Question

Let p be the sum of all possible determinants of order 2 having 0, 1, 2 and 3 as their four elements. Then the common root a of the equations

x² + ax + [m + 1] = 0

x² + bx + [m + 4] = 0

x² – cx + [m + 15] = 0

Such that a > p, where a + b + c = 0 and m =$\lim_{n \rightarrow \infty}$$\sum_{r = 1}^{2n}\frac{r}{\sqrt{n^{2} + r^{2}}}$ .(where[.] denotes the greatest integer function)

• A. ± 3
• B. 3
• C. –3
• D. None of these

Answer 1

Answer: (B)

3

Answer 2

Let a be the common root, then

a² + aa + [m + 1] = 0 …(1)

a² + ab + [m + 4] = 0 …(2)

a² – ca + [m + 15] = 0 …(3)

Apply (1) + (2) – (3)

a² + [m] – 10 = 0 …(4)

But m = $\lim_{n \rightarrow \infty}$ $\frac { 1 } { \mathrm { n } }$ $\sum_{r = 1}^{2n}\frac{r}{\sqrt{n^{2} + r^{2}}}$=$\lim_{n \rightarrow \infty}$ $\frac { 1 } { \mathrm { n } }$ $\sum_{r = 1}^{2n}\frac{r/n}{\sqrt{1 + \left( \frac{r}{n} \right)^{2}}}$

= $\int_{0}^{2}\frac{x}{\sqrt{1 + x^{2}}}$dx = $\left( \sqrt{1 + x^{2}} \right)_{0}^{2}$= $\sqrt{5}$– 1

Now [m] = [$\sqrt{5}$ – 1] = 1

Now from (4), a² + 1 – 10 = 0 Ž a = ± 3 …(5)

Now number of determinants of order 2 having 0, 1, 2, 3 = 4! = 24. Let D₁ = $\left| \begin{matrix} a_{1} & a_{2} \\ a_{3} & a_{4} \end{matrix} \right|$be one such determinant and there exists another determinant.

D₂ = $\left| \begin{matrix} a_{3} & a_{4} \\ a_{1} & a_{2} \end{matrix} \right|$ (obtained on interchanging R₁ and R₂)

such that D₁ + D₂ = 0.

p = sum of all the 24 determinants = 0

since a > p Ž a > 0

from (5), a = 3.

Hence (2) is the correct answer