Question

(3) 6.2
2
12 Ify
sine v3 cose
then the minimum value of y is
(2) 2
1
1
() 73.1
2.
made being hented Hea Alin at any​

Answer 1

Answer:

 

2

sin 3 cos

y 

  

y will be minimum when the denominator m will be maximum.



m     sin 3 cos 

must be maximum to get minimum value of y.

Maximum or minimum value of any function [m=f(x)] can be found by putting

0

dm

dx



If

2

2

0; d m

dx



then the value of x found by putting

0

dm

dx



will give minimum value of m.

If

2

2

0; d m

dx



then the value of x found by putting

0

dm

dx



will give maximum value of m.

So, we have;

m     sin 3 cos 

cos 3 sin dm

d

   



 

2

2

– sin – 3 cos – sin 3 cos d m

d

      



As

2

2

0

d m

dx



(i.e. negative); so putting

0

dm

d

 



0

cos 3 sin 0

cos 3 sin

1

tan

3

30

1

sin

2

3

cos

2

   

  

 

 

 

 

Putting these value of sin and cos in m will give maximum value of m.

min.

2 4 1

1 3 (1 3)

3

2 2

y      

  

   