InterviewSolution
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InterviewSolution
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समाकलनों `A=int_(0)^(pi)(sinxdx)/(sinx+cosx)` और `B=int_(0)^(pi)(sinxdx)/(sinx-cosx)` पर विचार कीजिए। B का मान क्या है ?A. `pi//4`B. `pi//2`C. `3pi//4`D. `pi` |
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Answer» Correct Answer - B माना `I=A=int_(0)^(pi)(sinx)/(sinx+cosx)dx`. . . (i) तथा `I=int_(0)^(pi)(sinx)/(sinx-cosx)dx` . . . (ii) `[becauseint_(0)^(a)f(x)dx=int_(0)^(pi)f(a-x)dx]` समी (i) व (ii) को जोड़ने पर, `2I=int_(0)^(pi)((sinx)/(sinx+cosx)+(sinx)/(sinx-cosx))dx` `rArr2I=` `int_(0)^(pi)(sinx(sinx-cosx+sinx+cosx))/(sin^(2)x-cos^(2)x)` `rArr2I=int_(0)^(pi)(2sin^(2)x)/(sin^(2)x-cos^(2)x)dx` `rArr2I=4int_(0)^(pi//2)(sin^(2)x)/(sin^(2)x-cos^(2)x)dx` . . . (iii) `[becauseint_(0)^(2a)f(x)dx=2int_(0)^(a)f(x)dx]` `rArr2I=4int_(0)^(pi//2)(cos^(2)x)/(cos^(2)x-sin^(2)x)dx` . . . (iv) समी (iii) व (iv) को जोड़ने पर, `4I=4int_(0)^(pi//2)((sin^(2)x-cos^(2)x)/(sin^(2)x-cos^(2)x))dx` `rArr4I=4[x]_(0)^(pi//2)rArr4I=4xx(pi)/(2)rArrI=(pi)/(2)` |
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