To je drugo vprašanje. Obkrožil je n napisano kot dvom. Ali mi lahko kdo pomaga, da pridem skozi to?

Odgovor:

Prosimo, da se sklicujete na Razlaga.

Pojasnilo:

Glede na to, # e ^ (f (x)) = ((10 + x) / (10-x)), x v (-10,10).

#:. ln ^ (f (x)) = ln ((10 + x) / (10-x)) #.

#:. f (x) * lne = ln ((10 + x) / (10-x)), #

# ie, f (x) = ln ((10 + x) / (10-x)) …………………….. (ast_1)) #.#, # ali, f (x) = ln (10 + x) -ln (10-x) #.

Priključitev # (200x) / (100 + x ^ 2) # namesto # x #, dobimo, # f ((200x) / (100 + x ^ 2)) #, # = ln {10+ (200x) / (100 + x ^ 2)} - ln {10- (200x) / (100 + x ^ 2)} #, # = ln {(1000 + 10x ^ 2 + 200x) / (100 + x ^ 2)} - ln {(1000 + 10x ^ 2-200x) / (100 + x ^ 2)} #, # = ln {10 (100 + x ^ 2 + 20x)} / (100 + x ^ 2) - ln {10 (100 + x ^ 2-20x)} / (100 + x ^ 2) #, # = ln {10 (100 + x ^ 2 + 20x)} / (100 + x ^ 2) -: {10 (100 + x ^ 2-20x)} / (100 + x ^ 2) #,

# = ln {(100 + x ^ 2 + 20x) / (100 + x ^ 2-20x)} #, # = ln {((10 + x) / (10-x)) ^ 2} #.

Tako #f ((200x) / (100 + x ^ 2)) = ln {((10 + x) / (10-x)) ^ 2} ……….. (ast_2) #.

Zdaj, z uporabo # (ast_1) in (ast_2) # v

#f (x) = k * f ((200x) / (100 + x ^ 2)) ………………….. "Given" #, dobimo, #ln ((10 + x) / (10-x)) = k * ln {((10 + x) / (10-x)) ^ 2} #, # t.j., ln ((10 + x) / (10-x)) = ln ((10 + x) / (10-x)) ^ (2k) #.

#:. 1 = 2k, ali, k = 1/2 = 0.5, "kar je možnost" (1).