The equation \(\displaystyle{{\sin}^{{4}}{x}}-{\left({k}+{2}\right)}{{\sin}^{{2}}{x}}-{\left({k}+{3}\right)}={0}\) possesses a solution if: A)

London Douglas

London Douglas

Answered question

2022-03-31

The equation sin4x(k+2)sin2x(k+3)=0 possesses a solution if:
A) k3
B) k<2
C) 3k2
D) k is any (+ve) value

Answer & Explanation

Charlie Haley

Charlie Haley

Beginner2022-04-01Added 14 answers

Explanation:
sin4x(k+2)sin2x(k+3)=0≈∼(1)
Let sin2x=z
f(z)=z2(k+2)z(k+3)=0≈∼(2)
there are two cases note thta for this quadratic (2) has
B24AC=(k+4)20
Case-1: exactly one root should be in (0,1)f(0)f(1)<03<k<2
Case-2: Both roots in (0,1)0<z0=k+22<1andf(1)>1andf(0)>0. we get 2<k<0,andk<3andk<2 whose overlap is null. So only case (1) is possible. Finally Eq. (1) will exactly one solution if 3<k<2 which is option (C).

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