Question

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

• A. $5$
• B. $6$
• C. $3$
• D. $7$

Answer 1

The correct option is D $7$

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

Applying $D=b^2-4ac$

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

$⟹$16−λ>0

$⟹$λ<16 (i)

∴ sum of the roots =α+β=$\frac{16}{4}=4$

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

Now, 1<α<2 (ii)

2<β<3 (iii)

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

2<αβ<6

$⟹2<\frac{\lambda }{4}<6$

Multiplying 4, we get,

8<λ<24 (iv)

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

8<λ<16

∴λ can have 7 integral solutions