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 sum of max and min values of function f(x)=sin-12x+cos-12x+sec-12x is
a) π
b) π/2
c) 2π
d) 3π/2

```
8 years ago

8 Points
```
```
8 years ago
10 Points
```										Hi Anurag,
f(x)=sin-12x+cos-12x+sec-12x
sin-12x+cos-12x = π/2 for all x
sec-12x belongs to [0,π ]-{π/2}
therefore min of sec-12x is 0
max of sec-12x is π
therefore min of f(x) is π/2
max of f(x) is 3π/2
sum of min and max  values of f(x) is (π/2 + 3π/2) = 2π
Ans: (c)
All the best
```
8 years ago
8 Points
```										As Sin-1x  & Cos-1x is defined for x belongs to[-1,1] . Sec-1x is defined for x≤-1 & x≥1.Also Sin-1x+Cos-1x=¶/2 in the domain specified.
Domain of above function f(x) is x={-1/2 , 1/2}   (By solving inequalities -1≤x≤1,x≤-1 & x≥1)
Hence max{f(x)}=¶/2+Sec-1 (2*1/2)=¶/2+¶/2=¶
while min{f(x)}=0.So their sum=¶       ;   Answer is (a).

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

## Other Related Questions on Algebra

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 1,272 off
USE CODE: Neerajwin
Get extra Rs. 187 off
USE CODE: Neerajwin