Question

Let the function $f\left(x\right)$, defined as
$f\left(x\right)=\{\begin{array}{cc}3ax+b& x<1\\ 11& x=1\\ 5ax-2b& x>1\end{array}$
be continuous at $x=1$. $a$ and $b$ are the roots of a quadratic equation, then the equation is

• A. $x^2-5x+6=0$
• B. $x^2+5x+6=0$
• C. $x^2-5x-6=0$
• D. ${3x}^2-5x+2=0$
• E. ${2x}^2-5x+3=0$

Answer 1

The correct option is A $x^2-5x+6=0$

Since $f\left(x\right)$ is continuous at $x=1$
$\therefore \begin{array}{c}lim\\ x\to 1^{-}\end{array}f\left(x\right)=\begin{array}{c}lim\\ x\to 1^{-}\end{array}f\left(x\right)=f\left(1\right)$
$\begin{array}{c}lim\\ x\to 1^{+}\end{array}(5ax-2b)=\begin{array}{c}lim\\ x\to 1^{-}\end{array}(3ax+b)=11$
$\Rightarrow 5a-2b=3a+b=11$
$\therefore 5a-2b=11....\left(i\right)$
and $3a+b=11....\left(ii\right)$
Solving (i) and (ii), we get
$5a-2b=11$
$6a+2b=22$
$11a=33$ $\Rightarrow a=3$
Substituting the value of a in (ii), we get
$3\left(3\right)+b=11$ $\therefore b=2$
Hence $a=3,b=2$

Therefore, the equation is

$(x-2)(x-3)=x^2-5x+6=0$