all values of 'k' for which the eq: sin^2x - (k-3)sinx+k=0 has real roots?

Question

A) (9,infinity)
B) (1,9)
C) (-infinity,1)
D) (-infinity,0)
chose correct option and explain

Answer 1

The discriminant must be >= 0 for real roots. The discriminant is

b^2 - 4ac. Here this is (-(k-3))^2 - 4(1)(k)

Do this out, it is k^2 - 6k + 9 - 4k or k^2 - 10k +9 >= 0

(k-9)(k-1) >=0

k >=9 or k <= 1 are the boundary points.

The problem is, if k = 9 you get
sin^2x - 6 sin x + 9 = 0 which means sin(x) = 3 which is impossible.

So only k <= 1 can work. That leaves C as best answer for now...

Answer 2

Let u=sin x. Then the equation becomes u^2 - (k-3)u + k = 0.

By the quadratic formula,

u = (k-3)/2 +/- sqrt((k-3)^2 - 4k)/2
= (k-3)/2 +/- sqrt((k-1)(k-9))/2

Therefore we must have either k>=1 and k>=9, or k<=1 and k<=9. Simplifying, we must have k>=9 or k<=1.

However, if k>=9, then (k-3)/2 >= 3, and
sqrt(k^2 - 10k + 9)/2 = sqrt((k-5)^2 - 16)/2 < (k-5)/2 = (k-3)/2 + 1,
so the smallest root is still greater than 1. We must have a root u between -1 and 1, since -1 <= sin x <= 1.

Therefore it must be (-infinity,1).

Answer 3

Quadratic formula x = (7 +/- sqrt(40 9 - 16k))/8 The roots would be actual while 40 9 - 16k >= 0 40 9 >= 16k 40 9/sixteen >= ok The function might have actual roots for ok <= 40 9/sixteen The roots would be non-actual while 40 9 - 16k < 0 The function might have imaginary roots for ok > 40 9/sixteen

Answer 4

C
the eq is right if k<=1 or k>=9
for k>=9 you get that sinx doesn't belong to [-1,1] that it should
therefore k<=1 (for which sinx belongs to [-1,1]) is the correct answer