#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+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

# a fxn is defined on[-1,1]and area of sq. is 4. with its vertices at (0,0) and (x,f(x)) then f(x) is...

147 Points
11 years ago

Dear Jauneet

one vertex of square is given (0,0) .then a square of area four can not be formed whose rest three  point  lie in [-1,1]

so such function is not possible,

if (x,f(x) is the vertix of other side of diagonal,and other two vertix can be outside the [-1,1] then square is possible

then length of diagonnal = √{f2(x) +x2}

so area =4 =(digonal)2/2

4= {f2(x) +x2}/2

so f(x) =±√(8-x2)

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.

All the best.

Regards,