InterviewSolution
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InterviewSolution
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दिया गया है कि `a_(n)=int(sin^(2){(n+1)x})/(sin2x)dx` `a_(n-1)-a_(n-4)` किसके बराबर है ?A. `-1`B. 0C. 1D. 2 |
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Answer» Correct Answer - B `a_(n-1)=int_(0)^(pi)(sin^(2){(n-1+1)}x)/(sin2x)dx` `=int_(0)^(pi)(sin^(2){nx})/(sin2x)dx` . . . (i) `int_(0)^(pi)f(x)dx=int_(0)^(pi)f(pi-x)dx` के प्रगुण से , `a_(n-1)=int_(0)^(pi)(sin^(2){n(pi-x)})/(sin2(pi-x))dx` `=int_(0)^(pi)(sin^(2){nx})/(-sin2x)dx` . . . (ii) समी (i) और (ii) को जोड़ने पर, `2a_(n-1)=0rArra_(n-1)=0` अब, `a_(n-4)=int_(0)^(pi)(sin^(2){(n-4+1)x})/(sin2x)dx` `=int_(0)^(pi)(sin^(2){(n-3)x})/(sin2x)dx` . . . (iii) `int_(0)^(pi)f(x)dx=int_(0)^(pi)f(pi-x)dx` के प्रगुण से, `a_(n-4)=int_(0)^(pi)(sin^(2){(n-3)(pi-x)})/(sin2(pi-x))dx` `=int_(0)^(pi)(sin^(2)(n-3)x)/(-sin2x)` . . . (iv) समी (iii) और (iv) को जोड़ने पर, `2a_(n-4)=0rArra_(n-4)=0` `thereforea_(n-1)-a_(n-4)=0-0=0` |
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