Question

If $A$ and $B$ are the two real values of $k$ for which the system of equations $x+2y+z=1$, $x+3y+4z=k.x+5y+10z=k^2$ is consistent, then $A+B=$

• A. 3
• B. 4
• C. $\underline{5}$
• D. 7

Answer 1

Answer: (A)

Given system of equation is,
$x+2y+z=1$
$x+3y+4z=k$
$x+5y+10z=k^2$
$\therefore D=\left|\begin{array}{ccc}1& 2& 1\\ 1& 3& 4\\ 1& 5& 10\end{array}\right|$
$=1(30-20)-2(10-4)+1(5-3)=10-12+2=0$
Since, $D=0$
$\therefore$ Given system of equation is consistent.
Therefore, $D_1=0$
$D_1=\left|\begin{array}{ccc}1& 2& 1\\ k& 3& 4\\ k^2& 5& 10\end{array}\right|$
$\Rightarrow 1(30-20)-2\left(10k-4k^2\right)+\left(5k-3k^2\right)=0$
$\Rightarrow 10-20k+8k^2+5k-3k^2=0$
$\Rightarrow 5k^2-15k+10=0$
$\Rightarrow k^2-3k+2=0$
$\Rightarrow (k-2)(k-1)=0$
$\Rightarrow k=2,1$
Hence, the real values of $k$ i.e.
$A=2$ and $B=1$
$\therefore A+B=2+1=3$