(1 + cot2 θ) (1 – cos θ) (1 + cos θ) = …………………. (a) tan2 θ – sec2 θ (b) sin2 θ – cos2 θ (c) sec2 θ – tan2 θ (d) cos2 θ – sin2 θ

Answer

`2 le alpha le 14,` No

Explanation

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.