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Consider the system of equations cos^(-1)x + (sin^(-1) y)^(2) = (p pi^(2))/(4) and (cos^(-1) x) (sin^(-1) y)^(2) = (pi^(4))/(16), p in Z The value of p for which system has a solution is |
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Answer» 1 and `sin^(-1) y = b rArr b in [-pi//2, pi//2]` We have `a + b^(2) = (p pi^(2))/(4)`...(i) and `ab^(2) = (pi^(4))/(16)`..(ii) SINCE `b^(2) in [0, pi^(2)//4]`, we GET `a + b^(2) in [0, pi + pi^(2)//4]` So, from Eq. (i) we get `0 le (p pi^(2))/(4) le pi + (pi^(2))/(4)` i.e., `0 le p le (4)/(pi) + 1` Since `p in Z, " so " p = 0, 1 " or " 2` Substituting the value of `b^(2)` from Eq. (i) Eq. (ii), we get `a((p pi^(2))/(4) -a) = (pi^(4))/(16)` `rArr 16a^(2) - 4p pi^(2) a + pi^(4) = 0`...(III) Since `a in R, " we have " D ge 0` i.e., `16^(2) ge 4- 64 pi^(4) ge 0` or `p^(2) ge 4 " or " p ge 2 " or " p =2` Substituting `p =2` in Eq. (iii), we get `16a^(2) -8pi^(2) a + pi^(4) = 0` or `(4a - pi^(2))^(2) = 0` or `a = (pi^(2))/(4) = cos^(-1) x " or " x = cos.(pi^(2))/(4)` From Eq. (ii), we get `(pi^(2))/(4) b^(2) = (pi^(4))/(16)` or `b = +- (pi)/(2) = sin^(-1) y " or " y = +- 1` |
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