NTSE Stage 2 SAT 2015 Question Paper

The given equation is
$7\left(\mathrm{y}+\frac{1}{\mathrm{y}}\right)-2\left({\mathrm{y}}^2+\frac{1}{{\mathrm{y}}^2}\right)=9$
Let $\mathrm{y}+\frac{1}{\mathrm{y}}=\mathrm{m}$ ... (i)
Now squaring both side, we get
${\mathrm{y}}^2+\frac{1}{{\mathrm{y}}^2}+2={\mathrm{m}}^2$
∴ ${\mathrm{y}}^2+\frac{1}{{\mathrm{y}}^2}={\mathrm{m}}^2-2$ ... (ii)
Now putting the value of equation (i) and (ii) in original equation
$7\left(\mathrm{y}+\frac{1}{\mathrm{y}}\right)-2\left({\mathrm{y}}^2+\frac{1}{{\mathrm{y}}^2}\right)=9$
$7\mathrm{m}-2\left({\mathrm{m}}^2-2\right)=9$
$7\mathrm{m}-2\left({\mathrm{m}}^2-2\right)=9$
∴ $7\mathrm{m}-2{\mathrm{m}}^2+4=9$
∴ $2{\mathrm{m}}^2-7\mathrm{m}-4+9=0$
∴ $2{\mathrm{m}}^2-7\mathrm{m}+5=0$
∴ $2{\mathrm{m}}^2-5\mathrm{m}-2\mathrm{m}+5=0$
∴ $\mathrm{m}\left(2\mathrm{m}-5\right)-1\left(2\mathrm{m}-5\right)=0$
∴ $\left(\mathit{m}-1\right)\left(2\mathrm{m}-5\right)=0$
∴ $\mathrm{m}=1$ or $\mathrm{m}=\frac{5}{2}$
Now $\mathrm{y}+\frac{1}{\mathrm{y}}=\mathrm{m}=1$ is not possible as the value of $\mathrm{y}+\frac{1}{\mathrm{y}}$ will be always greater then 1.
∴ $\mathrm{y}+\frac{1}{\mathrm{y}}=\frac{5}{2}$
∴ ${\mathrm{y}}^2+1=\frac{5}{2}\mathrm{y}$
∴ ${\mathrm{y}}^2-\frac{5}{2}\mathrm{y}+1=0$
∴ $2{\mathrm{y}}^2-5\mathrm{y}+2=0$
∴ $2{\mathrm{y}}^2-4\mathrm{y}-\mathrm{y}+2=0$
∴ $2\mathrm{y}\left(\mathrm{y}-2\right)-1\left(\mathrm{y}-2\right)=0$
∴ $\left(2\mathrm{y}-1\right)\left(\mathrm{y}-2\right)=0$
∴ $\mathrm{y}=\frac{1}{2}\mathrm{or}\mathrm{y}=2$
Hence number of integral solution is 1.
∴ Option (b) is correct