JEE Advance - Mathematics Hindi (2016 - Paper 2 Offline - No. 13)
मान लें
$$f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {\left( {{{{n^n}\left( {x + n} \right)\left( {x + {n \over 2}} \right)...\left( {x + {n \over n}} \right)} \over {n!\left( {{x^2} + {n^2}} \right)\left( {{x^2} + {{{n^2}} \over 4}} \right)....\left( {{x^2} + {{{n^2}} \over {{n^2}}}} \right)}}} \right)^{{x \over n}}},$$
सभी $$x>0$$ के लिए। तब
$$f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {\left( {{{{n^n}\left( {x + n} \right)\left( {x + {n \over 2}} \right)...\left( {x + {n \over n}} \right)} \over {n!\left( {{x^2} + {n^2}} \right)\left( {{x^2} + {{{n^2}} \over 4}} \right)....\left( {{x^2} + {{{n^2}} \over {{n^2}}}} \right)}}} \right)^{{x \over n}}},$$
सभी $$x>0$$ के लिए। तब
$$f\left( {{1 \over 2}} \right) \ge f\left( 1 \right)$$
$$f\left( {{1 \over 3}} \right) \le f\left( {{2 \over 3}} \right)$$
$$f'\left( 2 \right) \le 0$$
$$\,{{f'\left( 3 \right)} \over {f\left( 3 \right)}} \ge {{f'\left( 2 \right)} \over {f\left( 2 \right)}}$$
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