How to find the value of k give that equation has no solution?

Question

Can someone please explain to me how to do question 13? Thanks!

2016 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-06T12:25:20

#A#

#sin^2(x)-2sin(x)+k=0#

Let #u=sin(x)#

Then:

#u^2-2u+k=0#

Using the discriminant of a quadratic:

#b^2-4ac#

IF:

#b^2-4ac>0# ( real and different roots )

#b^2-4ac=0# ( real and repeated roots )

#b^2-4ac<0# ( complex roots )

For no real solution we need:

#b^2-4ac<0#

Plugging in values:

#(-2)^2-4(1)(k)<0#

#4-4k<0#

#k>1#

#k in (1, oo)# ( open interval, endpoints not included )

i.e. #color(white)(88)1< k< oo#