Question

(a) If x be real, then the expression $\frac{2a(x-1){\sin }^2\theta }{x^2-{\sin }^2\theta }$ does not lie between $2a{\sin }^2\left(\theta /2\right)$ and $2a{\cos }^2\left(\theta /2\right)$.
(b) A function $f:R\to R$, where R is the set of real numbers, is defined by $f\left(x\right)=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$ Find the interval of values of $\alpha$ for which f is onto.
Is the function one-to-one for $\alpha =3?$ Justify yout answer.

Answer 1

(a) If the given expression be y, then
$y(x^2-{sin}^2\theta )=2a(x-1){sin}^2\theta$
$x^{2}y-2ax{sin}^2\theta -{sin}^2\theta (y-2a)=0$
Since x is real $\therefore \Delta \ge 0$
$4a^2{sin}^4\theta +4y{sin}^2\theta (y-2a)\ge 0$
Cancel $4{sin}^2\theta ,a+ive$ quantity.
or $y^2-2ay+a^2{sin}^2\theta =+ive$
${(y-a)}^2-a^2{cos}^2\theta =+ive$
$(y-a+acos\theta )(y-a-acos\theta )=+ive$
$\therefore$ y does not lie between p and q where p<q
or y does not lie between
$a(1-cos\theta )$ and $a(1+cos\theta )$
or $2a{sin}^2\frac{\theta }{2}$ and $2a{cos}^2\frac{\theta }{2}$
(b) Refer Q.46(a)
Let $y=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$; then
$x^2(\alpha +8y)+6x(1-y)-(8+\alpha y)=0$
Since x is real $\Delta \ge 0$
$\therefore$ $36{(1-y)}^2+4(\alpha +8y)(8+\alpha y)\ge 0$
or $9(1-2y+y^2)+\{8\alpha y^2+(64+{\alpha }^2)y+8\alpha \}\ge 0$
or $(9+8\alpha )y^{21}+y(46+{\alpha }^2)+(9+8\alpha )\ge 0$ i.e. $+ive$
The above quadratic expression has to be $+ive$ for all real values of y as f is onto.
The condition as in last part is $\Delta \le 0$ and $9+8\alpha >0$
or ${(46+{\alpha }^2)}^2-4{(9+8\alpha )}^2\le 0$
$(46+{\alpha }^2+18+16\alpha )(46+{\alpha }^2-18-16\alpha )\le 0$
or $({\alpha }^2+16\alpha +64)({\alpha }^2-16\alpha +28)\le 0$
${(\alpha +8)}^2(\alpha -2)(\alpha -14)\le 0$
or $2\le \alpha \le 14$ and for this $9+8\alpha$ is $+ive$.
When $\alpha =3$
$y=\frac{3x^2+6x-8}{3+6x-8x^2}$
when $y=0$ we get
$3x^2+6x-8=0$
$\therefore$ $x=\frac{1}{3}(-3\pm \sqrt{33}$
Thus for $y=0 we get two values of x.
Hence the function is not one-one in this case.