1984

True / False (with explanation)

If a<b<c<d, then the roots of the equation,

(x-a)(x-c) + 2(x-b)(x-d) = 0 are real and distinct

let (x-a)(x-b)(x-c) = y

then x² - (a+b) x + ab = xy - cy

thus, x² - (a+b+y) x + ab +cy = 0 .....1

as x is real, then, discriminant ≥ 0
then,
(a+b+y)² - 4ab - 4cy ≥ 0
a²+b²+2ab + y²+ 2ay + 2by -4ab - 4cy≥0
y² + 2(a+b-2c) y +(a-b)² ≥ 0 ....2

as expression 2 is positive,
discriminant should be negetive,
thus,
4(a+b-2c)² - 4(a-b)² ≤ 0
or, (a+b-2c+a-b) (a+b-2c-a+b) ≤ 0
0r, (a-c)(b-c)≤0

thus, c lies between a and b.
thus, (c) and (d) should be tha answers.. :)

(x-a)(x-c) + 2(x-b)(x-d) = 0
or, 3x² - (a+c+2d+2b)x + ac +2bd =0

from, this find the discriminant..
see what are the conditions u get for, b²-4ac≥0
would be a lenghty one.

true.

1984-2

The correct way of doing this would be--

consider f(x) = (x-a)(x-c)+2(x-b)(x-d)

f(t₁) >0 ; t₁>d

f(c) <0

f(t₂) >0 ; t₂<a

By Intermediate Value theorem there is one root of f(x)=0 in the interval (t₁,c)

and another root in the interval (c,t₂)

so f(x) = 0 has real and distinct roots.

Correct!