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
```        if f(x) is real valued function discontinuous at all integral points lying in [0,n] and if (f(x))2 =1 for x ε [0,n] then number of functions f(x) are
1]2n+1
2]6*3n
3]2*3n-1
4]3n+1```
8 years ago

147 Points
```										You can use short cut to solve this problem
put n=1
now our domain is [0,1]
then f(x) should be discontinuous at 0 and 1
given (f(x))2 =1
f(x) =1 or -1
case 1
let f(x)=1 at x=0
then  f(x) =-1 for x ε (0,1)
and f(x) = 1  at x=1

second case
let f(x)= -1 at x=0
then  f(x) =1 for x ε (0,1)
and f(x) = -1  at x=1

so there are only 2 possibilities
noe check the option for n=1
3rd option is correct

```
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