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JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Continuity and Differentiability

If the function

f (x) = {\frac{l o g _{e} ( 1 - x + x ^{2} ) + l o g _{e} ( 1 + x + x ^{2} )}{s e c x - c o s x}, k, x \in (\frac{- π}{2}, \frac{π}{2}) - {0} x = 0

is continuous at

x = 0

, then k is equal to :

(A)1✓ Correct
(B)-1
(C)e
(D)0

Answer: (A)

Solution

For continuity at

x = 0

, we need

k = lim_{x \to 0} f (x)

.

Note that

(1 - x + x^{2}) (1 + x + x^{2}) = 1 + x^{2} + x^{4}

, so:

x \to 0 lim \frac{ln ( 1 + x ^{2} + x ^{4} )}{sec x - cos x} = x \to 0 lim \frac{ln ( 1 + x ^{2} + x ^{4} ) \cdot cos x}{1 - cos ^{2} x}

= x \to 0 lim \frac{ln ( 1 + x ^{2} + x ^{4} ) \cdot cos x}{sin ^{2} x}

Using

ln (1 + u) \sim u

as

u \to 0

and

sin x \sim x

:

= x \to 0 lim \frac{( x ^{2} + x ^{4} ) cos x}{x ^{2}} = x \to 0 lim (1 + x^{2}) cos x = 1

Therefore

k = 1

.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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