1. If $$x=(9+4\sqrt{5})^{48} = [x] +f$$, where [x] is defined as integral part of x and f is a faction, then x (1 - f) equals






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  • By: anil on 05 May 2019 03.32 am
    It is given that $$x=(9+4sqrt{5})^{48}$$ ... (1) Let us assume that $$y=(9-4sqrt{5})^{48}$$ ... (2) We can see that  0 < y < 1.  Also, x + y = 2($$48C0*(9)^{48}$$+$$48C2*(9)^{46}*(4sqrt{5})^2$$+$$48C4(9)^{44}*(4sqrt{5})^4$$+...+$$48C48*(4sqrt{5})^{48}$$) We can see that x + y is an integer therefore we can say that y + f = 1. Hence, y = (1 - f) We can see that = x(1 - f) = x*y $$Rightarrow$$ $$(9+4sqrt{5})^{48}(9-4sqrt{5})^{48}$$ $$Rightarrow$$ $$(9^2-(4sqrt{5})^2)^{48}$$
    $$Rightarrow$$ $$(81-80)^{48}$$ = 1 Hence, option A is the correct answer.
  • By: anil on 05 May 2019 03.32 am
    It is given that $$x=(9+4sqrt{5})^{48}$$ ... (1) Let us assume that $$y=(9-4sqrt{5})^{48}$$ ... (2) We can see that  0 < y < 1.  Also, x + y = 2($$48C0*(9)^{48}$$+$$48C2*(9)^{46}*(4sqrt{5})^2$$+$$48C4(9)^{44}*(4sqrt{5})^4$$+...+$$48C48*(4sqrt{5})^{48}$$) We can see that x + y is an integer therefore we can say that y + f = 1. Hence, y = (1 - f) We can see that = x(1 - f) = x*y $$Rightarrow$$ $$(9+4sqrt{5})^{48}(9-4sqrt{5})^{48}$$ $$Rightarrow$$ $$(9^2-(4sqrt{5})^2)^{48}$$
    $$Rightarrow$$ $$(81-80)^{48}$$ = 1 Hence, option A is the correct answer.
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