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The equation `(x/(x+1))^2+(x/(x-1))^2=a(a-1)`hasFour real roots if `a >2`Four real roots if `a |
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Answer» Correct Answer - D We have, `((x)/(x+1))^(2)+((x)/(x-1))^(2)=a(a-1)` `implies((x)/(x+1)+(x)/(x-1))^(2)-(2x^(2))/(x^(2)-1)=a(a-1)` `implies ((2x^(2))/(x^(2)-1))^(2)-((2x^(2))/(x^(2)-1))=a(a-1)` `implies ((2x^(2))/(x^(2)-1))((2x^(2))/(x^(2)-1)-1)=a(a-1)` ` implies y(y-1)=a(a-1)," where "y=(2x^(2))/(x^(2)-1)` `implies y^(2)-a^(2)-y+a=0` `implies (y-a)(y+a-1)=0 impliesy=a,y=1-a` When y=a, we have `(2x^(2))/(x^(2)-1)=a impliesx=+-sqrt((a)/(a-2))` Clearly, `x in R`if `a in (-oo,0) cup(2,oo)` When y=1-a, we have `(2x^(2))/(x^(2)-1)=a impliesx=+-sqrt((a-1)/(a-2))` Clearly, `x in R` if `a in (-oo,-1)cup (1,oo)`. Thus,the given equation has four real rootsif `a gt2` or a lt-1` and exactly two real roots if `1 lt a lt 2`. Hence,all the options are true . |
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