InterviewSolution
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InterviewSolution
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A function `f: IR ->IR`, where `IR`, is the set of real numbers, is defined by `f(x) = (ax^2 + 6x - 8)/(a+6x-8x^2)` Find the interval of values of a for which is onto. Is the functions one-to-one for `a =3 ?` Justify your answer. |
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Answer» Correct Answer - `2 le alpha le 14,` No Let, `y=(alpha x^(2)+6x-8)/(alpha +6x-8x^(2))` `rArr alpha y + 6xy-8x^(2)y=alpha x^(2)+6x-8` `rArr -alpha x^(2)-8x^(2)y+6xy-6x+alpha y +8 =0` `rArr alpha x^(2)+8x^(2)y-6xy+6x-alpha y-8=0` `rArr x^(2)(alpha +8y)+6x(1-y)-(8+alpha y)=0` Since, x is real. `rArr B^(2)-4AC ge 0` `rArr 36(1-y)^(2)+4(alpha +8y)(8+alpha y) ge 0` `rArr 9(1-2y+y^(2))+[8alpha +(64 +alpha^(2))y+8alpha y^(2)] ge 0` `rArr y^(2)(9+8alpha)+y(46+alpha^(2))+9+8alpha ge 0 " ...(i)" ` `rArr A ge 0, D le 0, rArr 9+8alpha gt 0` `and (46+alpha^(2))^(2)-4(9+8alpha)^(2) le 0` `rArr alpha gt 9//8` ` and [46 + alpha^(2) -2(9+8alpha)][46 +alpha^(2)+2(9+8alpha)] le 0` `rArr a gt -9//8` `and (alpha^(2)-16alpha+28)(alpha^(2)+16alpha+64) le 0` `rArr alpha gt -9//8` `and [(alpha -2)(alpha-14)](alpha-8)^(2) le 0` `rArr alpha gt -9//8` `and (alpha -2)(alpha -14) le 0 " " [because (alpha +8)^(2) ge 0]` `rArr alpha gt -9//8` `and 2 le alpha le 14` `rArr 2 le alpha le 14` Thus, `f(x)=(alpha x^(2)+6x-8)/(alpha+6x-8x^(2))` will be onto, if `2 le alpha le 14` Again, when `alpha =3` `f(x)=(3x^(2)+6x-8)/(3+6x-8x^(2))`, in this case `f(x)=0` `rArr 3x^2)+6x-8=0` `rArr x=(-6pm sqrt(36+96))/(6)=(-6pm sqrt(132))/(6)=(1)/(3)(-3pm sqrt(33))` This shows that `f[(1)/(3)(-3+ sqrt(33))]=f[(1)/(3)(-3- sqrt(33))]=0` Therefore, f is not one-to-one. |
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