Question

If $\alpha$ and $\beta$ be the roots of $a{x}^2+bx+c=0$, then $\underset{x\to a}{lim}\frac{1-cos(a{x}^2+bx+c)}{(x-\alpha {)}^2}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\because a{x}^2+bx+c=a(x-a)(x-\beta )\therefore \underset{x\to \alpha }{lim}\frac{1-cos(a{x}^2+bx+c)}{(x-{\alpha }^2)}=\underset{x\to \alpha }{lim}\frac{1-cos\left\{a\right(x-\alpha \left)\right(x-\beta \left)\right\}}{(x-{\alpha }^2)}=\underset{x\to \alpha }{lim}\frac{(1-cos\{a(x-\alpha )(x-\beta )\left\}\right)(x+cos\{a(x-\alpha )(x-\beta )\left\}\right)}{(x-\alpha {)}^2(1+cos\left\{a\right(x-\alpha \left)\right(x-\beta \left)\right\})}=\underset{x\to \alpha }{lim}\frac{si{n}^2\left(a\right(x-\alpha \left)\right(x-\beta \left)\right)}{a^2(x-\alpha {)}^2(x-\beta {)}^2}.\frac{a^2.(x-\beta {)}^2}{(1+cos\{a(x-\alpha )(x-\beta )\left\}\right)}=1^2.\frac{a^2(\alpha -\beta {)}^2}{(1+1)}=\frac{a^2}{2}(\alpha -\beta {)}^2$