InterviewSolution
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InterviewSolution
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if `cos^(-1)((2)/(3x))+cos^(-1)((3)/(4x))=(pi)/2, x>3/4` then `x=` (a) `(sqrt(145))/(11)` (b) `(sqrt(145))/(12)` (c) `(sqrt(146))/(10)` (d) `(sqrt(146))/(11)`A. `(sqrt ( 145))/( 10 )`B. `( sqrt ( 146))/( 12)`C. ` (sqrt ( 145))/( 12)`D. `(sqrt ( 145))/( 11)` |
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Answer» Correct Answer - C We have ` cos ^(-1) ((2)/(3x)) + cos ^(-1) ((3)/(4x)) = (pi)/(2)` `rArr cos ^(-1) ((2)/(3x). ( 3)/(4x) - sqrt (1 - (4)/( 9x ^(2))) sqrt (1 - ( 9)/( 16 x ^(2))) = (pi)/(2)` `[ because cos ^(-1) x + cos ^(-1) y = cos ^(-1) ( xy - sqrt (1 - x ^(2)) sqrt (1 -y ^(2)))]` `rArr cos^(-1) ((1)/(2x^2)) - (sqrt ( 9x ^(2) - 4) sqrt ( 16 x ^(2) -9 ))/( 12 x ^(2))) = (pi)/(2)` `rArr = ( 6 - sqrt(9x ^(2) - 4 ) sqrt ( 16x ^(2) -9))/( 12 x ^(2) ) = cos ""(pi )/(2) = 0 ` `rArr sqrt (9x^(2) - 4) sqrt ( 16x ^(2) - 9 ) = 6` On squaring both sides, `rArr (9x ^(2) - 4) ( 16 x ^(2) -9) = 36` `rArr 144 x ^(4) - 81 x ^(2) - 64 x ^(2) + 36 = 36` `rArr " " x ^(2)( 144 x ^(2) - 145) =0` `rArr " " x =0 or x= pm sqrt ((145)/( 144)) = pm =( sqrt ( 145))/( 12)` But ` x gt (3)/(4), " " x = ( sqrt ( 145))/( 12)` |
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