Let f(X) = ∫01 |t - x| tdt.
1) Number of common tangent to curve x2 + y2 = 4 and f(x)
(A) 1 (B) 2
(c) 4 (D) infinite
1hey r u sure of teh options
f(x) comes out to be
-x2+ 13 , x≤0
x33-x2+13 , 0<x<1
x2-13 , x≥1
on ploting des i dont think der will be any common tangent with teh given circle
1wen x≤0
t-x is +ve hence mod(t-x) =t-x so given integral is f(x)=∫01(t-x)tdt
wen 0<x<1
now since x and t lie in teh same interval and so to decide sign of (t-x) we need to break 0,1) futher into (0,x] and [x,1)
so f(x)=∫0xmod(t-x)tdt+∫x1mod(t-x)tdt
=∫0x(x-t)tdt+∫x1(t-x)tdt
wen x≥1
f(x)=∫01mod(t-x)tdt=∫01(x-t)tdt
4REQUEST :Plzzzz dont keep changing ur name ya
It's difficult to identify whose posting