1.

If `tanalpha`is equal to the integral solution of the inequality `4x^2-16 x+15

Answer» It is given that slope of bisector is`cos beta.`
`:. cos beta = tan45^@ =>cos beta = 1->(1)`
Now, `4x^2-16x+15 lt 0`
`=>4x^2-10x-6x+15 lt 0`
`=>2x(2x-5)-3(2x-5) lt 0`
`=>(2x-3)(2x-5) lt 0`
`:. 3/2 lt x lt 5`
As, `tan alpha` is an integral solution for this inequality,
`:. tanalpha = 2`
`=>cotalpha = 1/2->(2)`
Now, `sin(alpha+beta)sin(alpha-beta) = sin^2alpha-sin^2beta`
`=1/(cosec^2alpha) - (1-cos^2beta)`
`=1/(1+cot^2alpha)-1+cos^2beta`
From (1) and (2),
`1/(1+1/4) -1+1 = 4/5`
So, option `d` is the correct option.


Discussion

No Comment Found

Related InterviewSolutions