QJEEM22Maf36c: JEE Main 2022 Mathematics - Class 12 JEEM22Maf36c Solution: If , then

Question

Verified Solution • 2025–26 Session moderate

If ${b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N}$ , then

A ${b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4}$ are in A.P. with common difference $-$ 2
B ${1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}}$ 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 ${1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}}$ are in an A.P. with common difference $-$ 2

Accepted Answer

${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx}$

$= \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx}$

$= \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$

So, ${b_3} - {b_2}$ , ${b_4} - {b_3}$ , ${b_5} - {b_4}$ are in H.P.

$\Rightarrow {1 \over {{b_3} - {b_2}}},\,{1 \over {{b_4} - {b_3}}},\,{1 \over {{b_5} - {b_4}}}$ are in A.P. with common difference $-$ 2.

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