If $b_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} n x}{\sin x} d x, n \in N$, then

(D) $\frac{1}{\mathrm{~b}_{3}-\mathrm{b}_{2}}, \frac{1}{\mathrm{~b}_{4}-\mathrm{b}_{3}}, \frac{1}{\mathrm{~b}_{5}-\mathrm{b}_{4}}$ are in an A.P. with common difference -2

(A) $\mathrm{b}_{3}-\mathrm{b}_{2}, \mathrm{~b}_{4}-\mathrm{b}_{3}, \mathrm{~b}_{5}-\mathrm{b}_{4}$ are in an A.P. with common difference -2

(B) $\frac{1}{b_{3}-b_{2}}, \frac{1}{b_{4}-b_{3}}, \frac{1}{b_{5}-b_{4}}$ are in an A.P. with common difference 2

(C) $\mathrm{b}_{3}-\mathrm{b}_{2}, \mathrm{~b}_{4}-\mathrm{b}_{3}, \mathrm{~b}_{5}-\mathrm{b}_{4}$ are in a G.P.

Definite Integrals — JEE Main · 2022 · 25 Jun 2022 · Shift 2 Question with Solution

b_{n} = \int_{0}^{\frac{π}{2}} \frac{c o s ^{2} n x}{s i n x} d x, n \in N

, then

(A) $b_{3} - b_{2}, b_{4} - b_{3}, b_{5} - b_{4}$ are in an A.P. with common difference -2
(B) $\frac{1}{b _{3} - b _{2}}, \frac{1}{b _{4} - b _{3}}, \frac{1}{b _{5} - b _{4}}$ are in an A.P. with common difference 2
(C) $b_{3} - b_{2}, b_{4} - b_{3}, b_{5} - b_{4}$ are in a G.P.
(D) $\frac{1}{b _{3} - b _{2}}, \frac{1}{b _{4} - b _{3}}, \frac{1}{b _{5} - b _{4}}$ are in an A.P. with common difference -2✓ Correct

Answer: (D)

Solution

Taking the difference of consecutive terms,

b_{n + 1} - b_{n} = \int_{0}^{π /2} \frac{cos ^{2} ( n + 1 ) x - cos ^{2} n x}{sin x} d x

, and the identity

cos^{2} A - cos^{2} B = - sin (A + B) sin (A - B)

turns the numerator into

- sin (2 n + 1) x sin x

. The

sin x

cancels, leaving

b_{n + 1} - b_{n} = - \int_{0}^{π /2} sin (2 n + 1) x d x = \frac{1}{2 n + 1} [cos (2 n + 1) x]_{0}^{π /2}

. Since

2 n + 1

is odd,

cos (2 n + 1) \frac{π}{2} = 0

, so the bracket is

0 - 1 = - 1

and

b_{n + 1} - b_{n} = - \frac{1}{2 n + 1}

. Its reciprocal is therefore

\frac{1}{b _{n + 1} - b _{n}} = - (2 n + 1)

, which for

n = 2, 3, 4

gives

- 5, - 7, - 9

; these reciprocals form an arithmetic progression whose common difference is

- 2

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Definite Integrals

Solution

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