get 3/( 6 * 2 * 3) = 1/12

Q13/046) 16x^4−mx^3+(2m+17)x^2−mx+16=0 has four distinct roots forming a geometric progression. Find all values of m.

Let the roots be a/r^3, a/r, ar and ar^3 (the common ratio = r^2)

product of roots = a^4 = 16/16 = 1

so the polynomial is q( x +1/r^3)(x+1/r)(x)(x+r)(x+ r^3)

or n( r^3 x + 1)(r x + 1) ( x + r) (x+r^3)

as the coefficients are symmetric we have n =1

now take the product to get r^4 = 16

r = 2 or -2(only possible values)

by putting 2 and then -2 we can get the result

r = 2 => (8x+1)(2x+1)(x+2)(x+8)

= ( 16x^2 + 10 x + 1)(x^2 + 10 x + 16)

= 16x^4 + 170 x^3 + 357x^2 + 170 x + 16

It does not satisfy

r = -2 => (8x-1)(2x-1)(x-2)(x-8)

= ( 16x^2 - 10 x + 1)(x^2 - 10 x + 16)

= 16x^4 - 170 x^3 + 357x^2 - 170 x + 16

m = 170 satisfies the original equation and hence m = 170 and solutions are 1/8. 1/2, 2, 8

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