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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 |
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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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