1.

Find the minimum value of sin e+cos 9

Answer»

Sin^2x

Also Cos^2x >= Cos^4x

Add sin^2x both sides

Cos^2x + sin^2x>= Cos^4x+sin^2x

1>= Cos^4x+sin^2x

Therefore max of A is 1

Also both terms

Cos^4x+sin^2x

= (1 –sin^2x)^2 +sin^2x

= 1 + sin^4x – 2sin^2x + sin^2x

=Sin^4x – sinn^2x + 1

= (sin^2x – 1/2)^2 + 3/4

now Min =3/4 and Max = 1


Discussion

No Comment Found

Related InterviewSolutions