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

JEE · Math · previous-year question

  1. A.$${b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4}$$ are in A.P. with common difference $$-$$2
  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
  3. C.$${b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4}$$ are in a G.P.
  4. 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 $$-$$2correct

Answer

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

Explanation

$${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx} $$ $$ = \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx} $$ $$ = \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$$ 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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