Question

If $\alpha ,\beta$ be the roots of $4x^2-16x+\lambda =0$, where $\lambda ϵR$, such that $1<\alpha <2$ and $2<\beta <3$, then the number of integral solutions of $\lambda$ is

• A. $7$
• B. $6$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$7$

Given: $4x^2-16x+\lambda =0$

Applying $D=b^2-4ac$

${16}^2-16\lambda >0$

$⟹16-\lambda >0$

$⟹\lambda <\mathrm{16........}\left(i\right)$

$\therefore$ sum of the roots $=\alpha +\beta =\frac{16}{4}=4$

and Product of the roots $=\alpha \beta =\frac{\lambda }{4}$

Now, $1<\alpha <\mathrm{2.....}\left(ii\right)$

$2<\beta <\mathrm{3.....}\left(iii\right)$

Multiplying (ii) and (iii), we get,

$2<\alpha \beta <6$

$⟹2<\frac{\lambda }{4}<6$

Multiplying 4, we get,

$8<\lambda <\mathrm{24......}\left(iv\right)$

Cobining (i) and (iv), we get,

$8<\lambda <16$

$\therefore \lambda$ can have $7$ integral solutions