ExamsMy practice
Ques. 1JEE Main 2024, 9 April (Morning Shift)
Let n ( n n4 + 1 - 2n (n2 + 1 ) n4 + 1 . . + n n4 + 16 - 8n (n2 + 4 ) n4 + 16 + . . + n n4 + n4 - 2n n2 (n2 + n2 ) n4 + n4 ) be k, using only the principal values of the inverse trigonometric functions. Then k2 is equal to .
Answer
32
Solution
The rth pair of terms is n n4 + r4 - 2nr2(n2 + r2) n4 + r4 = 1n [ 1 - (r/n)2 (1 + (r/n)2 ) 1 + (r/n)4 ]
So, as a limit of a sum (r/n x, 1n dx):
Limit = 01 1 - x2(1 + x2) 1 + x4 dx = 01 ( 1x2 - 1 )dx (x + 1x ) (x + 1x )2 - 2
Put x + 1x = t, so ( 1x2 - 1 )dx = -dt; as x goes 0 1, t goes 2:
Limit = 2 dtt t2 - 2 = 1 2 [ -1 t 2 ]2
= 1 2 ( 2 - 4 ) = 4 2
Hence k = 4 2 and k2 = 32.
Answer: 32