Sequences and Series — JEE Main · 2021 · 24 Feb 2021 · Shift 1 Question with Solution

e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + \dots .\infty) l o g_{e} 2}

satisfies the equation

t^{2} - 9 t + 8 = 0

, then the value of

\frac{2 s i n x}{s i n x + 3 c o s x} (0 < x < \frac{π}{2})

Answer: (B)

Solution

The infinite geometric series:

cos^{2} x + cos^{4} x + cos^{6} x + \dots = \frac{c o s ^{2} x}{1 - c o s ^{2} x} = \frac{c o s ^{2} x}{s i n ^{2} x}

(for

∣ cos^{2} x ∣ < 1

).

So the expression becomes:

e^{\frac{c o s ^{2} x}{s i n ^{2} x} l n 2} = 2^{\frac{c o s ^{2} x}{s i n ^{2} x}} = 2^{c o t^{2} x}

This satisfies

t^{2} - 9 t + 8 = 0

, giving

t = 1

t = 8

.

Since

0 < x < \frac{π}{2}

, we have

cot^{2} x > 0

, so

2^{c o t^{2} x} > 1

. Thus

t = 8

2^{c o t^{2} x} = 8 = 2^{3} \Rightarrow cot^{2} x = 3 \Rightarrow cot x = 3

(positive in first quadrant)

So

tan x = \frac{1}{3}

, which means

x = 30°

\frac{π}{6}

.

Now compute:

\frac{2 sin x}{sin x + 3 cos x} = \frac{2 \cdot \frac{1}{2}}{\frac{1}{2} + 3 \cdot \frac{3}{2}} = \frac{1}{\frac{1}{2} + \frac{3}{2}} = \frac{1}{2}

Wait, let me recalculate. At

x = \frac{π}{6}

sin x = \frac{1}{2}

cos x = \frac{3}{2}

\frac{2 \cdot \frac{1}{2}}{\frac{1}{2} + 3 \cdot \frac{3}{2}} = \frac{1}{\frac{1}{2} + \frac{3}{2}} = \frac{1}{2}

But option (B) is

\frac{1}{2}

. However, if

cot x = 3

, then

x = \frac{π}{6}

. Let me verify by dividing numerator and denominator by

cos x

\frac{2 tan x}{tan x + 3} = \frac{2 \cdot \frac{1}{3}}{\frac{1}{3} + 3} = \frac{\frac{2}{3}}{\frac{1 + 3}{3}} = \frac{2}{4} = \frac{1}{2}

Actually, the answer should be

\frac{1}{2}

, but if the source says (A)

3

, there may be an error.

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option