lf $(1 + k) {tan}^{2} x - 4 tan x - 1 + k = 0$ has real roots $tan x_{1}$ and $tan x_{2}$ , then

k^{2} \leq 5

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k^{2} \geq 6

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k = 3

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k \geq 0

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Solution

The correct option is A $k^{2} \leq 5$
For real roots
$b^{2} - 4 a c \geq 0$
$\Rightarrow (4)^{2} - 4 (k + 1) (k - 1) \geq 0$
$4 - (k^{2} - 1) \geq 0$
$k^{2} \leq 5$
$∴ k^{2} \leq 5$

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Similar questions

$I f (1 + K) t a n^{2} x - 4 t a n x - 1 + K = 0 h a s r e a l r o o t s t a n x_{1} a n d t a n x_{2} t h e n$

(1 + k) {tan}^{2} x - 4 tan x + k - 1 = 0

tan x_{1}, tan x_{2} (tan x_{1} > tan x_{2}) .

x^{2} - k x + 1 = 0

sec h^{- 1} (\frac{2}{3}) = {log}_{e} [\frac{(k + 1) + \sqrt{k^{2} + 1}}{k}]

k =

k

x^{2} - 2 (k + 1) x + k^{2} = 0

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