Question

If $\alpha ,\beta$ are the roots of the quadratic equation $4x^2-16x+\lambda =0,\lambda \in R$ such that $1<\alpha <2$ and $2<\beta <3$, then the number of integral values of $\lambda$ are

• A. $5$
• B. $6$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$3$

Given quadratic equation $4x^2-16x+\lambda =0$
$\Rightarrow x=\frac{4\pm \sqrt{16-\lambda }}{2}$
Let $\alpha =\frac{4-\sqrt{16-\lambda }}{2}$ and $\beta =\frac{4+\sqrt{16-\lambda }}{2}$
For real roots $16-\lambda >0$
Since, $1<\alpha <2$
$1<\frac{4-\sqrt{16-\lambda }}{2}<2$
$\Rightarrow 12<\lambda$
Since, $2<\beta <3$
$2<\frac{4+\sqrt{16-\lambda }}{2}<3$
$\Rightarrow 12<\lambda$$\therefore 12<\lambda <16$

Therefore, integral values of $\lambda$ are $13,14,15$.
Hence, option 'D' is correct.