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JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Differentiation

The value of

lo g_{e} 2 \frac{d}{d x} (lo g_{c o s x} cosec x)

at

x = \frac{π}{4}

is

(A) $- 22$
(B) $22$
(C) $- 4$
(D) $4$ ✓ Correct

Answer: (D)

Solution

Let

y = lo g_{c o s x} cosec x = \frac{l n ( cosec x )}{l n ( c o s x )} = - \frac{l n ( s i n x )}{l n ( c o s x )}

.

Differentiating:

\frac{d y}{d x} = - \frac{c o t x \cdot l n ( c o s x ) + t a n x \cdot l n ( s i n x )}{( l n ( c o s x ) ) ^{2}}

At

x = \frac{π}{4}

:

sin x = cos x = \frac{1}{2}

,

cot x = tan x = 1

,

ln (sin x) = ln (cos x) = - \frac{l n 2}{2}

.

\frac{d y}{d x}_{x = π /4} = - \frac{1 \cdot ( - \frac{l n 2}{2} ) + 1 \cdot ( - \frac{l n 2}{2} )}{( \frac{l n 2}{2} ) ^{2}} = \frac{l n 2}{( \frac{l n 2}{2} ) ^{2}} = \frac{4}{l n 2}

Therefore,

lo g_{e} 2 \cdot \frac{4}{l n 2} = ln 2 \cdot \frac{4}{l n 2} = 4

.

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option

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