InterviewSolution
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InterviewSolution
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To the equation `2^2pi//cos^((-1)x)-(a+1/2)2^pi//cos^((-1)x)-a^2=0`has only one real root, then`1lt=alt=3`(b) `ageq1``alt=-3`(d) `ageq3`A. `1 le a le 3`B. `a ge 1`C. `a le -3`D. `a ge 3` |
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Answer» Correct Answer - B::C `1 le (pi)/(cos^(-1) x) lt oo` `rArr 2 le 2^((pi)/(cos^(-1)x)) lt oo` Hence, 2 should lie between or on the roots of `t^(2) - (a + (1)/(2)) t - a^(2) = 0, " where " t = 2^(pi//cos^(-1)x)` `rArr f(2) le 0 rArr a^(2) + 2a -3 ge 0` `rArr a in (-oo, -3] uu [1, oo)` |
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