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The equation (cosp-1)^x^2+(cos p)x+s in ...

The equation `(cosp-1)^x^2+(cos p)x+s in p=0` in the variable `x` has real roots. The `p` can take any value in the interval `(0,2pi)` (b) `(-pi)` (c) `(-pi/2,pi/2)` (d) `(,pi)`

A

`(0,2pi)`

B

`(-pi,0)`

C

`(-(pi)/(2),(pi)/(2))`

D

`(0,pi)`

Text Solution

Verified by Experts

The correct Answer is:
d
Since, the given quadratic equation
`" " (cos p - 1)x^(2) + (cos p) x + sin p = 0 `
has real roots.
` therefore ` Discriminant, `cos ^(2) p - 4 sin p(cos p - 1 ) ge 0 `
` rArr (cos p - 2 sin p ) ^(2) - 4 sin p(1 - sin p ) ge 0 `
` because " " 4 sin p ( 1- sin p ) gt 0 for 0 lt p lt pi `
and `" " (cos p - 2 sin p )^(2) ge 0 `
Thus, ` (cos p - 2 sin p ) ^(2) + 4 sin p (1 - sin p) ge 0 `
for `" " 0 lt p lt pi `.
Hence, the equations has real for `0 lt p lt pi `
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