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JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Limits

If

α > β > 0

are the roots of the equation

a x^{2} + b x + 1 = 0

, and

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

, then

k

is equal to

(A) $2 β$
(B) $2 α$ ✓ Correct
(C) $α$
(D) $β$

Answer: (B)

Solution

Since

α, β

are roots of

a x^{2} + b x + 1 = 0

, by transformation

\frac{1}{α}, \frac{1}{β}

are roots of

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

. Thus

x^{2} + b x + a = (x - \frac{1}{α}) (x - \frac{1}{β})

. The limit becomes

lim_{x \to \frac{1}{α}} [\frac{1 - c o s (( x - \frac{1}{α} ) ( x - \frac{1}{β} ))}{2 ( 1 - α x ) ^{2}}]^{1/2}

. Using

1 - cos θ \approx \frac{θ ^{2}}{2}

for small

θ

and

1 - α x = α (\frac{1}{α} - x)

, we get

[\frac{( \frac{1}{α} - \frac{1}{β} ) ^{2}}{4 α ^{2}}]^{1/2} = \frac{\frac{1}{β} - \frac{1}{α}}{2 α}

. Comparing with

\frac{1}{k} (\frac{1}{β} - \frac{1}{α})

gives

k = 2 α

.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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