Let $\alpha$ and $\beta$ be the distinct roots of $a x^{2}+b x+c=0$, then $\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^{2}+b x+c\right)}{(x-\alpha)^{2}}$ is equal to

(A) $\frac{a^{2}}{2}(\alpha-\beta)^{2}$

(C) $-\frac{a^{2}}{2}(\alpha-\beta)^{2}$

(D) $\frac{1}{2}(\alpha-\beta)^{2}$

JEE Main · 2005 · 08 May 2005 · Offline · Limits

Let

α

and

β

be the distinct roots of

a x^{2} + b x + c = 0

, then

lim_{x \to α} \frac{1 - c o s ( a x ^{2} + b x + c )}{( x - α ) ^{2}}

is equal to

Answer: (A)

Since

α

and

β

are the roots,

a x^{2} + b x + c = a (x - α) (x - β)

, so near

x = α

the argument

a (x - α) (x - β)

is small. Using

1 - cos θ = 2 sin^{2} \frac{θ}{2}

, the expression becomes

\frac{2 sin ^{2} [ \frac{a ( x - α ) ( x - β )}{2} ]}{( x - α ) ^{2}}

. As

x \to α

, with

u = \frac{a ( x - α ) ( x - β )}{2} \to 0

we may replace

sin u

u

, giving

2 \cdot \frac{a ^{2} ( x - α ) ^{2} ( x - β ) ^{2}}{4 ( x - α ) ^{2}} \to \frac{a ^{2} ( α - β ) ^{2}}{2}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option