1.

if `(1+k)tan^2x-4tanx-1+k=0` has real roots `tanx_1` and `tanx_2` thenA. `k^(2) le 5`B. `tan(x_(1) + x_(2)) = 2`C. for `k = 2, x_(1) = pi//4`D. all of these

Answer» Correct Answer - D
We have, `tan x_(1) and tan x_(2)` as the roots of the equation `(1+k) tan^(2) x - 4 tan x + (k-1) = 0`
`therefore" "tan x_(1)+tan x_(2)=(4)/(1+k)and tan x_(1) tan x_(2)=(k-1)/(k+1)`
`rArr" "tan(x_(1) + x_(2))=(tan x_(1) + tan x_(2))/(1-tan x_(1) tan x_(2)) = 2`
So, option (b) is correct.
For k = 2, the equation reduces to `3 tan^(2) x -4 tan x+1 = 0`
`rArr" "(tan x-1)(3 tan x-1)=0`
`rArr" "tan x = 1 or tan x=(1)/(3)`
`rArr" "x_(1)=pi//4 and x_(2) = tan^(-1) (1)/(3)`
Since the equation has real roots. Therefore,
Disc `ge 0 rArr 16 - 4 (k^(2) -1) ge 0 rArr k^(2) le 5`
Hence, all options are correct.


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