InterviewSolution
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InterviewSolution
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For a positive integer n , `f_n(theta)=(tantheta/2)(1+sectheta)(1+sec2theta)(1+sec4theta)....(1+sec2^ntheta.)`, thenA. `f_(2) ((pi)/(16))=1`B. `f_(3) ((pi)/(32))=1`C. `f_(4) ((pi)/(64))=1`D. `f_(5) ((pi)/(128))=1` |
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Answer» Correct Answer - A::B::C::D NOTE Multiplicative loop is very important approach in IIT Mathematics `(tan.(theta)/(2))(1+sectheta)=(sintheta//2)/(sintheta//2)[1+(1)/(costheta)]` `=((sintheta//2)2cos^(2)theta//2)/((costheta//2)costheta)` `=((2sintheta//2)cos theta//2)/(costheta)=(sintheta)/(costheta)=tan theta` `therefore" " f_(n)(theta) = (tantheta//2)(1+sectheta)` `(1+sec 2 theta)(1+sec2^(2)theta)...(1+sec2^(n)theta)` `=(tan theta)(1+sec 2 theta)(1+sec2^(2) theta)...(1+sec2^(n)theta)` `=tan2 theta.(1 + sec2^(2)theta)...(1+sec2^(n)theta)` `=tan (2^(n)theta)` Now, `f_(2)((pi)/(16))=tan(2^(2).(pi)/(16))=tan((pi)/(4))=1` Therefore, (a) is the answer. `f_(3)((pi)/(32))=tan(2^(3).(pi)/(32))=tan((pi)/(4))=1` Therefore, (b) is the answer. `f_(4)((pi)/(64))=tan(2^(4).(pi)/(64))=tan((pi)/(4))=1` Therefore, (c) is the answer. `f_(5)((pi)/(128))=tan(2^(5).(pi)/(128))=tan((pi)/(4))=1` Therefore, (d) is the answer. |
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