NTSE Stage 2 SAT 2015 Question Paper

As the degree of given polynomial is ‘3’, hence it will have three zeroes
Let α β γ be the zeroes of the given polynomial
Now α β γ are in A.P.
∴ Let $\mathrm{\alpha }=\mathrm{a}-\mathrm{d}$ $\mathrm{\beta }=\mathrm{a}$ and $\mathrm{\gamma }=\mathrm{a}+\mathrm{d}$
Now $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=\frac{144}{64}$
∴ $3\mathrm{a}=\frac{144}{64}$
∴ $\mathrm{a}=\frac{144}{64\times 3}=\frac{3}{4}$
and $\mathrm{\alpha }\mathrm{\beta }\mathrm{\gamma }=\frac{15}{64}$
∴ $\left(\mathrm{a}-\mathrm{d}\right)\left(\mathrm{a}\right)\left(\mathrm{a}+\mathrm{d}\right)=\frac{15}{64}$
∴ $\mathrm{a}\left({\mathrm{a}}^2-{\mathit{d}}^2\right)=\frac{15}{64}$
∴ $\frac{3}{4}\left(\frac{9}{16}-{\mathrm{d}}^2\right)=\frac{15}{64}$
∴ $\frac{9}{16}-{\mathrm{d}}^2=\frac{5}{16}$
∴ ${\mathrm{d}}^2=\frac{9}{16}-\frac{5}{16}=\frac{4}{16}$
∴ ${\mathrm{d}}^2=\frac{1}{4}$
∴ $\mathrm{d}=\pm \frac{1}{2}$
Now difference between largest and smallest zeroes of polynomial
$=\left(\mathrm{a}+\mathrm{d}\right)-\left(\mathrm{a}-\mathrm{d}\right)$
$=2\mathrm{d}=2\left(\frac{1}{2}\right)$
(∴ In option all the values are +ve) =1
∴ Option (a) is correct