If x = a cos θ and y = b sin θ, then b2x2 + a2y2 = …………… (1) a2b2 (2) ab (3) a4b4 (4) a2 + b2

Answer

`t in [-(pi)/(2), -(pi)/(10)]cup [(3pi)/(10),(pi)/(2)]`

Explanation

Given, `2sint=(1-2x+5x^(2))/(3x^(2)-2x-1),t in [-(pi)/(2),(pi)/(2)]`
Put `2 sint=y rArr -2 le y le 2`
` therefore y=(1-2x+5x^(2))/(3x^(2)-2x-1)`
`rArr (3y-5)x^(2)-2x(y-1)-(y+1)=0`
Since, `x in R -{1, -1//3}`
`["as, "3x^(2)-2x-1 ne 0 rArr (x-1)(x+1//3) ne 0]`
` :. D ge 0`
`rArr 4(y-1)^(2)+4(3y-5)(y+1) ge 0`
`rArr y^(2)-y-1 ge 0`
`rArr (y-(1)/(2))^(2)-(5)/(4) ge 0`
`rArr (y-(1)/(2)-(sqrt(5))/(2))(y-(1)/(2)+(sqrt(5))/(2)) ge 0`
`rArr y le (1-sqrt(5))/(2)`
`or y ge (1+sqrt(5))/(2)`
`rArr 2 sin t le (1-sqrt(5))/(2) `
`or 2 sint ge (1+sqrt(5))/(2)`
`rArr sin t le sin(-(pi)/(10))`
`or sin t ge sin((3pi)/(10))`
`rArr t le -(pi)/(10) or t ge (3pi)/(10)`
Hence, range of t is `[-(pi)/(2), -(pi)/(10)]cup [(3pi)/(10),(pi)/(2)]`.