Question

If (1 + k) tan²x - 4 tan x - 1 + k = 0 has real roots tan x₁ and
tan x₂, then
(a) k² ≤ 5 (b) k² ≥ 26 (c) k = 3 (d) none of these

Answer 1

Question:

If (1 + k)•tan²x - 4•tanx - 1 + k = 0 has real roots tanx₁ and tanx₂, then –

(a) k² ≤ 5 (b) k² ≥ 26 (c) k = 3 (d) none of these

Answer:

a). k² ≤ 5

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

Here ,

The given quadratic equation is :

(1 + k)•tan²x - 4•tanx - 1 + k = 0

Let ,

y = tanx

Then ,

The given quadratic equation will reduce to :

(1+k)y² - 4y - 1 + k = 0

Clearly, here we have ;

a = 1 + k

b = - 4

c = - 1 + k

Thus,

The determinant of the quadratic equation will be

=> D = b² - 4ac

=> D = (-4)² - 4(1+k)(-1+k)

=> D = 16 - 4(k+1)(k-1)

=> D = 16 - 4(k² - 1)

=> D = 4•[ 4 - (k² - 1) ]

=> D = 4•[ 4 - k² + 1 ]

=> D = 4•[5 - k²]

For real roots , the determinant must be greater than or equal to zero .

Thus,

=> D ≥ 0

=> 4•(5 - k²) ≥ 0

=> 5 - k² ≥ 0

=> 5 ≥ k²

=> k² ≤ 5

Hence,

The required answer is : a) k² ≤ 5