Question

If $\alpha$ is repeated root of $a{x}^2+bx+c=0$ then $\underset{x\to \alpha }{lim}\frac{\tan \left(a{x}^2+bx+c\right)}{{\left(x-\alpha \right)}^2}$ is

• A. $a$
• B. $b$
• C. $c$
• D. $d$

Answer 1

Answer: (A)

$a$

Given,$\alpha$ is repeated root of $a{x}^2+bx+c=0$ then we've, $a{x}^2+bx+c=a(x-\alpha {)}^2$......(1)

Now,

$\underset{x\to \alpha }{lim}\frac{\tan (a{x}^2+bx+c)}{(x-\alpha {)}^2}$

$=\underset{x\to \alpha }{lim}\frac{\tan \left\{a\right(x-\alpha {)}^2\}}{(x-\alpha {)}^2}$ [ Using (1)]

$=\underset{x\to \alpha }{lim}\frac{\sin \left\{a\right(x-\alpha {)}^2\}}{a(x-\alpha {)}^2}\times a\times \underset{x\to \alpha }{lim}\frac{1}{{\cos }^2\left\{a\right(x-\alpha {)}^2\}}$

$=1\times a\times \frac{1}{1}$

$=a$.