Question

If $t$ is real and $\lambda =\frac{t^2-3t+4}{t^2+3t+4}$, then the number of solutions of the system of equations $3x-y+4z=3,x+2y-3z=-2,6x+5y+\lambda z=-3$ is

• A. One
• B. Two
• C. Zero
• D. Infinite

Answer 1

The correct option is A One

Given: $\lambda =\frac{t^2-3t+4}{t^2+3t+4}$

$\Rightarrow (\lambda -1)t^2+3(\lambda +1)t+4(\lambda -1)=0$

$t$ is real $\therefore$ Discriminant$\ge 0$

$\Rightarrow \left(3\right(\lambda +1){)}^2-4(\lambda -1\left)\right(4(\lambda -1))\ge 0$

$\Rightarrow -7{\lambda }^2+50\lambda -7\ge 0$

$\Rightarrow 7{\lambda }^2-50\lambda +7\le 0$

$\Rightarrow (7\lambda -1)(\lambda -7)\le 0$

$\lambda ϵ[\frac{1}{7},7]$

By Cramer's rule we can find no. of solutions,

$D=\left|\begin{array}{ccc}3& -1& 4\\ 1& 2& -3\\ 6& 5& \lambda \end{array}\right|$

$=3(2\lambda +15)+(\lambda +18)+4(5-12)$

$D=7\lambda +35$

Now $\lambda ϵ[\frac{1}{7},7]\Rightarrow D>0$

And as $D>0$ so unique solution of the given system of equations.

Hence, $\left(A\right)$