Q-1: Prove that has no integral solution
Solution: Let then the equation is
As u, v are integers, >1402
But 372 = 1369, 382 = 1444
Minimum possible value of v=38.
v must be even.
Let v=38+2k where k N
i.e. which is incorrect.
u, v cannot have positive integral solutions.
So, x, y cannot have integral solutions.
Q-2: If a, b are the roots of and also of and if are the roots of then prove that n must be even integer
Sincea, b are the roots of
This is true only if n is an even integer.
Q-3: Show that the equation
has no imaginary roots,
Where A, B, C---------K, a, b, c------k and l are real.
Solution: Assume a+ib is an imaginary root of the given equation then conjugate of this root a-ib is also root of this equation.
Putting x = a+ib and x = a-ib in the given equation then,
The expression in bracket 0
2ib = 0 b = 0 (because i 0)
Hence all roots of the given equation are real.
Q-4: Solve in
The A.M of x+3, x-1 is , i.e. (x+1)
Put x+1 = y then x+3 = y+2, x-1 = y-2
The equation becomes
The corresponding equation =0 has roots 1, -1.
The sign scheme of y R is as follows.
y2-1, 0 holds if y -1 or y 1
Now y -1 x+1 -1 or x -2
y 1 x+1 1 or x 0
Hence x -2 or x 0
Therefore the solution set = .
Q-5: let a, b, c be real, if has two real roots a and b where a< -1 and
b >1 then show that
Q-6: If a<b<c<d prove that the equation (x-a) (x-c) + k (x-b) (x-d) = 0 has real roots for all k R.
Equation (1) will have real roots if,
Now the discriminant of the equation corresponding (2) is,
Also (b-d)2>0 so (2) is true for all k R.
Hence given equation have real roots.
Q-1: If p be the first of n arithmetic means between two numbers and q be the first of n harmonic means between the same two numbers, prove that q cannot lie between
Solution: Let the numbers be x, y. Let the n AMs between them be A1, A2, -------- An
Then y = (n+2) th term = x + (n+1) d (d being the common difference)
Equation (1) Þ y = (n+1) p-nx putting this in (2).