Question

If $x_1$ and $x_2$ are the real and distinct roots of $a{x}^2+bx+c=0$, then
$e^{\underset{x\to x_1}{lim}\frac{(1+\sin (a{x}^2+bx+c))-1}{x-x_1}}$ is equal to:

• A. $e^{x_1-x_2}$
• B. $e^{x_2-x_1}$
• C. $e^{a(x_1-x_2)}$
• D. $e^{a(x_2-x_1)}$

Answer 1

The correct option is C $e^{a(x_1-x_2)}$

Since, $x_1$ and $x_2$ are the real and distinct roots of $a{x}^2+bx+c=0$
Then , $a(x^2-(x_1+x_2)x+x_1{x}_2)=0$
$\Rightarrow a(x-x_1)(x-x_2)=0$
Now, $e^{\underset{x\to x_1}{lim}\frac{(1+\sin (a{x}^2+bx+c))-1}{x-x_1}}$
${=e}^{{lim}_{x\to x_1}\frac{1+\sin \left(a\right(x-x_1\left)\right(x-x_2\left)\right)-1}{x-x_1}}$
$=e^{{lim}_{x\to x_1}\frac{\sin \left(a\right(x-x_1\left)\right(x-x_2\left)\right)}{x-x_1}}$
$=e^{{lim}_{x\to x_1}\frac{\sin \left(a\right(x-x_1\left)\right(x-x_2\left)\right)}{a(x-x_2)(x-x_1)}a(x-x_2)}$
$=e^{a(x_1-x_2)}$