Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        The minimum value of the function 2cos2x-cos4x in 0`
8 years ago

Ramesh V
70 Points
```										We have f(x) = 2cos2x - cos4xf ''(x) = -4sin2x + 4sin4x= 4(sin4x - sin2x)= 8cos3x.sinx    Now,  f ''(x) = 0 8cos3xsinx= 0cos3xsinx = 0cos3x = 0 or sinx = 0or x = 0, .or x = 0, .    (As 0  x  )Now,f(0) = 2cos0 - cos0 = 2 - 1 = 1   f = 2cos- cos2= -2 - 1 = -3      and f() = 2cos2- cos4= 2 - 1 = 1Thus, the maximum value of f(x) is  and the minimum value of f(x) is -3.
--
Regards
Ramesh
```
8 years ago
147 Points
```										Hi pallavi

2cos2x-cos4x
=2 cos2x -2cos22x +1
=-2(cos2x-1/2)2 +3/2
minimum value occur when cos2x-1/2 will be maximum
=-2(-1-1/2)2 +3/2
=-3

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best pallavi.   Regards, Askiitians Experts Badiuddin

```
8 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Differential Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details