If 3 d = 1 2 k | k+ k- |+c , then k=

Step-by-step solution

(A) aligned 3 d &= 3 -4 ² d = (3-4 ² ) d &= 1 3-4 ² d aligned Dividing both numerator and denominator by ² , we get aligned I &= ² 3 ² -4 ² d = ² 3 (1+ ² )-4 ² d I &= ² 3- ² d aligned Put =t ² d =d tI= 1 ( 3 )²-t² d t= 1 2 3 | 3 +t 3 -t |= 1 2 3 | 3 + 3 - | Comparing with given data, we get k = 3