If b_n=∫0(π)/2(cos^2nx)/sinxdx,n∈ℕ, then

Correct answer: (D) 1/(b_3-b_2),1/(b_4-b_3),1/(b_5-b_4) are in an A.P. with common difference -2. Given, b_n=∫0(π)/2(1+cos2nx)/sinxdx b_(n+1)-b_n=∫0(π)/2(cos^2n+1x-cos^2nx)/sinxdx =∫0(π)/2(-sin2n+1xsinx)/sinxdx =(cos(2n+1)x)/(2n+1)0(π)/2=(-1)/(2n+1) So,1/(b_3-b_2)=-5 1/(b_4-b_3)=-7 1/(b_5-b_4)=-9 So,1/(b_3-b_2),1/(b_4-b_3),1/(b_5-b_4) are in A.P. with c.d=-2

Q42 / 90 · JEE Main 2022 (25 Jun Shift 2) · Mathematics

If $b_{n} = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} n x}{\sin x} d x, n \in ℕ$ , 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.
Correct answer
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 $- 2$

Solution

Given,

$b_{n} = \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 n x}{\sin x} d x$

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

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

$= {(\frac{\cos (2 n + 1) x}{2 n + 1})}_{0}^{\frac{π}{2}} = \frac{- 1}{2 n + 1}$

So, $\frac{1}{b_{3} - b_{2}} = - 5$

$\frac{1}{b_{4} - b_{3}} = - 7$

$\frac{1}{b_{5} - b_{4}} = - 9$

So, $\frac{1}{b_{3} - b_{2}}, \frac{1}{b_{4} - b_{3}}, \frac{1}{b_{5} - b_{4}}$ are in A.P. with $c . d = - 2$

Hence, the correct answer is option: 'D'.

If bn=∫0π2cos2nxsinxdx,n∈ℕ, then

Solution

Relevant Questions

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