Question

If alpha and beta are roots of the equation 6x 2 + 11x + 3 = 0, then

Answer 1

Given :

The quadratic equation as 6 x² + 11 x + 3 = 0

The roots of equation are α , β

To Find :

Which of the following are real

Solution :

As , quadratic equation is 6 x² + 11 x + 3 = 0

now, Solving the equation by mid-term break rule

i.e  6 x² + 11 x + 3 = 0

Or, 6 x² + 2 x + 9 x + 3 = 0

Or,  2 x (3 x + 1) + 3 (3 x + 1) = 0

Or,  (3 x + 1)  (2 x + 3) = 0

∴   (3 x + 1) = 0        and        (2 x + 3) = 0

i.e  3 x = - 1                            2 x = - 3

Or,   x = $\dfrac{-1}{3}$                            x = $\dfrac{-3}{2}$

Now,

As  The roots of equation are α , β

So, α = $\dfrac{-1}{3}$    and   β = $\dfrac{-3}{2}$

Thus

(a ) $cos-^{1}$ α  = $cos-^{1}$ ( $\dfrac{-1}{3}$ )

∴ The range = [ - 1 , 1 ]

(b) $Sin^{- 1}\beta$ = $Sin^{-1}$ ( $\dfrac{-3}{2}$ )

∴ The range = [ - 1 , 1 ]

(c) $cosec-^{1}$ α = $cosec-^{1}$( $\dfrac{-1}{3}$ )

 ∴ The range = R -  [ - 1 , 1 ]

(d)  $cot-^{1}$ α , $Cot^{- 1}\beta$

For, $cot-^{1}$ α =  $cot-^{1}$ ( $\dfrac{-1}{3}$ )

And  $Cot^{- 1}\beta$ = $cot^{-1}$ ( $\dfrac{-3}{2}$ )

Both  $cot-^{1}$ x is define d for both these values of roots as both define R

Hence, $cot-^{1}$ x is define d for both these values of roots as both define Real . Answer