SATHEE: Theory of Equations 1 Question 1

Theory of Equations 1 Question 1

1. If $α$ and $β$ are the roots of the quadratic equation, $x^{2} + x \sin θ - 2 \sin θ = 0, θ \in 0, \frac{π}{2}$ , then $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) (α - β)^{24}}$ is equal to

(a) $\frac{2^{12}}{(\sin θ + 8)^{12}}$

(b) $\frac{2^{6}}{(\sin θ + 8)^{12}}$

(d) $\frac{2^{12}}{(\sin θ - 8)^{6}}$

(2019 Main, 10 April I)

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Answer:

Correct Answer: 1. (a)

Solution:

Given quadratic equation is

$x^{2} + x \sin θ - 2 \sin θ = 0, θ \in 0, \frac{π}{2}$

and its roots are $α$ and $β$ .

So, sum of roots $= α + β = - \sin θ$

and product of roots $= α β = - 2 \sin θ$

$\Rightarrow α β = 2 (α + β)$

Now, the given expression is $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) (α - β)^{24}}$

$\begin{aligned} = \frac{α^{12} + β^{12}}{\frac{1}{α^{12}} + \frac{1}{β^{12}} (α - β)^{24}} = \frac{α^{12} + β^{12}}{\frac{β^{12} + α^{12}}{α^{12} β^{12}} (α - β)^{24}} \\ = \frac{α β}{(α - β)^{2}} = \frac{α β}{(α + β)^{2} - 4 α β} \\ = \frac{2 (α + β)}{(α + β)^{2} - 8 (α + β)} [from Eq. (i)] \\ = \frac{2}{(α + β) - 8} = \frac{2}{- \sin θ - 8} [∵ α + β = - \sin θ] \\ = \frac{2^{12}}{(\sin θ + 8)^{12}} \end{aligned}$

Theory of Equations 1 Question 1

1. If $α$ and $β$ are the roots of the quadratic equation, $x^{2} + x \sin θ - 2 \sin θ = 0, θ \in 0, \frac{π}{2}$ , then $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) (α - β)^{24}}$ is equal to

Answer:

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Theory of Equations 1 Question 1

1. If α and β are the roots of the quadratic equation, x2+xsin⁡θ−2sin⁡θ=0,θ∈0,π2, then α12+β12(α−12+β−12)(α−β)24 is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. If $α$ and $β$ are the roots of the quadratic equation, $x^{2} + x \sin θ - 2 \sin θ = 0, θ \in 0, \frac{π}{2}$ , then $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) (α - β)^{24}}$ is equal to