See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

            2 sin 7   2 7  7 2  50. If 2 n 2 2 2 2 (2n 1)(2n 2n 1) T ,n N,n 2 (n 1) n (n 1) (n 2)          . If n r n r 2 K lim T    then 27 k is equal to AITS-FT-I (Paper-2)-PCM-JEE(Advanced)/22 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 25 Ans. 3 Sol. 2 2 r 2 2 2 2 2 2 2 2 (2r 1)(2r 2r 1) (2r 1)(2r 2r 1) T (r 1) (r 2r) (r 1) ((r 1) 1)               r 2 2 2 2 1 1 T (r 1) ((r 1) 1)             2 2 2 2 2 2 2 2 2 2 2 2 n 1 1 1 1 1 1 K lim .... (2 1) (3 1) (3 1) (4 1) (n 1) ((n 1) 1)                                          1 k 9  51. Let f : R R  be a function defined by 3 2 f(x) x 3x 9x 27.     If x (a,b) (c,d)   then 3 2 3 2 f (x) 3f (x) 9f(x) 27 f(x 4x 3x 19).        Then (b a) (c d)    is Ans. 10 Sol. 2 f '(x) 3(x 2x 3) 3(x 1)(x 3)       Case-I f '(x) 0  i.e. f(x) is increasing the x 3 or x 1   3 2 3 2 f(f(x) f(x 4x 3x 19) f(x) x 4x 3x 19          2 x 6x 8 0 x (2,4)      x (3,4)  Case-II f '(x) 0  i.e. f(x) is decreasing then 1 x 3   3 2 f(x) x 4x 3x 19     2 x 6x 8 0 (x 2)(x 4) 0         x 2 or x 4    x ( 1,2)  x ( 1,2) (3,4)   b a c d 2 1 3 4 10         Section – C (Maximum Marks: 12) This section contains THREE (03) question stems. There are TWO (02) questions corresponding to each question stem. The answer to each question is a NUMERICAL VALUE. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places. Question Stem for Question Nos. 52 and 53 Question Stem Let f be a real valued function from N to N satisfying the relation f(m + n) = f(m) + f(n) for all m, n  N. 52. If range of f contains all the even numbers, then the possible value of f(1) are A & B then 3A 2B 5  is equal to Ans.