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JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Continuity and Differentiability

If the function

f (x) = ⎩ ⎨ ⎧ \frac{1}{x} lo g_{x} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}), x < 0 k, x = 0 \frac{c o s ^{2} x - s i n ^{2} x - 1}{x ^{2} + 1 - 1}, x > 0

is continuous at

x = 0

, then

\frac{1}{a} + \frac{1}{b} + \frac{4}{k}

is equal to:

(A)4
(B)-5✓ Correct
(C)5
(D)-4

Answer: (B)

Solution

For

x > 0

,

\frac{cos ^{2} x - sin ^{2} x - 1}{x ^{2} + 1 - 1} = \frac{cos 2 x - 1}{x ^{2} + 1 - 1}

; with

cos 2 x - 1 \approx - 2 x^{2}

and

x^{2} + 1 - 1 \approx \frac{x ^{2}}{2}

, the right-hand limit is

- 4

. For

x < 0

,

\frac{1}{x} ln (\frac{1 + x / a}{1 - x / b}) \to \frac{1}{a} + \frac{1}{b}

as

x \to 0

. Continuity makes

k = \frac{1}{a} + \frac{1}{b} = - 4

, so

\frac{1}{a} + \frac{1}{b} + \frac{4}{k} = - 4 + \frac{4}{- 4} = - 5

.

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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